Unwinding confusion about (bi)rational maps

I am often confused with birational maps, or just rational maps in general, because they are not quite maps in the usual sense. Let me try to undo my confusion by rewriting some materials on these confusing objects in this posting. I am following Vakil's online book.

Given schemes $X, Y,$ we define a rational map $X \dashrightarrow Y$ to be an equivalence class of scheme maps $\alpha : U \rightarrow Y,$ where $U \subset X$ is a dense open subset with the following equivalence relation: $(\alpha, U) \sim (\beta, V)$ means there is a dense open subset $W \subset U \cap V$ such that $\alpha|_{W} = \beta|_{W}.$

Remark. We often assume that $X$ is reduced because then dense open subsets are precisely scheme-theoretically dense open subschemes of $X,$ the notion that works better than the usual (topological) density in various situations. Using the latter notion gives an analogous notion to that of rational maps on $X,$ which seems to be studied by some people.

A rational map $X \dashrightarrow Y$ is said to be dominant if there is any representative $U \rightarrow Y$ whose image is dense in $Y.$ A dominant rational map $\pi : X \dashrightarrow Y$ is called birational if there is another dominant rational map $\phi : Y \dashrightarrow X$ such that

• $\pi \circ \phi \sim \mathrm{id}_{Y}$ and
• $\phi \circ \pi \sim \mathrm{id}_{X},$

on some dense open subsets (of $X$ and $Y,$ respectively).

When $X$ and $Y$ are reduced, this complicated looking notion means some thing simple: that is $X$ and $Y$ are birational if and only if there are some isomorphic dense open subsets $U \subset X$ and $V \subset Y.$

Let $k$ be a field. Then integral finite type $k$-schemes with dominant rational maps form a category. If we have a morphism $X \dashrightarrow Y,$ it sends the generic point of $X$ to that of $Y,$ so we get a corresponding function field extension $K(Y) \rightarrow K(X).$ It turns out that this establishes an equivalence between the category of integral finite type $k$-schemes with dominant rational maps over $k$ and the category of field extensions that are finitely generated over $K.$ (See Proposition 6.5.7 and 6.5.D in Vakil.)

Separatedness makes rational maps easier. If the target of a rational map is separated (over some base scheme), then it is easier to study the rational map, essentially due to the following fact (from 10.2.2. of Vakil):

Theorem (Reduced-to-Separated). Let $X, Y$ be schemes over a scheme $S.$ Suppose that $X$ is reduced and $Y$ is separated over $S.$ If two $S$-scheme maps from $X$ to $Y$ agree on a dense open subset of $X,$ then they must be identical.

How does this help us when studying rational maps? Let $X$ be reduced, defined over a field $k,$ and $Y$ separated over $k.$ A rational map $X \dashrightarrow Y$ defined over $k$ is giving a $k$-scheme map $U \rightarrow Y,$ where $U \subset X$ is a dense open subset. If there is another representative $V \rightarrow Y,$ the above theorem now tells us that the two representatives agree on $U \cap V.$ This implies that we can glue these two maps to get another representative $U \cup V \rightarrow Y.$ Note that we can glue arbitrarily many representatives this way, and get a scheme map $W \rightarrow Y,$ where $W$ is the union of all dense open subsets coming from the representatives. This $W$ is hence unique and deserves a name. It is called the domain of definition of the rational map $X \dashrightarrow Y.$

This is great because in this situation (where the source is reduced and the target is separated), we can get an actual map that contains the information about a rational map. This is much less confusing!

Upper semicontinuity

This is a notion that I had to look up whenever I face it. Hence, I decided to write it down in my own terms so that I don't have to look it up as many times. In Vakil (11.4), this notion is defined as follows: given a topological space $X,$ a function $f : X \rightarrow \mathbb{R}$ is called upper semi-continuous if $f^{-1}((-\infty, a)) \subset X$ is open for any $a \in \mathbb{R}.$ This is a formal definition, and experts don't seem to think this way. This is a condition on $f$ so that it can only jump up when taking limits.

Theorem. The following are equivalent:

• $f$ is upper semi-continuous;
• for any sequence $(x_{n})$ in $X$ (with possibly uncountable indices $n$) such that $f(x_{n}) \geq f(x),$ if $x_{n} \rightarrow x$ in $X,$ then $f(x_{n}) \rightarrow f(x)$ in $\mathbb{R}.$

Proof. Let $f$ be upper semi-continuous. Given a sequence $(x_{n})$ in $X$ such that $f(x_{n}) \geq f(x),$ if $x_{n} \rightarrow x$ in $X,$ fix any $\epsilon > 0.$ Then $f^{-1}((-\infty, f(x) + \epsilon)) \subset X$ is open by upper semi-continuity of $f$ and since it contains $x,$ knowing that $x_{n} \rightarrow x$ in $X,$ we may find some $x_{n} \in f^{-1}((-\infty, f(x) + \epsilon)).$ This implies that $$f(x_{n}) \in (-\infty, f(x) + \epsilon) \subset \mathbb{R}.$$ Since $f(x_{n}) \geq f(x) > f(x) - \epsilon,$ we have $$f(x_{n}) \in (f(x) - \epsilon, f(x) + \epsilon) \subset \mathbb{R}.$$ This shows that $f(x_{n}) \rightarrow f(x)$ in $\mathbb{R}.$

Conversely, suppose the second condition above. Fix $a \in \mathbb{R},$ and we aim to show that $f^{-1}((-\infty, a)) \subset X$ is open. Fix any $x \in f^{-1}((-\infty, a)).$ We want to show that there is an open neighborhood $U \ni x$ in $X$ such that $U \subset f^{-1}((-\infty, a)).$ We now work by contradiction and say that there is no such $U.$ In other words, for any neighborhood $U_{n} \ni x$ in $X,$ there exists $x_{n} \in U_{n}$ such that $f(x_{n}) \geq a > f(x).$ However, using the second condition, this implies that $$f(x) = \lim_{n \rightarrow \infty}f(x_{n}) \geq a > f(x),$$ a contradiction. This finishes the proof. $\Box$

Sheaves on small étale sites (possibly to be extended)

We follow Milne's book. I think Patrick Kelley for summarizing this material before when we studied together.

We have explained the definition of a site in this posting. The target example is $\textbf{Et}_{X},$ the site of étale maps into a scheme $X,$ whose maps between them are (étale) maps over $X.$

A site is a category $\mathcal{C}$ together with coverings for each of its object. A presheaf valued in $\textbf{Set}$ on $\mathcal{C}$ is a functor $\mathscr{F} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}.$ We may change the target category $\textbf{Set}$ into other categories to define analogous definitions. Maps between two presheaves are given by natural transformations, and thus, presheaves on $\mathcal{C}$ form a category.

A presheaf $\mathscr{F} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ is said to be a sheaf if it satsifies the following two extra axioms for every object $U$ of $\mathscr{C}.$

1. Given any covering $\{U_{i} \rightarrow U\}_{i \in I}$ of $U$ and $s, t \in \mathscr{F}(U),$ if $s|_{U_{i}} = t|_{U_{i}}$ for all maps $U_{i} \rightarrow U$ in the covering, then $s = t.$
2. Given any covering $\{U_{i} \rightarrow U\}_{i \in I}$ of $U$ and $s_{i} \in \mathscr{F}(U_{i})$ for each map $U_{i} \rightarrow U$ in the covering, if $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$ for all maps $U_{i} \rightarrow U$ and $U_{j} \rightarrow U$ in the covering, then there is $s \in \mathscr{F}(U)$ such that $s|_{U_{i}} = s_{i}$ for all maps $U_{i} \rightarrow U$ in the covering.

Note that sheaves on $\mathcal{C}$ form a full subcategory of the category of presheaves on $\mathcal{C}.$ We now focus on the case $\mathcal{C} = \textbf{Et}_{X}$ for the following proposition, which we learn from Milne (II. Proposition 1.5). We write $\textbf{Zar}_{X}$ to mean the small Zariski site of a scheme $X.$ We will take the values of sheaves in $\textbf{Ab}.$

Proposition. Let $\mathscr{F}$ be a presheaf on $\textbf{Et}_{X}.$ Then $\mathscr{F}$ is a sheaf if and only if the following two properties are satisfied.

1. For any object $U$ of $\textbf{Et}_{X},$ the restriction of $\mathscr{F}$ to $\textbf{Zar}_{U}$ is a sheaf.
2. For any covering $\{V \rightarrow U\},$ consisting of a single map of $\textbf{Et}_{X}$ such that $V$ and $U$ are affine, the sequence $\mathscr{F}(U) \rightarrow \mathscr{F}(V) \rightrightarrows \mathscr{F}(V \times_{U} V)$ given by restrictions is an equalizer.

Sketch of proof. It is evident that sheaf axioms of $\mathscr{F}$ implies the two given conditions, noting that open embeddings are of étale. 

For the converse, our goal is to check that the sequence $$\mathscr{F}(U) \rightarrow \prod_{i \in I}\mathscr{F}(U_{i}) \rightrightarrows \prod_{i,j \in I}\mathscr{F}(U_{i} \times_{U} U_{j})$$ given by the restrictions is an equalizer.

Fix any covering $\{U_{i} \rightarrow U\}_{i \in I}$ in $\textbf{Et}_{X}$ and consider $V := \bigsqcup_{i \in I}U_{i}.$ The first condition implies that we have $$\mathscr{F}(V) \simeq \prod_{i \in I}\mathscr{F}(U_{i})$$ induced by the restrictions. Note that we have $$V \times_{U} V \simeq \bigsqcup_{i, j \in I} (U_{i} \times_{U} U_{j}),$$ given by showing that the right-hand side satisfies the universal property of the left-hand side. Having this in mind, one may check the our goal reduces to checking that the sequence $$\mathscr{F}(U) \rightarrow \mathscr{F}(V) \rightrightarrows \mathscr{F}(V \times_{U} V)$$ is an equalizer. The second condition tells us that this holds when $I$ is finite and each $U_{i}$ for $i \in I$ and $U$ are affine.

For the general case, denote by $\pi : V \rightarrow U$ be the morphism induced by maps $U_{i} \rightarrow U$ from the fixed covering of $U.$ We may write $U = \bigcup_{j \in J} W_{j},$ where each $W_{j} \subset U$ is affine open. Then $\pi^{-1}(W_{j}) \subset V$ is open, so we may write $$\pi^{-1}(W_{j}) = \bigcup_{k \in K_{j}} W_{j,k},$$ where $W_{j,k} \subset V$ are affine opens. Note that $\{W_{j,k} \rightarrow W_{j}\}_{k \in K_{j}}$ forms an étale cover of $W_{j},$ where the maps are given by $$W_{j,k} \hookrightarrow \pi^{-1}(U_{j}) \xrightarrow{\pi} W_{j}.$$ Since $\pi(W_{j,k}) \subset W_{j}$ is open and $W_{j}$ is quasi-compact, we may assume that $K_{j}$ is finite for such étale covering of $W_{j}.$ We have $$V = \bigcup_{j \in J}\pi^{-1}(W_{j}) = \bigcup_{j \in J, k \in K_{j}} W_{j,k}.$$

So far, we know that the following sequences are equalizers:

• $\mathscr{F}(U) \rightarrow \prod_{j \in J}\mathscr{F}(W_{j}) \rightrightarrows \prod_{j,j' \in J}\mathscr{F}(W_{j} \cap W_{j'})$ by the first condition;
• $\mathscr{F}(V) \rightarrow \prod_{j \in J}\prod_{k \in J_{j}}\mathscr{F}(W_{j,k}) \rightrightarrows \prod_{j,j' \in J}\prod_{k \in K_{j}, k' \in K_{j'}}\mathscr{F}(W_{j,k} \cap W_{j',k'})$ by the first condition;
• $\mathscr{F}(W_{j}) \rightarrow \prod_{k \in K_{j}}\mathscr{F}(W_{j,k}) \rightrightarrows \prod_{k,l \in K_{j}} \mathscr{F}(W_{j,k} \times_{U} W_{j,l})$ since $K_{j}$ is a finite set and $W_{j}, W_{j,k}$ are affines.

We are now ready to show that $$\mathscr{F}(U) \rightarrow \mathscr{F}(V) \rightrightarrows \mathscr{F}(V \times_{U} V)$$ is an equalizer. Given any $s \in \mathscr{F}(U),$ suppose that $s|_{V} = 0 \in \mathscr{F}(V).$ This means that $s|_{W_{j,k}} = 0 \in \mathscr{F}(W_{j,k}),$ for all $j \in J$ and $k \in K_{j}.$ Hence, we have $s|_{W_{j}} = 0 \in \mathscr{F}(W_{j})$ for all $j \in J,$ which implies that $s = 0 \in \mathscr{F}(U).$ This already shows that $\mathscr{F}$ is a separated presheaf (meaning a presheaf with the first condition for being a sheaf), and we will use this observation soon.

Denote by $p : V \times_{U} V \rightarrow V$ the projection on the first component and $q : V \times_{U} V \rightarrow V$ the second. Given any $u \in \mathscr{F}(V),$ suppose that $$p^{*}u = q^{*}u\in \mathscr{F}(V \times_{U} V).$$ Then $$(p^{*}u)|_{W_{j,k} \times_{U} W_{j,l}} = (q^{*}u)|_{W_{j,k} \times_{U} W_{j,l}} \in \mathscr{F}(W_{j,k} \times_{U} W_{j,l})$$ for all $j \in J$ and $k, l \in K_{j}.$ Define $(t_{j,k}) \in \prod_{k \in K_{j}}\mathscr{F}(W_{j,k})$ by $$t_{j,k}|_{W_{j,k} \times_{U} W_{j,l}} := (p^{*}u)|_{W_{j,k} \times_{U} W_{j,l}}$$ and $$t_{j,l}|_{W_{j,k} \times_{U} W_{j,l}} := (q^{*}u)|_{W_{j,k} \times_{U} W_{j,l}}.$$ Then we have a unique $t_{j} \in \mathscr{F}(W_{j})$ such that $t_{j}|_{W_{j,k}} = t_{j,k}$ for all $j, k.$

We will be done as soon as we can show that $t_{j}|_{W_{j} \cap W_{j'}} = t_{j'}|_{W_{j} \cap W_{j'}}$ for all $j, j' \in J.$ The reason is that this would give us a unique $t \in \mathscr{F}(U)$ such that $t|_{W_{j}} = t_{j}$ for all $j \in J$ and since $$(t|_{V})|_{W_{j,k}} = (t_{j})|_{W_{j,k}} = t_{j,k} = u|_{W_{j,k}}$$ for all $j, k,$ as we can check by applying one more map, we must have $t|_{V} = u,$ as desired.

To check $t_{j}|_{W_{j} \cap W_{j'}} = t_{j'}|_{W_{j} \cap W_{j'}},$ it is enough to check $$t_{j}|_{W_{j,k} \cap W_{j',k'}} = t_{j'}|_{W_{j,k} \cap W_{j',k'}},$$ where $k \in K_{k}$ and $k' \in K_{j'}.$ This is the same as checking $$t_{j,k}|_{W_{j,k} \cap W_{j',k'}} = t_{j',k'}|_{W_{j,k} \cap W_{j',k'}},$$ which we already know. This finishes the proof. $\Box$

Computing differential 1-forms of an algebra

We follow Chapter 2 of Bosch, Lütkebohmert, and Raynaud and notes by Hochster, as well as other references such as Vakil. By a ring, we will mean a commutative unital ring.

Let $R$ be a ring and $A$ an $R$-algebra. Given an $A$-module $M,$ an $R$-linear map $D : A \rightarrow M$ such that $$D(fg) = fD(g) + gD(f)$$ for all $f, g \in A$ is called an $R$-derivation of $A$ into $M$.

Remark. Note that $D(1) = D(1) + D(1)$ so that $D(1) = 0.$ Hence, for any $r \in R,$ we have $D(r) = D(r \cdot 1) = rD(1) = 0.$ That is, the derivative of any constant (i.e., element of $R$) is zero!

Warning. The only $R$-derivation $D : A \rightarrow M$ that is $A$-linear is the zero map, due to a similar argument as above. Whenever I get confused, I call $D$ an $R$-linear derivation instead.

We denote by $\mathrm{Der}_{R}(A, M)$ the set of all $R$-derivations of $A$ into $M,$ which is also an $A$-module given by the $A$-module structure of $M.$ The module of relative differential forms (of degree 1) of $A$ over $R$ is an $A$-module $\Omega^{1}_{A/R}$ with an $R$-derivation $d = d_{A/R} : A \rightarrow \Omega^{1}_{A/R},$ which is called the exterior differential, such that $$\mathrm{Hom}_{A}(\Omega^{1}_{A/R}, M) \simeq \mathrm{Der}_{R}(A, M)$$ in $\textbf{Set}$ given by $\phi \mapsto \phi \circ d.$ That is, the functor $\mathrm{Der}_{R}(A, -) : \textbf{Mod}_{A} \rightarrow \textbf{Set}$ is representable. Of course, this functor can be seen as $\textbf{Mod}_{A} \rightarrow \textbf{Mod}_{A},$ but we do not need this. It is abstract nonsense to show that such $d$ is unique, and one may construct it as follows.

Construction of the exterior differential. A formal construction of $d : A \rightarrow \Omega^{1}_{A/R}$ for the sake of existence is quite easy. That is, we take $$\Omega^{1}_{A/R} = \frac{\bigoplus_{a \in A} A da}{(d(a + b) - da - db, d(ab) - adb - bda, d(ra) - rda)_{a, b \in A, r \in R}}.$$ Taking $d : A \rightarrow \Omega^{1}_{A/R}$ by $a \mapsto da,$ we get what we need.

In practice, such a formal construction is quite useless, so we give better descriptions of $d : A \rightarrow \Omega^{1}_{A/R}.$

Lemma. Let $A := R[t_{i}]_{i \in I},$ a free $R$-algebra. Then we have $\Omega^{1}_{A/R} = \bigoplus_{i \in I} A dt_{i},$ where $d : A \rightarrow \bigoplus_{i \in I} A dt_{i}$ is given by $$df = \sum_{i \in I}\frac{\partial f}{\partial t_{i}} dt_{i}.$$ That is, the exterior derivative is given by taking the gradient.

Proof. Using the product rule for partial derivatives of polynomials, we check that $d$ is indeed an $R$-derivation. To check the universal property, consider any $R$-linear derivation $D : A \rightarrow M,$ the only way to define $A$-module map $\phi : \Omega^{1}_{A/R} \rightarrow M$ such that $D = \phi \circ d$ is to send $df$ to $Df,$ so in particular, we send $dt_{i}$ to $Dt_{i},$ which actually constructs such $\phi$ as $\Omega^{1}_{A/R}$ is free $A$-algebra over such elements $dt_{i}.$ This finishes the construction for this case. $\Box$

In general, an $R$-algebra $A$ is isomorphic to an $R$-algebra of the form $$R[t_{i}]_{i \in I}/\mathfrak{b},$$ where $\mathfrak{b} \subset R[t_{i}]_{i \in I}$ is an ideal. The next statement computes $d : A \rightarrow \Omega^{1}_{A/R}$ with respect to such a presentation. Now, the following, together with Lemma computes the exterior derivation in this case.

Theorem.  Let $$A := B/\mathfrak{b},$$ where $B$ is an $R$-algebra and $\mathfrak{b} \subset B$ is an ideal. Then we can construct $$\Omega^{1}_{A/R} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}},$$ where $d_{A/R} : A \rightarrow \Omega^{1}_{A/R}$ is given by $\bar{b} \mapsto \overline{d_{B/R}b}.$

Proof. Let $\Omega^{1}_{A/R}$ be as described in the statement. Then $\mathfrak{b}\Omega^{1}_{A/R} = 0,$ so $\Omega^{1}_{A/R}$ is a module over $A = B/\mathfrak{b}.$ The map $B \rightarrow \Omega^{1}_{A/R}$ given by $b \mapsto \overline{d_{B/R}b}$ is a $B$-linear map that kills $\mathfrak{b},$ so the induced map $d_{A/R} : \bar{b} \mapsto \overline{d_{B/R}b}$ is a well-defined module map over $A = B/\mathfrak{b}.$

To check the universal property, let $D : A \rightarrow M$ be any $R$-derivation. This is the same as saying that $M$ is a $B$-module such that $\mathfrak{b}M = 0$ and that we have an $R$-derivation $\tilde{D} : B \rightarrow M$ such that $\tilde{D}\mathfrak{b} = 0,$ given by $\tilde{D}b = D\bar{b}.$ Hence, there is a unique $B$-module map $\tilde{\phi} : \Omega^{1}_{B/R} \rightarrow M$ such that $d_{R/B}b \mapsto \tilde{D}b = D\bar{b}.$ This map kills $\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b},$ which gives an $A$-module map $\phi : \Omega^{1}_{A/R} \rightarrow M$ such that $\phi \circ d_{A/R} = D.$ The uniqueness of $\phi$ easily follows, which finishes the proof. $\Box$

Application. If $A = B/\mathfrak{b}$ with $B = R[x_{i}]_{i \in I},$ then we have $$\Omega^{1}_{B/R} = \bigoplus_{i \in I} B dx_{i}$$ so that $$\frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R}} = \frac{\bigoplus_{i \in I} B dx_{i}}{\bigoplus_{i \in I} \mathfrak{b}B dx_{i}} \simeq \bigoplus_{i \in I} A dx_{i}.$$ We can write $\mathfrak{b} = (f_{j})_{j \in J}$ for some family $f_{j} \in B$ so that $$d_{B/R}\mathfrak{b} = \{df_{j}\}_{j \in J} = \left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ This implies that under the above isomorphism, we have $$\frac{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}}{\mathfrak{b}\Omega^{1}_{B/R}} \simeq A\left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ Therefore, we may write $$\Omega^{1}_{A/R} = \bigoplus_{i \in I} A dx_{i} \Biggm/ A\left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ This is the content of Key fact 21.2.3 in Vakil. The author uses the last description to construct $\Omega^{1}_{A/R}.$

Jacobian description of differential $1$-forms. This is from 21.2.E in Vakil. Again, when $A = B/\mathfrak{b},$ where $B$ is an $R$-algebra and $\mathfrak{b} \subset B$ is an ideal, we have $$\Omega_{A/R}^{1} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}},$$ and we saw this formally, but let's try to imbue some context now.

Let $B = k[x_{1}, \dots, x_{n}],$ where $k$ is a field so that we may write $\mathfrak{b} = (f_{1}, \dots, f_{r})$ for some $f_{j} \in B.$ That is, we consider the case $$A = \frac{k[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}$$ a finitely presented algebra over $k.$ We have $$\Omega_{B/R}^{1} = B dx_{1} \oplus \cdots \oplus B dx_{n},$$ so $$\begin{align*} \frac{\Omega_{B/R}^{1}}{\mathfrak{b}\Omega_{B/R}^{1}} &= \frac{B dx_{1} \oplus \cdots \oplus B dx_{n}}{\mathfrak{b} dx_{1} \oplus \cdots \oplus \mathfrak{b} dx_{n}} \\ &\simeq (B/\mathfrak{b}) dx_{1} \oplus \cdots \oplus (B/\mathfrak{b}) dx_{n} \\ &= A dx_{1} \oplus \cdots \oplus A dx_{n}. \end{align*}$$ Since $$\Omega_{A/R}^{1} = \frac{\Omega_{B/R}^{1}}{\mathfrak{b}\Omega_{B/R}^{1} + Bd_{B/R}\mathfrak{b}} \simeq \frac{\Omega_{B/R}^{1}/\mathfrak{b}\Omega_{B/R}^{1}}{d_{B/R}\mathfrak{b}(\Omega_{B/R}^{1}/\mathfrak{b}\Omega_{B/R}^{1})},$$ we may write $$\Omega_{A/R}^{1} = \frac{A dx_{1} \oplus \cdots \oplus A dx_{n}}{(df_{1}, \dots, df_{r})}.$$ Note that $$df_{j} = \frac{\partial f_{j}}{\partial x_{1}} dx_{1} + \cdots + \frac{\partial f_{j}}{\partial x_{n}} dx_{n}$$ in $A dx_{1} \oplus \cdots \oplus A dx_{n}.$

As a result, we have the exact sequence $$A dy_{1} \oplus \cdots \oplus A dy_{r} \xrightarrow{J} A dx_{1} \oplus \cdots \oplus A dx_{n} \rightarrow \Omega_{A/R}^{1} \rightarrow 0,$$ where $J$ can be described as an $A$-linear map $J : A^{\oplus r} \rightarrow A^{\oplus n}$ given by the following Jacobian matrix $$J = \left[ \frac{\partial f_{j}}{\partial x_{i}} \right]_{1 \leq i, j \leq n}$$ whose entries are in $A = k[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}).$

Localization. If we have any ring map $B \rightarrow A$ and $S \subset A$ is a mulitplicative submonoid, then we have $$\Omega^{1}_{S^{-1}A/B} \simeq S^{-1}\Omega^{1}_{A/B},$$ as $S^{-1}A$-module. Moreover, one may check that the map $$S^{-1}A \rightarrow S^{-1}\Omega^{1}_{A/B}$$ given by $$\frac{a}{s} \mapsto \frac{sd_{A/B}(a) - ad_{A/B}(s)}{s^{2}}$$ is a well-defined $B$-linear derivation that is compatible with $$d_{S^{-1}A/B} : S^{-1}A \rightarrow \Omega^{1}_{S^{-1}A/B}$$ and the isomorphism given above.

Example. Let $k$ be a field. We have $$\Omega^{1}_{k(t_{1}, \dots, t_{n})/k} = k(t_{1}, \dots, t_{n})dt_{1} \oplus \cdots \oplus k(t_{1}, \dots, t_{n})dt_{n},$$ whose exterior derivative can be given by the quotient rule.

Example. Again, let $k$ be a field and let $A := k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}].$ We want to compute $\Omega^{1}_{A/k}.$ We have $$A = k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}] = S^{-1}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}],$$ where $$S = k[t_{1}, \dots, t_{n}] \setminus (0) \subset k[t_{1}, \dots, t_{n}] \subset k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}].$$ Hence, we have $$\begin{align*}\Omega^{1}_{A/k} &= S^{-1}\Omega^{1}_{k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]/k} \\ &= S^{-1}\left(\bigoplus_{i=1}^{m}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]dt_{i} \oplus \bigoplus_{j=1}^{n}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]dx_{j}\right) \\ &= \bigoplus_{i=1}^{m}k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}]dt_{i} \oplus \bigoplus_{j=1}^{n}k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}]dx_{j}, \end{align*}$$ which is quite concrete.

Example. We keep using $k$ to denote the base field. Then consider $$A = \frac{k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{n}]}{(f_{1}(\boldsymbol{t}, x_{1}, \dots, x_{n}), \cdots, f_{r}(\boldsymbol{t}, x_{1}, \dots, x_{n}))}.$$ Write $$B = k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{n}]$$ and $$\mathfrak{b} = (f_{1}(\boldsymbol{t}, x_{1}, \dots, x_{n}), \cdots, f_{r}(\boldsymbol{t}, x_{1}, \dots, x_{n})).$$ We have $$\Omega^{1}_{B/k} = \bigoplus_{i=1}^{m}k(\boldsymbol{t})[\boldsymbol{x}]dt_{i} \oplus \bigoplus_{j=1}^{n}k(\boldsymbol{t})[\boldsymbol{x}]dx_{j}$$ and $$\begin{align*}d_{B/k}\mathfrak{b} &= \{df_{1}, \dots, df_{r}\} \\ &= \left\{ \sum_{i=1}^{m}(f_{l})_{t_{i}} dt_{i} + \sum_{j=1}^{n}(f_{l})_{x_{j}} dx_{i} \right\}_{l=1}^{r}\end{align*}.$$ This lets us compute $$\begin{align*} \Omega^{1}_{A/k} &= \frac{\Omega^{1}_{B/k}}{\mathfrak{b}\Omega^{1}_{B/k} + Bd_{B/k}\mathfrak{b}} \\ &\simeq \frac{\Omega^{1}_{B/k}/\mathfrak{b}\Omega^{1}_{B/k}}{(\mathfrak{b}\Omega^{1}_{B/k} + Bd_{B/k}\mathfrak{b})/\mathfrak{b}\Omega^{1}_{B/k}} \\ &\simeq \frac{\bigoplus_{i=1}^{m}Adt_{i} \oplus \bigoplus_{j=1}^{n}Adx_{j}
}{A\left\{ \sum_{i=1}^{m}(f_{l})_{t_{i}} dt_{i} + \sum_{j=1}^{n}(f_{l})_{x_{j}} dx_{j} \right\}_{l=1}^{r}}, \end{align*}$$ which is very explicit.

Diagonal description. Consider a scheme map $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(R)$ between affine schemes. The diagonal map $\Delta_{A/R} : \mathrm{Spec}(A) \rightarrow \mathrm{Spec}(A) \times_{R} \mathrm{Spec}(A)$ is the scheme map induced by the ring map $A \otimes_{R} A \rightarrow A$ given by $a \otimes a' \mapsto aa'$ (as also remarked the proof of Proposition 10.1.3 in Vakil.) Denote by $I$ the kernel of this ring map, and consider the map $d : A \rightarrow I/I^{2}$ given by $f \mapsto 1 \otimes f - f \otimes 1$ modulo $I^{2}.$

Remark. The map $A \otimes A \rightarrow A$ given above is called the multiplication map of $A$ over $R,$ because it is the $R$-linear map corresponding to the actual multiplication map $A \times A \rightarrow A.$ Indeed, we have $1 \otimes f - f \otimes 1 \in I$ because $$1 \otimes f - f \otimes 1 \mapsto f - f = 0$$ under the multiplication map.

Since $I$ kills $I/I^{2},$ we see that $I/I^{2}$ is a module over $(A \otimes_{R} A)/I.$ Since the map $A \otimes_{R} A \rightarrow A$ is surjective, we have $(A \otimes_{R} A)/I \simeq A.$ This makes $I/I^{2}$ an $A$-module. Note that $a \otimes 1 = 1 \otimes a$ in $(A \otimes_{R} A)/I$ both corresponding to $a \in A.$ Hence, we have $$a \cdot (1 \otimes f - f \otimes 1) = a \otimes f - (af) \otimes 1 =  1 \otimes (af) - f \otimes a$$ in $I/I^{2},$ as we admittedly omitted bar notations. The map $d : A \rightarrow I/I^{2}$ is an $R$-module map because $$\begin{align*}d(rf) &= 1 \otimes (rf) - (rf) \otimes 1 \\ &= r \otimes f - (rf) \otimes 1 \\ &= r \cdot (1 \otimes f) - r \cdot (f \otimes 1) \\ &= r \cdot (1 \otimes f - f \otimes 1) \\ &= r \cdot df\end{align*}$$ in $I/I^{2}.$

Moreover, we note that $d : A \rightarrow I/I^{2}$ is an $R$-derivation as we can check $$\begin{align*}d(fg) &= 1 \otimes (fg) - (fg) \otimes 1 \\ &= (1 \otimes f)(1 \otimes g) - (f \otimes 1)(g \otimes 1) \\ &= (1 \otimes f)(1 \otimes g) - (1 \otimes f)(g \otimes 1) + (g \otimes 1)(1 \otimes f) - (g \otimes 1)(f \otimes 1) \\ &= f \cdot (1 \otimes g -  g \otimes 1) + g \cdot (1 \otimes f - f \otimes 1) \\ &= f \cdot d(g) + g \cdot d(f), \end{align*}$$ in $I/I^{2},$ where again, we omitted many bars in the middle.

Theorem. The map $d : A \rightarrow I/I^{2}$ also describes the exterior derivative $d_{A/R} : A \rightarrow \Omega^{1}_{A/R}.$

Proof. We may assume $A = R[x_{i}]_{i \in S}/(f_{j})_{j \in T}.$ Then $$A \otimes_{R} A \simeq \frac{R[x_{i}, y_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{y}))_{j \in T}} = \frac{R[x_{i}, \Delta_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}},$$ where we used the following change of variables: $\Delta_{i} = y_{i} - x_{i}.$ In this presentation, the multiplication map $A \otimes_{R} A \rightarrow A$ is given by the map $$\frac{R[x_{i}, \Delta_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}} \rightarrow \frac{R[x_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}))_{j \in T}}$$ by $x_{i} \mapsto x_{i}$ and $\Delta_{i} \mapsto 0.$ Hence, by inspection, we can see that the kernel $I$ can be written as $$I = (\overline{\boldsymbol{\Delta}}) = \frac{(\boldsymbol{\Delta}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}},$$ so $$I/I^{2} \simeq \frac{(\boldsymbol{\Delta}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}{({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}.$$ Now, note that for any $f(\boldsymbol{x}) \in R[x_{i}]_{i \in S},$ we have $$f(\boldsymbol{x} + \boldsymbol{\Delta}) = f(\boldsymbol{x}) + \sum_{i \in S}\frac{\partial f(\boldsymbol{x})}{\partial x_{i}} \Delta_{i} + \sum_{i, i' \in S}\Delta_{i}\Delta_{i'}g_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta})$$ in $R[x_{i}, \Delta_{i}]_{i \in S}$ for suitable $g_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta}),$ which are in fact zero for all but finitely many $(i,i') \in S^{2}.$ In particular, this implies that $$({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T} = \left({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), \sum_{i \in I} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i}\right)_{j \in T}.$$ Write $B = R[x_{i}]_{i \in S} = R[\boldsymbol{x}]$ and $\mathfrak{b} = (f_{j}(\boldsymbol{x}))_{j \in T}.$ Then we can consider the surjective $B$-linear map $$\Omega^{1}_{B/R} = \bigoplus_{i \in I} B dx_{i} \twoheadrightarrow I/I^{2}$$ defined by $dx_{i} \mapsto \overline{\Delta_{i}}.$

It's time to compute its kernel. Consider a general element $$\sum_{i \in I}g_{i}(\boldsymbol{x}) dx_{i} \mapsto 0$$ under this map. This implies that $$\sum_{i \in I}g_{i}(\boldsymbol{x}) \Delta_{i} \in \left({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), \sum_{i \in S} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i}\right)_{j \in T}$$ in $R[\boldsymbol{x}, \boldsymbol{\Delta}],$ so we may write $$\sum_{i \in S}g_{i}(\boldsymbol{x}) \Delta_{i} = \sum_{i,i' \in S}h_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta}) \Delta_{i}\Delta_{i'} + \sum_{j \in S}h_{j}(\boldsymbol{x}, \boldsymbol{\Delta})f_{j}(\boldsymbol{x}) + \sum_{j \in S}\sum_{i \in S} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i},$$ so $$g_{i}(\boldsymbol{x}) = \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} + \text{ some element in } (f_{j}(\boldsymbol{x}))_{j \in T} = \mathfrak{b}$$ in $R[\boldsymbol{x}] = B.$ Conversely, for any such $g_{i}(\boldsymbol{x}),$ the sum $\sum_{i \in S}g_{i}(\boldsymbol{x}) dx_{i}$ is in the kernel. This implies that the kernel is precisely $$\mathfrak{b}\Omega_{B/R} + Bd_{B/R}\mathfrak{b},$$ so we now have the $B$-lienar isomorphism $$\Omega^{1}_{A/R} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega_{B/R} + Bd_{B/R}\mathfrak{b}} \simeq I/I^{2}.$$ We note that the $R$-linear derivation $A = R[\boldsymbol{x}]/(f_{j}(\boldsymbol{x}))_{j \in T} \rightarrow I/I^{2}$ is given by $g \mapsto 1 \otimes g - g \otimes 1$ can be explicitly described as $$g(\boldsymbol{x}) \mapsto g(\boldsymbol{x} + \boldsymbol{\Delta}) - g(\boldsymbol{x}) = \sum_{i \in I}\frac{\partial g(\boldsymbol{x})}{\partial x_{i}} \Delta_{i},$$ which shows that it must be the exterior derivative. $\Box$

Hodge theory: Lecture 15

In the past two lectures (Lecture 13 and Lecture 14), we used the facts (without proofs) that $\mathscr{A}^{p,q}_{M}$ (for any complex manifold $M$) and $\mathscr{A}^{p}_{M}$ (for any real manifold $M$) are acyclic. This let us have $$H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p, \bullet}_{M}))$$ for any complex manifold $M$ and $$H^{i}_{\mathrm{sing}}(M, \mathbb{R}) \simeq H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M}))$$ for any real manifold $M.$ The right-hand sides have more concrete for the sake of computations. For instance, the computation of $H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M}))$ on various examples of $M$ can be found in Section 28 of Tu's book.

Personal remark. I have not checked the isomorphism with the singular cohomology yet, but I should skip it for now and read these notes to fill this gap later. This is really great, because singular cohomology can be often flexible since we can use continuous deformations without changing the cohomology itself!

Our goal is to show that given any real manifold $M,$ any $\mathscr{C}^{\infty}_{M}$-module $\mathscr{F}$ is acyclic. In particular, this will show that the sheaves $\mathscr{A}^{p}$ and $\mathscr{A}^{p,q}$ (for complex manifolds) are acyclic. Our strategy is to show that

1. any $\mathscr{C}^{\infty}_{M}$-module is "soft", and
2. any soft sheaf is acyclic.

This posting will be more or less a regurgitation of these notes.

Restriction of sheaves to a subset. Let $\mathscr{F}$ be a sheaf valued in $\textbf{Ab}$ on a topological space $X.$ For any subset $S \subset X,$ its inclusion map $i_{S} : S \hookrightarrow X$ is continuous with the subspace topology. We denote by $\mathscr{F}|_{S} := i_{S}^{-1}\mathscr{F},$ which is a sheaf on $S,$ and write $\mathscr{F}(S) := \Gamma(S, \mathscr{F}|_{S}).$ From the last part of this posting, we have a concrete description of the set $\Gamma(S, \mathscr{F}|_{S})$: it consists of maps $$f : S \rightarrow \bigsqcup_{x \in S}\mathscr{F}_{x}$$ such that for any $x \in S,$ we may find an open $U \ni x$ in $X$ and $g \in \Gamma(U, \mathscr{F})$ such that $f(x) = g_{x}$ for all $x \in U \cap S.$ For convenience, we shall often write $$\Gamma(S, \mathscr{F}|_{S}) = \mathscr{F}(S),$$ which corresponds with our previous convention when $S$ is an open subset of $X.$ The global section of the sheaf map $$\mathscr{F} \rightarrow (i_{S})_{*}(\mathscr{F}|_{S})$$ (of sheaves on $X$) is given by $\Gamma(X, \mathscr{F}) \rightarrow \Gamma(S, \mathscr{F}|_{S}),$ which sends $h \mapsto h|_{S}.$ (This corresponds to the identity of $\mathscr{F}|_{S}$ in the adjunction formula in 2.7.B of Vakil.) We call this map the restriction map and note that this is the usual restriction map, if $S$ happens to be an open subset of $X.$

Personal remark. I think we should use pullbacks of modules (instead of inverse images of sheaves) to do similar activities as above when we are given ringed spaces, but I won't dig into these details, as we are not going to need this (as far as I reckon).

Remark. For our discussion, we only need to know restriction maps and technically do not need to know the inverse image sheaves. However, defining inverse image sheaves without the foundational story of sheaf theory seemed too ad-hoc and did not make musch sense to me, so I have summarized it in this posting. I will use the language in this summary because it seems impossible to avoid it for the discussions in this post.

Soft sheaves. A sheaf $\mathscr{F}$ valued in $\textbf{Ab}$ on a topological space $X$ is called soft if the restriction $\mathscr{F}(X) \rightarrow \mathscr{F}(Z)$ is surjective for every closed subset $Z \subset X.$

Paracompact spaces. A topological space $X$ is paracompact if

1. $X$ is Hausdorff and
2. every open cover $X = \bigcup_{i \in I}U_{i},$ there is a locally finite open cover $X = \bigcup_{j \in I}V_{j},$ refining the given cover.

Personal remark. The second condition is mouthful. Saying that $\{V_{j}\}_{j \in J}$ is locally finite in $X$ means that for any $x \in X,$ there are at most finitely many $V_{j}$ that contain $x.$ Saying that $\{V_{j}\}_{j \in J}$ refines $\{U_{i}\}_{i \in I}$ means that for each $V_{j}$ there is $U_{i} \supset V_{j}.$ This seems to me counter-intuitive at first, but I suppose I can think like "every small guy should refine some big guy" because if some $V_{j}$ does not lie in any $U_{i},$ it does not look like we are "refining" when we draw a picture. 

Example. If $X$ is compact, then $X$ is paracompact, because already we can find a finite open cover of $X.$

Example. If $X$ is a CW complex, then $X$ is paracompact. (I have not checked this yet, but I left the reference here.)

Example. Any smooth manifold is paracompact. More generally, any topological space that is locally compact, Hausdorff, and second-countable is paracompact, and the proof from Lee's book is written here.

Our goal is to prove the following two theorems:

Theorem 1. Let $X$ be a paracompact topological space. Given any sheaf $\mathscr{E}$ valued in $\textbf{Ab},$ we have $H^{i}(X, \mathscr{E}) = 0$ for all $i \geq 1.$ In particular, (say by the arguments in this posting) for any sheaf $\mathscr{F}$ valued in $\textbf{Ab},$ any resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{E}^{0} \rightarrow \mathscr{E}^{1} \rightarrow \mathscr{E}^{2} \rightarrow \cdots$$ with soft $\mathscr{E}^{i}$ (for all $i$) computes the cohomology of $\mathscr{F}$: $$H^{i}(X, \mathscr{F}) \simeq H^{i}(\Gamma(X, \mathscr{E}^{\bullet}))$$ for all $i \geq 0.$

Given a smooth real manifold $M$ (in particular, Hausdorff with a countable basis), consider the sheaf $\mathscr{C}_{M}^{\infty}$ of $\mathbb{R}$-valued smooth functions (on open subsets of $M$). Note that $(M, \mathscr{C}_{M}^{\infty})$ is a (locally) ringed space.

Theorem 2. Every $\mathscr{C}_{M}^{\infty}$-module is a soft sheaf.

Corollary. For any smooth real manifold $M,$ we have $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})) =: H^{i}_{\mathrm{dR}}(M)$$ for all $i \geq 0.$

Corollary. For any complex manifold $M,$ we have $$H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p,\bullet}_{M}))$$ for all $p, q \geq 0.$

Proof of Theorem 1. We will use one more word: a sheaf $\mathscr{A}$ on $X$ valued in $\textbf{Ab}$ is said to be flasque (or flabby) if for every pair of open subsets $U \hookrightarrow V$ in $X,$ the induced restriction map $\Gamma(V, \mathscr{A}) \rightarrow \Gamma(U, \mathscr{A})$ is surjective.

We proceed by induction on $i \geq 1.$ We use the following facts.

Fact 1. On a paracompact topological space, flasque sheaves are soft.

Fact 2. On any topological space, flasque sheaves are acyclic.

Fact 3. Given any soft sheaf $\mathscr{E}$ on a paracompact topological space $X,$ there is an exact sequence $0 \rightarrow \mathscr{E} \rightarrow \mathscr{A}$ in the category of sheaves on $X$ valued in $\textbf{Ab}$, where $\mathscr{A}$ is flasque.

Fact 4. On a paracompact topological space, given an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ if $\mathscr{E}$ and $\mathscr{A}$ are soft, then so is $\mathscr{B}.$

Fact 5. On a paracompact topological space, given any exact sequence $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0,$$ if $\mathscr{A}$ is soft, then the induced sequence on the global sections $$0 \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \rightarrow \Gamma(X, \mathscr{C}) \rightarrow 0$$ is exact.

Proof of Theorem 1. Now, using Fact 1, 2 and 4, we have an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ where $\mathscr{A}$ is flasque and $\mathscr{B}$ is soft.

Consider (coming from the sketchy argument from this posting) the long exact sequence $$0 \rightarrow \Gamma(X, \mathscr{E}) \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \xrightarrow{\alpha} H^{1}(X, \mathscr{E}) \rightarrow H^{1}(X, \mathscr{A}) \rightarrow \cdots.$$ By Fact 5, we note that $\alpha$ is surjective and by Fact 2, we see $H^{i}(X, \mathscr{A}) = 0$ for $i \geq 1.$ Using this and the long exact sequence, we see that $H^{1}(X, \mathscr{E}) = 0$ and $$H^{i}(X, \mathscr{B}) \simeq H^{i+1}(X, \mathscr{E})$$ for all $i \geq 1.$ Since $\mathscr{B}$ is soft, applying induction on $i \geq 1$ to the statement we want to prove, we are done. $\Box$

Proof of Theorem 2. We need the smooth version of Uryshon's lemma. We follow Tu's book (Problem 13.3 (a)).

Lemma (Smooth Uryshon). Let $M$ be a smooth real manifold. Fix any disjoint closed sets $A, B \subset M.$ There is $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(A) = \{0\}$ and $f(B) = \{1\}.$

Proof. Let $U := M \setminus A$ and $V := M \setminus B,$ which form an open cover of $M.$ Choose a $\mathscr{C}^{\infty}$ partition of unity $\varphi_{U}, \varphi_{V}$ subordinate to this open cover (e.g., Theorem 13.7 (b) in Tu's book). This means that $\varphi_{U}, \varphi_{V} \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that

• $\mathrm{supp}(\varphi_{U}) \subset U$ and $\mathrm{supp}(\varphi_{V}) \subset V,$ and
• $\varphi_{U} + \varphi_{V} = 1$ everywhere on $M.$

Since $\mathrm{supp}(\varphi_{V}) \subset V = M \setminus B,$ we note that $\varphi_{V} = 0$ on $B.$ Similarly, we see that $\varphi_{U} = 0$ on $A,$ so $\varphi_{V} = 1$ on $A.$ Thus, taking $f := \varphi_{V}$ will do the job. $\Box$

We also need another technical fact.

Fact 6. Let $X$ be any paracompact topological space and $\mathscr{F}$ be a sheaf on $X$ valued in $\textbf{Ab}.$ Then for every closed subset $Z \subset X$ and $s \in \mathscr{F}(Z),$ there is an open subset $U \supset Z$ in $X$ and $t \in \mathscr{F}(U)$ such that $t|_{U} = s.$

We now prove Theorem 2.

Proof of Theorem 2. Let $\mathscr{F}$ be any $\mathscr{C}_{M}^{\infty}$-module. Fix any closed subset $Z \subset M$ and let $s \in \mathscr{F}(Z).$ Our goal is to extend this to a section in $\mathscr{F}(M).$ Using Fact 6, we may first extend a little bit. Namely, there is an open subset $U \supset Z$ in $M$ and $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $M$ is paracompact, it is normal (e.g., see the last part of this posting) so that we can find open subsets $U_{1}, U_{2} \subset M$ such that $$Z \subset U_{1} \subset \bar{U_{1}} \subset U_{2} \subset \bar{U_{2}} \subset U.$$ Apply Smooth Uryshon to find $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(\bar{U_{1}}) = \{1\}$ and $f(M \setminus U_{2}) = \{0\}.$

We now crucially use the $\mathscr{C}_{M}^{\infty}$-module structure of on $\mathscr{F}.$ Namely, writing $f = f|_{U}$ as well, we have $ft \in \mathscr{F}(U).$ What's so special about this section? We have $(ft)|_{U \setminus \bar{U_{2}}} = 0$ because $$U \setminus \bar{U_{2}} \subset M \setminus \bar{U_{2}} \subset M \setminus U_{2},$$ where $f$ vanishes. This implies that $0 \in \Gamma(M \setminus \bar{U_{2}}, \mathscr{F})$ and $ft \in \Gamma(U, \mathscr{F})$ restrict to the same section on the intersection $(M \setminus \bar{U_{2}}) \cap U = U \setminus \bar{U_{2}}.$ Since $\mathscr{F}$ is a sheaf and $(M \setminus \bar{U_{2}}) \cup U = M,$ this means that there is a unique $g \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $g|_{M \setminus \bar{U_{2}}} = 0$ and $g|_{U} = ft.$ Since $f = 1$ on $U_{1} \supset Z,$ we must have $g|_{Z} = (ft)|_{Z} = t|_{Z} = s,$ which finishes the proof. $\Box$

Proof of technical facts. First, we note that Fact 6 implies Fact 1.

Fact 6 implies Fact 1. Let $\mathscr{F}$ be flasque. To show it is soft, let $Z \subset X$ be a closed subset. Given any $s \in \mathscr{F}(Z),$ we want to show that $s$ extends to an element in $\mathscr{F}(X).$ By Fact 1, we may find an open subset $U \supset Z$ in $X$ with $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $\mathscr{F}$ is flasque, we may find some $t' \in \mathscr{F}(X)$ such that $t'|_{U} = t,$ so $t'|_{Z} = s.$ Thus, we have shown that $\mathscr{F}$ is soft. $\Box$

Proof of Fact 6. Using the definition of $\mathscr{F}(Z),$ we may find a family $\{U_{i}\}_{i \in I}$ of open subsets of $X$ such that $Z \subset \bigcup_{i \in I}U_{i}$ with $s_{i} \in \mathscr{F}(Z)$ satisfying $s_{i}|_{U_{i} \cap Z} = s|_{U_{i} \cap Z}$ for each $i \in I.$ Then $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ forms an open cover of $X.$ Since $X$ is paracompact, this open cover must have a locally finite subcover. We may keep the open set $X \setminus Z$ in this subcover, and the union of other members cover $Z,$ so this procedure allows us to rechoose $U_{i}$ so that $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ is a locally finite open cover of $X.$

For every $x \in Z$, we have finitely many $U_{i_{1}}, \dots, U_{i_{r}}$ in this cover that contain $x$. Using paracompactness (or more precisely normality) of $X,$ we may find open $V_{i_{j}} \ni x$ in $U_{i_{j}}$ such that $\overline{V_{i_{j}}} \subset U_{i_{j}}.$ We note that the family $\{V_{i}\}_{i \in I} \cup \{X \setminus Z\}$ form an open cover of $X$ such that $$\overline{V_{i}} \subset U_{i}$$ and $Z \subset \bigcup_{i \in I}V_{i}.$

Personal remark. If we did not have local finiteness, we might still be okay in the situation above by some axiom of choice.

Fix $x \in X.$ Choose an open neighborhood $W_{x} \ni x$ in $X$ such that

• $W_{x}$ intersects only finitely many $U_{i}$ and
• $W_{x}$ is contained in some $V_{i'}.$

Write $s^{(x)} := s_{i}|_{W_{x}}.$ In particular, we have $s^{(x)}_{x} = (s_{i'})_{x} = s(x) \in \mathscr{F}_{x}.$ Whenever $x \notin \overline{V_{i}},$ we may replace $W_{x}$ by $W_{x} \setminus \overline{V_{i}}$ and modify $s^{(x)}$ accordingly, which would keep all the properties above. Since there can only be finitely many $\overline{V_{i}}$ intersecting $W_{x},$ this process will end after finitely many times, so we may assume that

• if $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ then $x \in \overline{V_{i}}.$

Whenever $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ we may replace $W_{x}$ with $W_{x} \cap U_{i},$ keeping all the properties given above. Since this process will end in finitely many times, we can also assume that

• if $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ then $W_{x} \subset U_{i}.$

Whenever $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ we have $s^{(x)}_{x} = s(x) = (s_{i})_{x},$ so by replacing $W_{x}$ with a small open neighborhood of $x$ in $W_{x}$ where $s^{(x)}$ and $s_{i}$ coincide, we may assume that $s^{(x)} = s_{i}|_{W_{x}}.$

Consider $U = \bigcup_{x \in Z}W_{x} \supset Z.$ Since $\mathscr{F}$ is a sheaf, to finish the entire proof, it is enough to show that $s^{(x)} = s^{(y)}$ on $W_{x} \cap W_{y}$ for all $x, y \in Z.$ Given any $z \in W_{x} \cap W_{y},$ we have some $V_{i} \ni x$ due to the above setting. This implies that $\overline{V_{i}} \neq W_{x} \neq \emptyset$ and $\overline{V_{i}} \neq W_{y} \neq \emptyset,$ so $W_{x}, W_{y} \subset U_{i},$ while $s_{i}|_{W_{x}} = s^{(x)}$ and $s_{i}|_{W_{y}} = s^{(y)}$ from the setting. This implies that $s^{(x)} = s_{i} = s^{(y)}$ on $W_{x} \cap W_{y},$ as desired. $\Box$

Proof of Fact 2. We follow these notes. Fix a topological space $X.$ First, we have a few facts.

Fact 2.1. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves (valued in $\textbf{Ab}$) on $X,$ if $\mathscr{F}$ is flasque, then we get an exact sequence $$0 \rightarrow \Gamma(X, \mathscr{F}) \rightarrow \Gamma(X, \mathscr{G}) \rightarrow \Gamma(X, \mathscr{H}) \rightarrow 0$$ when we take the global section functor.

Fact 2.2. Any injective sheaf on $X$ is flasque.

Fact 2.3. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves, if $\mathscr{F}$ and $\mathscr{G}$ are flasque, then $\mathscr{H}$ is flasque.

Proof of Fact 2. Let $\mathscr{F}$ be a flasque sheaf. Since $\textbf{Sh}_{X}$ has enough injectives (e.g., see Stacks Project), we may form an exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{I} \rightarrow \mathscr{Q} \rightarrow 0.$$ Consider the induced long exact sequence $$0 \rightarrow \Gamma(\mathscr{F}) \rightarrow \Gamma(\mathscr{I}) \rightarrow \Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F}) \rightarrow H^{1}(\mathscr{I}) \rightarrow \cdots$$ of abelian groups, where we suppressed $X$ in our notations. By Fact 2.1, the map $\Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F})$ is the zero map and since $\mathscr{I}$ is injective, we have $H^{1}(\mathscr{I}) = 0.$ Thus, we have $H^{1}(\mathscr{F}) = 0.$ For $i \geq 2,$ we have the exact sequence $$0 = \mathscr{H}^{i-1}(\mathscr{I}) \rightarrow H^{i-1}(\mathscr{H}) \rightarrow H^{i}(\mathscr{F}) \rightarrow \mathscr{H}^{i}(\mathscr{I}) = 0,$$ so $$H^{i-1}(\mathscr{H}) \simeq H^{i}(\mathscr{F}).$$ By Fact 2.2 and Fact 2.3, we note that $\mathscr{H}$ is also flasque, so arguing by induction, we may finish the proof. $\Box$

Proof of Fact 2.1. Let $s \in \Gamma(X, \mathscr{H}).$ Since taking the stalk at any point gives rise to an exact sequence, we may find an open cover $X = \bigcup_{i \in I}U_{i}$ with $f_{i} \in \Gamma(X, \mathscr{G})$ such that $f_{i} \mapsto s|_{U_{i}}$ for each $i \in I.$ Now, the pairs $(U_{i}, f_{i})$ form a partial ordering given by the inclusion of open subsets. Every chain has an upper bound, because $\mathscr{G}$ is a sheaf. Using Zorn's lemma, we may choose a maximal element $(U, f)$ among these pairs. In particular, we have $f \mapsto s|_{U}.$

To finish the proof, it is enough to show that $U = X.$ Suppose not. Then since $X = \bigcup_{i \in I}U_{i},$ there must be a nonempty $U_{i} \not\subset U.$ We have $$(f - f_{i})|_{U_{i} \cap U} \mapsto 0 \in \Gamma(U_{i} \cap U, \mathscr{H}),$$ so there must be (unique) $h \in \Gamma(U_{i}, \cap U, \mathscr{F})$ such that $h \mapsto (f - f_{i})|_{U_{i} \cap U}.$ But then since $\mathscr{F}$ is flasque, there is $\tilde{h} \in \Gamma(U_{i}, \mathscr{F})$ such that $\tilde{h}|_{U_{i} \cap U} = h.$ Now, consider $\tilde{h} \mapsto g \in \Gamma(U_{i}, \mathscr{G}).$ Then $f_{i} + g \in \Gamma(U_{i}, \mathscr{G})$ and $f \in \Gamma(U, \mathscr{G})$ coincide on $U_{i} \cap U,$ so we can glue them as a unique section $t \in \Gamma(U_{i} \cup U, \mathscr{G}),$ using the fact that $\mathscr{G}$ is a sheaf. We have $t \mapsto s|_{U_{i} \cup U},$ but then since $U \subsetneq U_{i} \cup U,$ we get a contradiction to the maximality of $(U, f).$ This finishes the proof. $\Box$

Proof of Fact 2.2. The proof is available in this posting. $\Box$

Proof of Fact 3. This immediately follows from Fact 2.2, because we may just take $\mathscr{A}$ to be an injective sheaf.

Proof that Fact 5 implies Fact 4. This is surprisingly simple. See Lemma 2.5 in these notes. $\Box$

It now remains to prove Fact 5.

Proof of Fact 5. We first prove a general lemma we will use.

Lemma. Let $X$ be any topological space. Say we are given a finite union $X = Z_{1} \cup \cdots \cup Z_{r}$ where each $Z_{i} \subset X$ is a closed subset. For any sheaf $\mathscr{F}$ on $X$ valued in $\textbf{Ab},$ the following gives an equalizer diagram (in $\textbf{Ab}$) $$0 \rightarrow \mathscr{F}(X) \rightarrow \prod_{i=1}^{r}\mathscr{F}(Z_{i}) \rightrightarrows \prod_{1 \leq i, j \leq r}\mathscr{F}(Z_{i} \cap Z_{j}),$$ where the map from $\mathscr{F}(X)$ is given by $h \mapsto (h|_{Z_{i}})_{1 \leq i \leq r}$ and the following pair of maps are given by

• $(s_{i})_{i=1}^{r} \mapsto (s_{i}|_{Z_{i} \cap Z_{j}})_{1 \leq i, j \leq r}$ and
• $(s_{i})_{i=1}^{r} \mapsto (s_{j}|_{Z_{i} \cap Z_{j}})_{1 \leq i, j \leq r}.$

Proof. We write $Z_{ij} := Z_{i} \cap Z_{j}.$ Let $(s_{i}) \in \prod_{i=1}^{r}\mathscr{F}(Z_{i})$ with $s_{i}|_{Z_{ij}} = s_{j}|_{Z_{ij}}$ for all $1 \leq i, j \leq r.$ Define $$s : X \rightarrow \bigsqcup_{x \in X}\mathscr{F}_{x}$$ by $s|_{Z_{i}} = s_{i}.$ This is a well-defined set map, and to show $s \in \Gamma(X, \mathscr{F}),$ or equivalently, we need to show that $s$ is continuous. (Note that uniqueness is already determined.) Since $s|_{Z_{i}} = s_{i}$ are continuous and $Z_{1}, \dots, Z_{r}$ form a finite cover of closed subsets of $X,$ it follows that $s$ is continuous. $\Box$

Finally, we prove Fact 5.

Proof of Fact 5. Write $$0 \rightarrow \mathscr{A} \xrightarrow{\phi} \mathscr{B} \xrightarrow{\psi} \mathscr{C} \rightarrow 0$$ to mean the given exact sequence. Let $c \in \Gamma(X, \mathscr{C}).$ Our goal is to show that there is $b \in \Gamma(X, \mathscr{B})$ such that $\psi(b) = c$ on $X.$ Exactness at $\mathscr{C}$ lets us choose an open cover $X = \bigcup_{i \in I}U_{i}$ with sections $b_{i} \in \Gamma(U_{i}, \mathscr{B})$ such that $\psi_{U_{i}}(b_{i}) = c|_{U_{i}}$ for each $i \in I.$ Since $X$ is paracompact, we may assume that $\{U_{i}\}_{i \in I}$ is locally finite. Since paracompactness implies normality, we may find an open cover $X = \bigcup_{i \in I}V_{i}$ such that $\overline{V_{i}} \subset U_{i}$ for each $i \in I.$ (We need closed subsets to utilize the softness of $\mathscr{A}.$)

Given any subset $J \subset I,$ we write $Z_{J} := \bigcup_{j \in J} \overline{V_{j}}.$ Note that the local finiteness of $\{V_{j}\}_{j \in J}$ implies that $$Z_{J} = \bigcup_{j \in J} \overline{V_{j}} = \overline{\bigcup_{j \in J} V_{j}},$$ so $Z_{J} \subset X$ is a closed subset. (See Exercise 13.7 of Tu.)

Consider the set of pairs $(J, b)$ where $J \subset I$ and $b \in \mathscr{B}(Z_{J})$ such that $\psi(b) = c$ on $Z_{J}.$ Such set has a partial order by declaring $(J, b) \leq (J', b')$ to mean $J \subset J'$ and $b'|_{Z_{J}} = b.$ Given any chain $\{(J_{\alpha}, b_{\alpha}) : \alpha \in \mathcal{L}\},$ let $J := \bigcup_{\alpha} J_{\alpha},$ and define $b : Z_{J} \rightarrow \mathscr{B}$ by locally defining it to be $b_{\alpha}$ on $Z_{J_{\alpha}}.$ For any closed subset $E \subset \mathscr{B},$ note that $$b^{-1}(E) \cap Z_{J_{\alpha}} = b_{\alpha}^{-1}(E)$$ is closed because $b_{\alpha}$ is continuous. Note that $\{Z_{J_{\alpha}} : \alpha \in \mathcal{L}\}$ is locally finite, so $$b^{-1}(E) = \bigcup_{\alpha}b_{\alpha}^{-1}(E) = \bigcup_{\alpha}\overline{b_{\alpha}^{-1}(E)} = \overline{\bigcup_{\alpha}b_{\alpha}^{-1}(E)},$$ so $b^{-1}(E) \subset X$ is closed, so $b$ is continuous. This gives us $b \in \mathscr{B}(Z_{J}),$ so we get an upper bound.

Personal remark. Even though I am following Wells (p. 52), I cannot understand the author's argument that we can find an upperbound using the gluing axiom of sheaves for $\mathscr{B}.$ This seems to be an important part of this proof, as now we are about to use Zorn's lemma.

Now, by Zorn's lemma, we may choose a maximal element $(J, b)$ in the partially ordered set. (Note that we reset the notation $(J, b).$) To finish the proof, it is enough to show that $J = I,$ because we then have $Z_{I} = X$ so that $\psi(b) = c$ on $X.$

Suppose not: $J \subsetneq I.$ Then there is $i \in I \setminus J,$ so in particular, we have $$Z_{J} \subsetneq Z_{J} \cup \overline{V_{j}} = Z_{J \cup \{i\}}.$$ Our strategy is to reach a contradiction to the maximality of $(J, b)$ by finding some $b'' \in \mathscr{B}(Z_{J \cup \{i\}})$ such that $\psi(b'') = c$ on $Z_{J \cup \{i\}}.$

We have $\psi(b_{i}) = c = \psi(b)$ on $\overline{V_{i}} \cap Z_{J},$ so$$\psi(b_{i} - b) = 0$$ on $\overline{V_{i}} \cap Z_{J}.$ With some thought, this implies that there is $a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J})$ such that $$b_{i} - b = \phi(a')$$ on $\overline{V_{i}} \cap Z_{J}.$ Since $\overline{V_{i}} \cap Z_{J} \subset X$ is closed, softness of $\mathscr{A}$ implies that there is $a \in \mathscr{A}(X)$ such that $$a|_{\overline{V_{i}} \cap Z_{J}} = a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J}).$$ Now, consider $b'_{i} := b_{i} - \phi(a'|_{U_{i}}) \in \mathscr{A}(U_{i}).$ Then $$\psi(b'_{i}) = \psi(b_{i}) - \psi(\phi(a')) = \psi(b_{i})$$ on $U_{i}.$ Moreover, we have $$b'_{i}|_{\overline{V_{i}} \cap Z_{J}} =  (b_{i} - \phi(a'))|_{\overline{V_{i}} \cap Z_{J}} = b|_{\overline{V_{i}} \cap Z_{J}}.$$ Hence by Lemma, we may find $b'' \in \mathscr{B}(\overline{V}_{i} \cup Z_{J}) = \mathscr{B}(Z_{J \cup \{i\}})$ such that $b''|_{\overline{V_{i}}} = b'_{i}$ and $b''|_{Z_{J}} = b.$ This implies that $\psi(b'') = c$ on $Z_{J \cup \{i\}} = \overline{V_{i}} \cup Z_{J},$ so we get $(J \cup \{i\}, b'') > (J, b),$ contradicting the maximality of $(J, b).$ This finishes the proof. $\Box$

Extension of an open embedding by zero (a.k.a lower-shriek functor)

Motivating problem. We would like to show the following statement. Let $\mathscr{I}$ be an injective sheaf on a topological space $X$ valued in $\textbf{Ab}.$ Then it is flasque, meaning that for every open $U \subset X,$ the restriction map $\Gamma(X, \mathscr{I}) \rightarrow \Gamma(U, \mathscr{I})$ is surjective.

Denote by $i : U \hookrightarrow X$ the corresponding open embedding. We define the functor $$i_{!} : \mathrm{Mod}_{\mathscr{O}_{U}} \rightarrow \mathrm{Mod}_{\mathscr{O}_{X}}$$ as follows: given an $\mathscr{O}_{U}$-module $\mathscr{F},$ we define $O(X)^{\mathrm{op}} \rightarrow \textbf{Ab}$ by $V \mapsto \Gamma(V, \mathscr{F})$ if $V \subset U$ and $V \mapsto 0$ if $V \not\subset U.$ This is a presheaf on $X$ because the restriction maps are either given by restriction maps of $\mathscr{F}$ if the relevant open subsets are within $V$ or zero maps if any of the relevant open subsets is not contained in $V.$ Functoriality is easy to check because if any bigger open subset is zero, then we would have to get zero maps.

We define $i_{!}\mathscr{F}$ the sheafification of this presheaf. Since the presheaf has $\mathscr{O}_{X}$ action on it, this gives an $\mathscr{O}_{X}$-module structure on $i_{!}\mathscr{F}.$ Given any $\mathscr{O}_{U}$-linear map $\mathscr{F}_{1} \rightarrow \mathscr{F}_{2},$ we can easily get the map between corresponding presheaves on $X,$ which gives us a sheaf map $i_{!}\mathscr{F}_{1} \rightarrow i_{!}\mathscr{F}_{2}.$ Functoriality for a composition $\mathscr{F}_{1} \rightarrow \mathscr{F}_{2} \rightarrow \mathscr{F}_{3}$ is also easy to check, which tells us that $i_{!}$ is indeed a functor.

What's important is the following.

Adjunction. We have $$\mathrm{Hom}_{\mathscr{O}_{X}}(i_{!}\mathscr{F}, \mathscr{G}) \simeq \mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{F}, i^{-1}\mathscr{G}),$$ functorial in $\mathscr{F}$ and $\mathscr{G},$ where the isomorphism is taken in $\textbf{Set},$ and it is important to remember how this map is given.

If we start with an $\mathscr{O}_{X}$-linear map $i_{!}\mathscr{F} \rightarrow \mathscr{G},$ for any open $V \subset X,$ we have $$\Gamma(U \cap V, \mathscr{F}) \rightarrow \Gamma(U \cap V, i_{!}\mathscr{F}) \rightarrow \Gamma(U \cap V, \mathscr{G}),$$ and this is compatible with inclusions in $X$ as we consider other open subsets. Hence, this gives rise to $\mathscr{F} \rightarrow \mathscr{G}|_{U} = i^{-1}\mathscr{G}.$

Conversely, if we are given $\mathscr{F} \rightarrow i^{-1}\mathscr{G},$ then for any open $V \subset X,$ if $V \subset U,$ consider $\Gamma(V, \mathscr{F}) \rightarrow \Gamma(V, \mathscr{G})$ and if $V \not\subset U,$ consider $0 \rightarrow \Gamma(V, \mathscr{G}).$ As we vary $V,$ with respect to inclusion, we get a map from the presheaf we discussed above whose sheafification is $i_{!}\mathscr{F}$ to $\mathscr{G}.$

It is not difficult to check that these two procedures are inverses to each other (partially because $\mathscr{G}$ is a sheaf, which lets us use the universal property of the sheafification).

Personal remark. Note that $$\begin{align*} i^{*}\mathscr{G} &= i^{-1}\mathscr{G} \otimes_{i^{-1}\mathscr{O}_{X}}\mathscr{O}_{U} \\ &= i^{-1}\mathscr{G} \otimes_{\mathscr{O}_{U}}\mathscr{O}_{U} \\ &= i^{-1}\mathscr{G} \end{align*}$$ and both of them are $\mathscr{G}|_{U}.$ Note that this is a special situation where $i$ is an open embedding.

Understanding local sections. For our main application, we take $\mathscr{F} = \mathscr{O}_{U}$ in the adunction formula so that we have $$\mathrm{Hom}_{\mathscr{O}_{X}}(i_{!}\mathscr{O}_{U}, \mathscr{G}) \simeq \mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{O}_{U}, \mathscr{G}|_{U}).$$ Now, note that $$\mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{O}_{U}, \mathscr{G}|_{U}) \simeq \Gamma(U, \mathscr{G}),$$ again where we consider the isomorphism only as a bijection given by $\phi \mapsto \phi_{U}(1).$

We are now ready to tackle our motivating problem. We are given a topological space $X$ and an injective sheaf $\mathscr{I}$ valued in $\textbf{Ab}.$ We can view $X$ as a ringed space $(X, \mathscr{O}_{X}),$ where $\mathscr{O}_{X} = \underline{\mathbb{Z}}.$ Sheaves on $X$ valued in $\textbf{Ab}$ are precisely $\underline{\mathbb{Z}}$-modules. Hence, we will be done if we show the following.

Theorem. Let $(X, \mathscr{O}_{X})$ be a ringed space. Any injective $\mathscr{O}_{X}$-module is flasque.

Proof. Consider any injective $\mathscr{O}_{X}$-module $\mathscr{I}$ and an open subset $U \subset X.$ Let $s \in \Gamma(U, \mathscr{I}).$ Our goal is to find $t \in \Gamma(X, \mathscr{I})$ such that $t|_{U} = s.$

We know $s$ corresponds to an $\mathscr{O}_{U}$-linear map $\mathscr{O}_{U} \rightarrow \mathscr{I}|_{U}$ whose local section $$\Gamma(U \cap V, \mathscr{O}_{U}) \rightarrow \Gamma(U \cap V, \mathscr{I})$$ is given by $f \mapsto f s|_{U \cap V}.$ By adjunction, this corresponds to an $\mathscr{O}_{X}$-linear map $i_{!}\mathscr{O}_{U} \rightarrow \mathscr{I}.$ Note that the sequence of $\mathscr{O}_{X}$-linear maps $$0 \rightarrow i_{!}\mathscr{O}_{U} \rightarrow \mathscr{O}_{X}$$ the second of which is corresponding to $\mathrm{id} : \mathscr{O}_{U} \rightarrow \mathscr{O}_{X}|_{U}$ is exact as we can check its stalks. Since $\mathscr{I}$ is an injective $\mathscr{O}_{X}$-module, the map $i_{!}\mathscr{O}_{U} \rightarrow \mathscr{I}$ extends to an $\mathscr{O}_{X}$-linear map $\psi : \mathscr{O}_{X} \rightarrow \mathscr{I}.$ Let $t = \psi_{X}(1) \in \Gamma(X, \mathscr{I})$ be the corresponding section. Then we have $t|_{U} = s$ by sending $1$ according to the following identical compositions:

• $\Gamma(U, \mathscr{O}_{U}) = \Gamma(U, \mathscr{O}_{X}) \rightarrow \Gamma(U, \mathscr{I}),$
• $\Gamma(U, \mathscr{O}_{U}) \rightarrow \Gamma(U, i_{!}\mathscr{O}_{U}) \rightarrow \Gamma(U, \mathscr{I}).$

This finishes the proof $\Box$

Paracompactness

This rather formal notion comes up when I deal with computing singular/de Rham cohomology, so I wanted to just put it separately. We are following Lee and Hatcher for this posting (as well as Wikipedia, etc).

Paracompact spaces. A topological space $X$ is paracompact if

1. $X$ is Hausdorff and
2. for every open cover $X = \bigcup_{i \in I}U_{i},$ there is a locally finite open cover $X = \bigcup_{j \in I}V_{j},$ refining the given cover.

Personal remark. The second condition is mouthful. Saying that $\{V_{j}\}_{j \in J}$ is locally finite in $X$ means that for any $x \in X,$ there are at most finitely many $V_{j}$ that contain $x.$ Saying that $\{V_{j}\}_{j \in J}$ refines $\{U_{i}\}_{i \in I}$ means that for each $V_{j}$ there is $U_{i} \supset V_{j}.$ This seems to me counter-intuitive at first, but I suppose I can think like "every small guy should refine some big guy" because if some $V_{j}$ does not lie in any $U_{i},$ it does not look like we are "refining" when we draw a picture.

Example. If $X$ is Hausdorff and compact, then $X$ is paracompact, because for any open cover of $X,$ we can find a finite open subcover, which is certainly a locally finite refinement that covers $X.$

Example. If $X$ is a CW complex, then $X$ is paracompact. To see $X$ is Hausdorff, note that if any two distinct points lie in a cell with dimension $> 0,$ then we can take disjoint open neighborhoods of those points in the cell. Otherwise, both are $0$-cells, which can be separated by open sets in $X.$

Of course, a finite CW complex (a CW complex with finitely many cells in total) is already compact, so it is paracompact.

For a general CW complex, we may have infinitely many cells, so it cannot be compact. For instance, the real line $\mathbb{R}$ with the Euclidean topology is a CW complex because we can take $\mathbb{Z}$ to be $0$-cells, and the segments $$\cdots, [-2, -1], [-1, 0], [0, 1], [1, 2], \dots$$ $1$-cells. However, we know that $\mathbb{R}$ is not compact because it is not bounded.

The proof that a general CW complex is paracompact seems to be considered "elementary" by the community, but Fritsch and Piccinini (Theorem 1.3.5) spends about two full pages, which I decided to skip for now.

Example. Any topological manifold is paracompact.

We shall spend the rest of this posting to prove this statement.

Remark. Topological manifolds are somewhat more general notion of spaces incorporating smooth manifolds. A topological space $M$ is a topological manifold if

• $M$ is Hausdorff,
• $M$ is second-countable (meaning it has a countable basis),
• $M$ is locally Euclidean of dimension $n$ for some $n \in \mathbb{Z}_{\geq 0}.$

The second-countablity axiom will be used as the following form:

Lemma. If $X$ is a second-countable topological space, any basis of $X$ has a countable subbasis.

Proof. Let $\mathcal{U} = \{U_{i}\}_{i \in I}$ be any basis and $\mathcal{B} = \{B_{j}\}_{j=1}^{\infty}$ be a countable basis of $X.$ We follow a clever observation from these notes: any collection of open subsets with which we can write each $B_{j}$ as a countable union is a countable basis. Since $U_{i}$ form a basis, we can write $$B_{j} = \bigcup_{i \in I_{j}}U_{i},$$ and we would have been done if $I_{j} \subset I$ was countable. If not, for each $x \in B_{j},$ choose any $j_{x} \in J$ such that $x \in B_{j_{x}} \subset U_{i_{x}}$ for some $i_{x} \in I_{j},$ which we choose and fix. There can be some repetitions for $B_{j_{x}}$ and $U_{i_{x}},$ but the size of the collection of all $B_{j}$ where $j$ can appear as $j = j_{x}$ is greater than equal to those $U_{i}$ where $i$ can appear as $i = i_{x}.$ Such $i$ form an at most countable subcollection of $I_{j},$ which finishes the proof. $\Box$

The last axiom for topological manifolds means that for any $x \in M,$ there is an open neighborhood $U \ni x$ in $M$ and a homeomorphic open embedding $U \hookrightarrow \mathbb{R}^{n}.$ The number $n$ appearing here is called the dimension of $M.$ Note that for a smooth manifold, the homeomorphic open embeddings are also need to be "smoothly compatible" (see $\mathscr{C}^{\infty}$-compatbility on Section 5.2 in Tu's book). Note that any topological manifold $M$ is locally compact, meaning that every $x \in M$ of it has an open neighborhood contained in a compact subset of $M.$

Lemma. Let $X$ be a topological space. If $X$ is Hausdorff, then the following are equivalent:

1. $X$ is locally compact;
2. Every $x \in X$ has a precompact (compact closure) open neighborhood in $X$;
3. $X$ has a basis consisting of precompact open subsets of $X.$

Proof. Assume Statement 1. For $x \in X,$ take $x \in U \subset K,$ where $U$ is open in $X$ and $K$ is compact. Since $X$ is Hausdorff, we see that $K \subset X$ is closed. Thus, we have $\bar{U} \subset K,$ so $U$ is precompact. This shows Statement 2.

Assume Statement 2. For any given basis $\mathcal{B}$ of $X,$ let $\mathcal{B}'$ be the subcollection of $\mathcal{B}$ consisting of only precompact open subsets. We show that $\mathcal{B}'$ is still a basis of $X.$ For any $x \in X,$ using the assumption, pick any precompact open $U \ni x.$ Then we may find $B \in \mathcal{B}$ such that $x \in B \subset U.$ This implies that $\bar{B} \subset \bar{U},$ and since $\bar{U}$ is compact, its closed subset $\bar{B}$ must be compact as well. This implies that $B \in \mathcal{B}',$ showing our claim that it is a basis of $X.$ This shows Statement 3.

Statement 3 evidently implies Statement 1. $\Box$

Lemma 1. Let $X$ be a topological space that is

• second-countable, 
• locally compact, and 
• Hausdorff.

Then $X$ admits an exhaustion by compact subsets, meaning an sequence $K_{1} \subset K_{2} \subset \cdots$ of compact subsets whose union covers $X$ such that $K_{i} \subset K_{i+1}^{\mathrm{o}}$ for all $i \geq 1.$

Proof. Since $X$ is locally compact Hausdorff, we may find a basis consisting of precompact open subsets of $X.$ As we have shown above, the hypothesis that $X$ is second countable allows us to assume that this basis is countable. In particular, we have a countable precompact open cover $X = \bigcup_{i=1}^{\infty}U_{i}.$ We define $K_{1} := \overline{U_{1}}$ in $X,$ which is compact because $U_{1}$ is precompact. Write $j_{1} := 1.$ Since $K_{1}$ is compact, there must be $j_{2} \in \mathbb{Z}_{\geq 1}$ such that $$K_{1} \subset U_{1} \cup U_{2}, \cdots \cup U_{j_{2}}.$$ We define $K_{2} := \overline{U_{1}} \cup \cdots \cup \overline{U_{j_{2}}},$ which is compact in $X$ due to precompactness of $U_{i}.$ This way, whenever we have  can construct $$K_{i} := \overline{U_{1}} \cup \cdots \cup \overline{U_{j_{i}}}$$ and pick $j_{i+1} > j_{i}$ such that $$K_{i} \subset U_{1} \cup \cdots \cup U_{j_{i+1}}.$$ This way, we construct $K_{1} \subset K_{2} \subset K_{3} \subset \cdots$ recursively such that $K_{i} \subset K_{i+1}^{\mathrm{o}}$ for all $i \geq 1.$ This finishes the proof $\Box$

Lemma 2. Let $X$ be a topological space that admits an exhaustion by compact subsets. Then for any open cover $\mathcal{U}$ and any basis $\mathcal{B}$ of $M,$ there is a refinement $\mathcal{U}'$ of $\mathcal{U}$ such that

• $\mathcal{U}'$ is still an open cover of $X,$
• $\mathcal{U}'$ is countable,
• $\mathcal{U}'$ is locally finite, and
• $\mathcal{U}' \subset \mathcal{B}.$

Proof. Let $K_{1} \subset K_{2} \subset K_{3} \subset \cdots$ be a compact exhaustion of $X.$ Define $$R_{i} := K_{i+1} \smallsetminus K_{i}^{\mathrm{o}}$$ and $$T_{i} := K_{i+2}^{\mathrm{o}} \smallsetminus K_{i-1} \supset R_{i},$$ where $K_{0} := \emptyset.$ Note that $R_{i}$ is compact because it is a closed subset of a compact set $K_{i+1}.$ For each $x \in R_{i},$ we may take $U_{x} \in \mathcal{U}$ and $B_{x} \in \mathcal{B}$ such that $x \in B_{x} \subset U_{x} \cap T_{i}.$ Since $R_{i}$ is compact, we can find finitely many $B_{x_{i,1}}, \dots, B_{x_{i,n_{i}}}$ such that $$R_{i} \subset B_{x_{i,1}} \cup B_{x_{i,2}} \cup \cdots \cup B_{x_{i,n_{i}}} \subset (U_{i,x_{1}} \cup U_{i,x_{2}} \cup \cdots \cup U_{i,x_{n_{i}}}) \cap T_{i}.$$ Now, we may take $$\mathcal{U}' := \{B_{i,j} : i \in \mathbb{Z}_{\geq 1}, 1 \leq j \leq n_{i}\},$$ and check that it is locally finite and covers $X.$ This finishes the proof $\Box$

Corollary. Every locally compact, Hausdorff, and second-countable topoloigical space is paracompact. In particular, topoloigcal manifolds are paracompact.

Paracompactness implies normality. Recall that a topological space $X$ is said to be normal if for any disjoint closed subsets $A, B$ of $X,$ we can find disjoint open subsets $V \supset A$ and $W \supset B$ in $X.$ Note that this condition is equivalent to say that for every closed $A \subset X$ and open $U \supset A$ in $X,$ we can find another open subset $V$ of $X$ such that $A \subset \bar{V} \subset U,$ where the closure is taken in $X.$

Proof. Let $X$ be normal. Given any closed subset $A \subset X$ and an open subset $U \supset A$ in $X,$ the complement $B := X \smallsetminus U$ is a closed subset of $X$ disjoint to $A.$ Hence, we may find disjoint open subsets $V \supset A$ and $W \supset B$ in $X$ using normality. But then $$A \subset V \subset X \smallsetminus W \subset X \smallsetminus B = U,$$ and since $X \smallsetminus W$ is closed in $X,$ it contains $\bar{V},$ which implies what we want.

Conversely, assume the condition on $X$ we hoped to be equivalent to nomarlity of $X.$ Fix any disjotint closed subsets $A, B \subset X.$ Then we have $A \subset U := X \smallsetminus B,$ so we may find an open subset $V \subset X$ such that $A \subset V \subset \bar{V} \subset U.$ This implies that $$W := X \smallsetminus \bar{V} \supset B$$ and $W$ is an open subset of $X$ disjoint to $W.$ This shows that $X$ is normal. $\Box$

Theorem. Any paracompact topological space is nomal.

Proof. Let $A, B \subset X$ be disjoint closed subsets. We want to find disjoint open $V, W \subset X$ such that $A \subset V$ and $B \subset W.$

Case 1. Assume that $A = \{a\},$ a singleton.

For each $x \in B,$ using Hausdorffness, choose open $V_{x} \ni a$ and $W_{x} \ni x$ such that $V_{x} \cap W_{x} = \emptyset.$ In particular, we have $$\overline{W_{x}} \subset X \smallsetminus V_{x},$$ so $a \notin \overline{W_{x}}.$

This way, we construct an open cover $\{W_{x}\}_{x \in B}$ for $B$ in $X.$ Then $\{W_{x}\}_{x \in B} \cup \{X \smallsetminus B\}$ is an open cover of $X,$ so by paracompactness of $X,$ there is a locally finite refinement that is an open cover of $X.$ We may assume that it is of the form $\{W'_{x}\}_{x \in B} \cup \{U_{B}\},$ where $W'_{x} \subset W_{x}$ and $U_{B} \subset X \smallsetminus B.$ Note that $\{W'_{x}\}_{x \in B}$ is an open cover of $B$ in $X$ that is locally finite and $a \notin \overline{W'_{x}}.$

The key observation is that due to local finiteness, we have $$\overline{\bigcup_{x \in B} W'_{x}} = \bigcup_{x \in B} \overline{W'_{x}}$$ because any limit point of $\bigcup_{x \in B} W'_{x}$ is necessarily a limit point of the union of some finitely many $W'_{x}.$ In particular, this shows that the left-hand side does not contain $a.$ Then taking $$V := X \smallsetminus \overline{\bigcup_{x \in B} W'_{x}} \ni a$$ and $$W := \bigcup_{x \in B} W'_{x} \supset B,$$ we finish the proof for this case.

Case 2. Now we consider the general case.

For each $a \in A,$ take an open subset $V_{a} \ni a$ and $W_{a} \supset B$ in $X$ such that $V_{a} \cap W_{a} = \emptyset,$ using Case 1. Note that $\overline{V_{a}} \subset X \smallsetminus B,$ so $\overline{V_{a}} \cap B = \emptyset.$

Then $\{V_{a}\}_{a \in A}$ gives an open cover of $A,$ so $\{V_{a}\}_{a \in A} \cup \{X \smallsetminus B\}$ is an open cover of $X.$ Using paracompactness, we may find a locally finite refinement of this cover that is also an open cover of $X.$ We can write it in the form $\{V'_{a}\}_{a \in A} \cup \{U_{B}\}$ where $V'_{a} \subset V_{a}$ and $U_{B} \subset X \setminus B.$ Again, the upshot is that we have $$\overline{\bigcup_{a \in A} V'_{a}} = \bigcup_{a \in A} \overline{V'_{a}}$$ due to local finiteness. In particular, the left-hand side is disjoint to $B,$ so can obtain an open subset $$W := X \smallsetminus \overline{\bigcup_{a \in A} V'_{a}} \supset B$$ in $X.$ Since $$V := \bigcup_{a \in A} V'_{a} \supset A$$ is open and disjoint to $V,$ this finishes the proof $\Box$

The space (espace étale) of a presheaf

We follow Chapter 2 of Vakil and Chapter 2 of Wells. The organization might be slightly different from either of the references.

Fix a topological space $X.$ Let $\mathscr{F}$ be a presheaf on $X$ valued in $\textbf{Set},$ the category of sets. We will later change $\textbf{Set}$ into the following categories:

• $\textbf{Ab},$ the category of abelian groups;
• $\textbf{Ring},$ the category of (commutative) rings (with unity).

Given a presheaf $\mathscr{O}$ on $X$ valued in $\textbf{Ring},$ a presheaf $\mathscr{F}$ on $X$ valued in $\textbf{Ab}$ is an $\mathscr{O}$-module if we have multiplication maps $$\mathscr{O}(U) \times \Gamma(U, \mathscr{F}) \rightarrow \Gamma(U, \mathscr{F}),$$ for open $U \subset X,$ compatible with inclusion maps among various $U.$

However, for now, if we do not mention anything, the notation $\mathscr{F}$ will be a presheaf on $X$ valued in $\textbf{Set}.$

The espace étale of $\mathscr{F}$ is a topological space $\tilde{\mathscr{F}}$ constructed by the following procedure.

1. As a set, we define $\tilde{\mathscr{F}} := \bigsqcup_{x \in X}\mathscr{F}_{x}.$ Note that this comes with a set map $\pi_{\mathscr{F}} : \tilde{\mathscr{F}} \rightarrow X$ merely by $\mathscr{F}_{x} \rightarrow \{x\}.$ 
2. We give $\tilde{\mathscr{F}}$ a topology as follows.

Topology on the espace étale of a presheaf. Let $U \subset X$ be any open subset and fix any $s \in \Gamma(U, \mathscr{F}).$ Consider the set map $\tilde{s} : U \rightarrow \tilde{\mathscr{F}}$ defined by $$x \mapsto s_{x} \in \mathscr{F}_{x}.$$ Note that we have $\pi_{\mathscr{F}} \circ \tilde{s} = \mathrm{id}_{U},$ which is philosophically satisfying because we want to think of $\tilde{s}$ (or even just $s$) as a "section".

Lemma. Denote by $\mathrm{O}(X)$ the set of all open subsets of $X.$ The collection $$\{\tilde{s}(U) : U \in \mathrm{O}(X), s \in \Gamma(U, \mathscr{F})\}$$ from a basis for a topology on $\tilde{\mathscr{F}}.$

Proof. Fix any $x \in X$ and $s_{x} \in \mathscr{F}_{x}.$ By definition, this is from some $s \in \Gamma(U, \mathscr{F})$ with some open $U \ni x$ in $X.$ Again by definition, we have $s_{x} \in \tilde{s}(U).$ Hence, the sets of $\tilde{s}(U)$ with $U \in \mathrm{O}(X)$ and $s \in \Gamma(U, \mathscr{F})$ cover $\tilde{\mathscr{F}}.$

Now, take any $u_{x} \in \tilde{s}(U) \cap \tilde{t}(V).$ (Note that $u_{x} \in \mathscr{F}_{x}.$) This means that we have $s \in \Gamma(U, \mathscr{F})$ and $t \in \Gamma(V, \mathscr{F})$ such that $s_{x} = u_{x} = t_{x}.$ By definition of stalks, we can find an open neighborhood $W \ni x$ in $U \cap V$ such that $s|_{W} = u|_{W} = t|_{W} \in \Gamma(W, \mathscr{F}).$ This implies that $\tilde{s}|_{W} = \tilde{u}|_{W} = \tilde{t}|_{W},$ and it follows that $$u_{x} \in \tilde{u}(W) \subset \tilde{u}(W) = \tilde{s}(W) = \tilde{t}(W) \subset \tilde{s}(U) \cap \tilde{t}(V).$$ Thus, we have shown that the collection $$\{\tilde{s}(U) : U \in \mathrm{O}(X), s \in \Gamma(U, \mathscr{F})\}$$ forms a basis for a topology of $\tilde{\mathscr{F}}.$ $\Box$

Upshot. Thus, we can give $\tilde{\mathscr{F}}$ the topology generated by the basis we have given.

Lemma. Fix any open $U \subset X$ and $s \in \Gamma(U, X).$ Then $\tilde{s} : U \rightarrow \tilde{\mathscr{F}}$ is continuous.

Proof. For any basic open $\tilde{t}(V) \subset \tilde{\mathscr{F}},$ we have $$\begin{align*}\tilde{s}^{-1}(\tilde{t}(V)) &= \{x \in U : s_{x} \in \tilde{t}(V)\} \\ &= \{x \in V : s_{x} \in \tilde{t}(V)\} \\ &= \{x \in V : s_{x} = t_{x}\} \\ &= \tilde{s}^{-1}(V) \cap \tilde{t}^{-1}(V). \end{align*}$$ Hence, for any $x \in \tilde{s}^{-1}(\tilde{t}(V)),$ namely a point $x \in X$ with $s_{x} = t_{x},$ we can find open $W \ni x$ in $U \cap V$ such that $s|_{W} = t|_{W}.$ This means that $$x \in W \subset \tilde{s}^{-1}(\tilde{t}(V)).$$ Thus $\tilde{s}^{-1}(\tilde{t}(V))$ is open in $U.$ $\Box$

Lemma. The map $\pi_{\mathscr{F}} : \tilde{\mathscr{F}} \rightarrow X$ is continuous.

Proof. Fix any open $U \subset X.$ Then $\pi_{\mathscr{F}}^{-1}(U)$ is the union of the sets of the form $\tilde{s}(U)$ with $s \in \Gamma(U, \mathscr{F}).$ Any such set is a member of the basis for $\tilde{\mathscr{F}}$ we use, so in particular it is open in $\tilde{\mathscr{F}}.$ The union of open subsets is open in any topology, so $\pi_{\mathscr{F}}^{-1}(U)$ must be open. This finishes the proof. $\Box$

Lemma. The map $\pi_{\mathscr{F}} : \tilde{\mathscr{F}} \rightarrow X$ is a local homeomorphism.

Proof. Fix any $s_{x} \in \mathscr{F}_{x}.$ This gives some $s \in \Gamma(U, \mathscr{F})$ and $\tilde{s} : U \rightarrow \tilde{\mathscr{F}},$ which we checked to be continuous. If we restrict the target, the map $\tilde{s} : U \rightarrow \tilde{s}(U)$ has the continuous inverse given by $\pi_{\mathscr{F}},$ so this finishes the proof. $\Box$


Continuous sections of a presheaf. Given any open $U \subset X,$ let $\Gamma(U, \tilde{\mathscr{F}})$ be the set of all continuous $f : U \rightarrow \tilde{\mathscr{F}}$ such that $\pi \circ f = \mathrm{id}$ (i.e., a section).

Theorem. The set $$\Gamma(U, \tilde{\mathscr{F}})$$ precisely consists of set maps $f : U \rightarrow \tilde{\mathscr{F}}$ such that for each $x \in U,$

1. $f(x) \in \mathscr{F}_{x}$ and
2. there is some open $V \ni x$ in $U$ and $s \in \Gamma(U, \mathscr{F})$ such that $f|_{V} = \tilde{s}|_{V}.$

Proof. Saying $\pi(f(x)) = x$ is equivalent to saying $f(x) \in \mathscr{F}_{x},$ so the first condition is equivalent to saying that $f$ is a section (i.e., $\pi \circ f = \mathrm{id}$). Hence, we will assume the first condition now throughout the proof.

If $f$ satisfies the given conditions, then for any $x \in U,$ we have $\tilde{s}(V) \ni f(x),$ which is open in $\tilde{\mathscr{F}}$ such that $f^{-1}(\tilde{s}(V)) = V,$ which is open. (Note that $f^{-1}(\tilde{s}(V))$ is already a subset of $V$ before thinking about the second condition.)

Conversely, let $f \in \Gamma(U, \tilde{\mathscr{F}}).$ By continuity, we may find $\tilde{t}(W) \ni f(x)$ such that $f^{-1}(\tilde{t}(W)) \subset U$ is open. This means that $W \ni x$ is a open neighborhood in $U$ and $t \in \Gamma(W, \mathscr{F})$ such that $f(x) = t_{x} \in W.$ Now, take any open $V \subset U$ such that $x \in V \subset f^{-1}(\tilde{t}(W)),$ using openness. Letting $s := t|_{V},$ we get the second condition. $\Box$

Continuous sections of a presheaf forms the sheafification. Compare the above theorem with 2.4.7 in Vakil. Yes, the continuous sections of $\mathscr{F}$ will construct a sheafification of $\mathscr{F}.$

Remark. Given any open subsets $V \subset U$ in $X,$ we have $\Gamma(U, \tilde{\mathscr{F}}) \rightarrow \Gamma(V, \tilde{\mathscr{F}}),$ defined by literal restrictions. It is an easy check that $\tilde{\mathscr{F}}$ forms a presheaf with these restrictions. We use the notation $\tilde{\mathscr{F}}$ to also mean the presheaf of continuous sections of it.

Lemma. In fact, the presheaf $\tilde{\mathscr{F}}$ is a sheaf.

Proof. Let $U \subset X$ be an open subset and $U = \bigcup_{I \in I}U_{i}$ an open covering. Let $f, g \in \Gamma(U, \tilde{\mathscr{F}})$ satisfy $f|_{U_{i}} = g|_{U_{i}}$ for all $U_{i}.$ Then $f_{x} = g_{x} \in \mathscr{F}_{x}$ for any $x \in U,$ so $f = g.$ Now, given $f_{i} \in \Gamma(U, \tilde{\mathscr{F}})$ for $i \in I$ such that $f_{i}| _{U_{ij}} = f_{j}|_{U_{ij}}$ for all $i, j \in I,$ we can define $f : U \rightarrow \tilde{\mathscr{F}}$ by $f(x) := f_{i}(x)$ for $x \in U_{i}.$ Continuity can be checked locally, so we showed that we can glue sections uniquely. This shows that $\tilde{\mathscr{F}}$ is a sheaf. $\Box$

Theorem. The map $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ defined by $\Gamma(U, \mathscr{F}) \rightarrow \Gamma(U, \tilde{\mathscr{F}})$ with $s \mapsto \tilde{s}$ is a sheafification. That is, it is a map of presheaves such that if there is any sheaf $\mathscr{G}$ and a map $\mathscr{F} \rightarrow \mathscr{G}$ it must factor uniquely as $\mathscr{F} \rightarrow \tilde{\mathscr{F}} \rightarrow \mathscr{G}.$

Proof. It is easy to check that what we have described gives a map of presheaves $\phi : \mathscr{F} \rightarrow \tilde{\mathscr{F}}.$ Now, fix any sheaf $\mathscr{G}$ and a map $\psi : \mathscr{F} \rightarrow \mathscr{G}.$

Fix $f \in \Gamma(U, \tilde{\mathscr{F}}).$ We can find an open cover $U = \bigcup_{i \in I}U_{i}$ with $s_{i} \in \Gamma(U_{i}, \mathscr{F})$ such that $\tilde{s}_{i} = f|_{U_{i}}.$ Thus, our desired map $\eta_{U_{i}} : \Gamma(U_{i}, \tilde{\mathscr{F}}) \rightarrow \Gamma(U_{i}, \mathscr{G})$ must assign $\tilde{s}_{i} \mapsto \psi_{U_{i}}(s_{i}).$ Since $$\psi_{U_{i}}(s_{i})|_{U_{ij}} = \psi_{U_{ij}}(s_{i}|_{U_{ij}}) = \psi_{U_{ij}}(s_{j}|_{U_{ij}}) = \psi_{U_{j}}(s_{j})|_{U_{ij}},$$ there is a unique $\eta_{U}(f)$ that restricts to $\psi_{U_{i}}(s_{i})$ on $U_{i}.$ This has to be our definition for $$\eta_{U} : \Gamma(U, \tilde{\mathscr{F}}) \rightarrow \Gamma(U, \mathscr{G}),$$ so the uniqueness (and the factorization) is already done. One can check that this gives a map of presheaf $\eta : \tilde{\mathscr{F}},$ which finishes the proof. $\Box$

Corollary. If $\mathscr{F}$ is a sheaf on $X,$ then $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ (given by $s \mapsto \tilde{s}$) is an isomorphism of sheaves.

Theorem. For any presheaf map $\eta : \mathscr{F} \rightarrow \mathscr{G},$ the induced sheaf map $\tilde{\eta} : \tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{G}}$ as a set map satisfies the following properties:

1. the map restricts as $\mathscr{F}_{x} \rightarrow \mathscr{G}_{x}$ and
2. $\tilde{\eta}$ is continuous.

Proof. The map $\tilde{\eta} : \tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{G}}$ is the set map given by the maps $\eta_{x} : \mathscr{F}_{x} \rightarrow \mathscr{G}_{x}$ as at stalks. Hence, the first condition is immediately satisfied. Next, for any $t \in \Gamma(U, \mathscr{G}),$ we have $$\eta^{-1}(\tilde{t}(U)) = \{s_{x} \in \mathscr{F} : \eta_{x}(s_{x}) = t_{x}\}.$$ For any $s_{x}$ in this set (i.e., $\eta_{x}(s_{x}) = t_{x}$), we may find some open $W \ni x$ in $U$ and $s \in \Gamma(W, \mathscr{F})$ such that $\eta_{W}(s) = t|_{W}.$ This implies that $$s_{x} \in \tilde{s}(W) \subset \eta^{-1}(\tilde{t}(U)),$$ which shows that $\eta^{-1}(\tilde{t}(U)) \subset \mathscr{G}$ is open. This shows that the set map $\tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{G}}$ is continuous. $\Box$

For sheaves, the conditions in the above theorem seems to classify maps between sheaves.

Theorem. Given any sheaves $\mathscr{F}, \mathscr{G},$ a set map $\eta : \mathscr{F} \rightarrow \mathscr{G}$ on their espaces étale induces a sheaf map if and only if

1. the map restricts as $\mathscr{F}_{x} \rightarrow \mathscr{G}_{x}$ and
2. $\eta$ is continuous.

Proof. Due to the previous theorem, we only need to prove one direction. Say the set map $\eta : \mathscr{F} \rightarrow \mathscr{G}$ satisfies the two conditions above. We want to show that $\eta$ induces a sheaf map. Fix any open subset $U \subset X.$ For any $s \in \Gamma(U, \mathscr{F}),$ the composition $$\eta \circ s : U \rightarrow \mathscr{F} \rightarrow \mathscr{G}$$ is a section (i.e., $\eta(s_{x}) \in \mathscr{G}_{x}$ for all $x \in U$) because of the first condition. Since $s$ is continuous (as $\mathscr{F}$ is a sheaf) and $\eta$ is continuous (the second condition), we note that $\eta \circ s$ is continuous, so we have $\Gamma(U, \mathscr{F}) \rightarrow \Gamma(U, \mathscr{G})$ by $s \mapsto \eta \circ s.$ This is compatible with varying $U$ with inclusions, so this finishes the proof. $\Box$

Remark. Note that in the above theorem, we only used the fact that $\mathscr{F}$ is a sheaf. This seems to correspond to the fact that to show the following equivalence, we only need to assume that $\mathscr{F}$ is a sheaf (2.4.E. Vakil):

• $\mathscr{F} \rightarrow \mathscr{G}$ is an isomorphism;
• (the induced map) $\mathscr{F}_{x} \rightarrow \mathscr{G}_{x}$ is an isomorphism for all $x \in X.$

Geometric viewpoint for sheaves. Hence, when a sheaf $\mathscr{F}$ on $X,$ we may identify $\mathscr{F} = \tilde{\mathscr{F}}.$ That is, we can think of $$\mathscr{F} = \bigsqcup_{x \in X} \mathscr{F}_{x}$$ with the topology we discussed and each $s \in \Gamma(U, \mathscr{F})$ can be thought as a continuous map $s : U \rightarrow \mathscr{F}$ such that $s_{x} \in \mathscr{F}_{x}$ for all $x \in U.$ (That is, we treat $s$ as a continuous section of $\mathscr{F}$ over $U$.)

However, recall checking whether a sheaf map $\mathscr{F} \rightarrow \mathscr{G}$ is an isomorphism requires a bit more care. It is an isomorphism if and only if the map of stalks $\mathscr{F}_{x} \rightarrow \mathscr{G}_{x}$ at every $x \in X$ is an isomorphism. (See 2.4.E Vakil.) For this statement, we change the target category (e.g., from $\textbf{Set}$ to $\textbf{Ab}$).

Abelian group structure for espace étale. Now, let $\mathscr{F}$ be as presheaf on $X$ valued in $\textbf{Ab}.$ Since each $\mathscr{F}_{x}$ is an abelian group, this gives $\Gamma(U, \tilde{\mathscr{F}})$ a structure of an abelian group. The zero element $0$ is the one that sends each $x \in U$ to the zero element of $\mathscr{F}_{x},$ which can be checked to be continuous: $0^{-1}(\tilde{s}(U)) = U$ for any open $U \subset X.$

Note that $$\Gamma(\emptyset, \tilde{\mathscr{F}}) = 0.$$ The subtraction is continuous, because for any $f, g \in \Gamma(U, \tilde{\mathscr{F}}),$ it is given as $$f - g = \mu \circ (f, g) : U \rightarrow \tilde{\mathscr{F}} \times_{X} \tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{F}},$$ where $$\begin{align*}\tilde{\mathscr{F}} \times_{X} \tilde{\mathscr{F}} &:= \{(a, b) \in \tilde{\mathscr{F}} \times \tilde{\mathscr{F}} : \pi_{\tilde{\mathscr{F}}}(a) = \pi_{\tilde{\mathscr{F}}}(b)\} \\ &= \{(s_{x}, t_{y}) \in \tilde{\mathscr{F}} \times \tilde{\mathscr{F}} : x = y\},\end{align*},$$ which is the fiber product taken in $\textbf{Top}$, the category of topological spaces with the continuous map $\mu : \tilde{\mathscr{F}} \times_{X} \tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{F}}$ given by $(s_{x}, t_{x}) \mapsto s_{x} - t_{x}.$

Proof of continuity.  We may find an open $U \subset X$ such that $s, t \in \Gamma(U, \mathscr{F})$ lift $s_{x}, t_{x}.$ This implies that $s_{x} - t_{x} \in \tilde{w}(U),$ where $w = s - t \in \Gamma(U, \mathscr{F}).$ For any $$(a_{x}, b_{x}) \in \mu^{-1}(\tilde{w}(U)),$$ namely $a_{x} - b_{x} = s_{x} - t_{x}$, we may find open $W \ni x$ in $U$ and lifts $a, b \in \Gamma(W, \mathscr{F})$ such that $a-b = s-t$ on $W.$ We have $$(a_{x}, b_{x}) \in \tilde{a}(W) \times_{X} \tilde{b}(W) \subset \mu^{-1}(\tilde{w}(U)),$$ a basic open subset of $\tilde{\mathscr{F}} \times_{X} \tilde{\mathscr{F}}.$ This finishes the proof. $\Box$


Ring structure for espace étale. Let $\mathscr{O}$ be a presheaf on $X$ valued in $\textbf{Ring}.$ For every open $U \subset X,$ the set $\Gamma(U, \tilde{\mathscr{O}})$ has a ring structure: $\Gamma(\emptyset, \tilde{\mathscr{O}})$ is the zero ring, because it can only have one element. For any nonempty open $U \subset X,$ we want to define the ring structure by those of $\mathscr{O}_{x}$ for $x \in U.$ That is, for $\alpha, \beta \in \Gamma(U, \tilde{\mathscr{O}}),$ we define $$(\alpha \cdot \beta)_{x} := \alpha_{x} \beta_{x} \in \mathscr{O}_{x}$$ for $x \in U.$

We still need to check $\alpha \cdot \beta \in \Gamma(U, \tilde{\mathscr{O}}),$ meaning $\alpha \cdot \beta$ is continuous. Note that $$\alpha \cdot \beta = \nu \circ (\alpha, \beta) : U \rightarrow \mathscr{O} \times_{X} \mathscr{O} \rightarrow \mathscr{O},$$ where $$\nu : \tilde{\mathscr{O}} \times_{X} \tilde{\mathscr{O}} \rightarrow \tilde{\mathscr{O}}$$ defined by $(\alpha_{x}, \beta_{x}) \mapsto \alpha_{x} \beta_{x}.$

Lemma. The map $\nu$ defined above is continuous.

Proof . We may find an open $U \subset X$ such that $s, t \in \Gamma(U, \mathscr{O})$ lift $s_{x}, t_{x}.$ This implies that $s_{x}t_{x} \in \tilde{w}(U),$ where $w = st \in \Gamma(U, \mathscr{O}).$ For any $$(a_{x}, b_{x}) \in \nu^{-1}(\tilde{w}(U)),$$ namely $a_{x}b_{x} = s_{x}t_{x}$, we may find open $W \ni x$ in $U$ and lifts $a, b \in \Gamma(W, \mathscr{O})$ such that $ab = st$ on $W.$ Thus, we have $$(a_{x}, b_{x}) \in \tilde{a}(W) \times_{X} \tilde{b}(W) \subset \nu^{-1}(\tilde{w}(U)),$$ a basic open subset of $\tilde{\mathscr{F}} \times_{X} \tilde{\mathscr{F}}.$ This finishes the proof. $\Box$

Corollary. We have $\alpha \cdot \beta$ (or write $\alpha\beta$ for brevity) inside $\Gamma(U, \mathscr{O}),$ which gives $\Gamma(U, \tilde{\mathscr{O}})$ a ring structure with the unity given by mapping $x \mapsto 1_{x} \in \mathscr{O}_{x}$ for $x \in U.$


Module structure for espace étale. Now, let $\mathscr{F}$ be a presheaf of $\mathscr{O}$-module. Namely, we have

• a presheaf $\mathscr{O}$ on $X$ valued in $\textbf{Ring}$;
• a presheaf $\mathscr{F}$ on $X$ valued in $\textbf{Ab}$;
• and a map $\mathscr{O} \times \mathscr{F} \rightarrow \mathscr{F}$ of presheaf such that on each open $U \subset X,$ it gives $\Gamma(U, \mathscr{F})$ an $\mathscr{O}(U)$-modules structure.

Note that $\tilde{\mathscr{O}} \times \tilde{\mathscr{F}}$ is the sheafification of $\mathscr{O} \times \mathscr{F},$ and the induced (necessarily continuous) sheaf map $$\tilde{\mathscr{O}} \times \tilde{\mathscr{F}} \rightarrow \tilde{\mathscr{F}}$$ gives $\tilde{\mathscr{F}}$ a structure of $\tilde{\mathscr{O}}$-module, because necessary axioms can be checked at the level of stalks. 

Remark. It follows that given any $\alpha \in \Gamma(U, \tilde{\mathscr{O}})$ and $f \in \Gamma(U, \tilde{\mathscr{F}}),$ the map $\alpha f : U \rightarrow \tilde{\mathscr{F}},$ defined point-wise, is continuous (i.e., $(\alpha f)_{x} = \alpha_{x}f_{x}$).

Vector bundles. Given a sheaf $\mathscr{O}$ valued in $\textbf{Ring}$ on $X,$ let $\mathscr{F}$ be a locally free $\mathscr{O}$-module sheaf on $X$ of rank $r \in \mathbb{Z}_{\geq 0}.$ Then, as a set, we may write $$\mathscr{F} = \bigsqcup_{x \in X}\mathscr{O}_{x}^{\oplus r}.$$ We have set maps $$\pi_{i} : \mathscr{F} \rightarrow \mathscr{O}$$ for $1 \leq i \leq r$ given by $r$ projections $\mathscr{O}_{x}^{\oplus r} \twoheadrightarrow \mathscr{O}_{x}$ for each $x \in X.$ Note that saying that the set $\mathscr{F}$ is locally free precisely says that all the maps $\pi_{i}$ for $1 \leq i \leq r$ are continuous.

Remark. This (espace étale) is probably not the best way to think about vector bundles, especially when $X$ has extra structure. This approach only remembers crude topology. For a better approach when $X$ is a scheme, see 17.1.G of Vakil. Better approaches when $X$ has various manifold structures are well-studied too (and still important topic).


Application: visualizing inverse image sheaf. The main reason I wanted to go through this espace étale construction is to understand inverse image sheaves better. Let $\phi : (X, \mathscr{O}_{X}) \rightarrow (Y, \mathscr{O}_{Y})$ be a map of ringed spaces. This means that $\mathscr{O}_{X}$ is a sheaf of rings on $X,$ and similarly for $Y.$ When we say modules over these sheaves, we mean sheaves, not merely presheaves.

Consider the presheaf defined by $$U \mapsto \mathrm{colim}_{V \supset \phi(U)} \mathscr{G}(V).$$ The sheaifification of this is $\phi^{-1}\mathscr{G},$ called the inverse image of $\mathscr{G}$ under $\phi$, which is a priori a sheaf on $Y$ valued in $\textbf{Ab}.$ The action of $\mathscr{O}_{X}(U)$ on $\mathrm{colim}_{V \supset \phi(U)} \mathscr{G}(V)$ is given using the presheaf $$\mathscr{O}_{X}(U) \rightarrow \mathrm{colim}_{V \supset \phi(U)} \mathscr{O}_{Y}(V).$$ If we sheafifiy this, the sheaf we get $\phi^{-1}\mathscr{O}_{Y}$ gives a module structure on $\phi^{-1}\mathscr{G}.$

This definition makes sense, but it seems very difficult to work with. Let's try to understand it with the notion of espace étale.

Lemma. Let $\phi : X \rightarrow Y$ be a continuous map and $\mathscr{G}$ a sheaf on $Y.$ For any $x \in X,$ we have $$(\phi^{-1}\mathscr{G})_{x} \simeq \mathscr{G}_{\phi(x)},$$ as abelian groups, and each side satisfies the universal property of the other side (as stalks with appropriate maps).

In particular, the structure of $\mathscr{O}_{X,x}$-module on the right-hand side given by the restriction of scalars under $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{Y,\phi(x)}$ matches with the structure of $\mathscr{O}_{X,x}$-module given by the left-hand side.

Proof. Recall that $$(\phi^{-1}\mathscr{G})_{x} = \mathrm{colim}_{U \ni x}\mathrm{colim}_{V \supset \phi(U)}\mathscr{G}(V).$$ Just based on this definition, for any open $V \ni \phi(x),$ by taking $U = \phi^{-1}(V) \ni x,$ we have $$\mathscr{G}(V) \rightarrow (\phi^{-1}\mathscr{G})_{x}.$$ This map is compatible with all $V \ni \phi(x)$ and inclusions among them, so it induces $$\mathscr{G}_{x} \rightarrow (\phi^{-1}\mathscr{G})_{x}.$$ On the other hand, for any open $U \ni x$ in $X$ and open $V \supset \phi(U)$ in $Y,$ we have $V \ni \phi(x),$ so it comes with $$\mathscr{G}(V) \rightarrow \mathscr{G}_{\phi(x)}.$$ Note that this process is compatible with all open $U \ni x$ and inclusions among them, so we get $$(\phi^{-1}\mathscr{G})_{x} \rightarrow \mathscr{G}_{x},$$ and one may check that the two maps we have constructed are inverses to each other. (Perhaps this can be checked using functorial arguments even without considering elements.) The reason for us to ensure that we can write that this isomorphism is the identity is that one can also check that each side satisfies the universal property of the other. $\Box$

Now, let's think about what $\phi^{-1}\mathscr{G}$ should be. By identifying the stalks using the above lemma, the espace étale description (as a set) must look like $$\phi^{-1}\mathscr{G} = \bigsqcup_{x \in X}\mathscr{G}_{\phi(x)},$$ so $\Gamma(U, \phi^{-1}\mathscr{G})$ must consist of continuous sections $s : U \rightarrow \phi^{-1}\mathscr{G}.$ Again, saying that $s$ is a section means that $s(x) \in \mathscr{G}_{\phi(x)}$ for each $x \in U.$ Saying that $s$ is continuous means that for each $x \in U,$ there is an open $U' \ni x$ and $$f \in \mathrm{colim}_{V \supset \phi(U')}\mathscr{G}(V)$$ such that $\tilde{f} = s|_{U'},$ meaning $f_{x} = s(x) \in \mathscr{G}_{\phi(x)}.$ Note that we can choose $f \in \mathscr{G}(V)$ for some open $\phi(U')$ in $Y.$ We summarize this as follows:

Theorem. The set $\Gamma(U, \phi^{-1}\mathscr{G})$ consists of maps $$s : U \rightarrow \bigsqcup_{x \in X}\mathscr{G}_{\phi(x)}$$ such that for each $x \in U,$ we have

1. $s(x) \in \mathscr{G}_{\phi(x)}$ and
2. there is an open $U' \ni x$ in $U$ and open $V \supset \phi(U')$ in $Y$ with $f \in \mathscr{G}(V)$ such that $\tilde{f}|_{U'} = s|_{U'}.$

Remark. It is easier to think about other algebraic structures on $\Gamma(U, \phi^{-1}\mathscr{G}),$ coming from those of $\mathscr{G},$ because they can be detected at the level of stalks, which look like $\mathscr{G}_{\phi(x)}$ for $x \in U.$ Anyways, I think this is what Vakil meant at the end of 2.7.C.

Example. Let $j : U \hookrightarrow X$ be an open embedding of topological spaces, and consider a sheaf $\mathscr{F}$ (valued in some concrete category) on $X.$ We know immediately from the colimit definition that $j^{-1}\mathscr{F} = \mathscr{F}|_{U},$ so our espace étale definition had better match this.

For any open $W \subset U$ and  the set $\Gamma(W, j^{-1}\mathscr{F})$ consists of $$s : W \rightarrow \bigsqcup_{x \in U}\mathscr{F}_{x}$$ such that for each $x \in W,$ we have

1. $s(x) \in \mathscr{F}_{x}$ and
2. there is an open $W' \ni x$ in $W$ with $f \in \mathscr{F}(W')$ such that $\tilde{f}|_{W'} = s|_{W'}$

by ranging various $x \in W,$ we will range various $W'$ and $f,$ to glue $s$ back as an element of $\Gamma(W, \mathscr{F}|_{U}),$ which shows our assertion.

Example. Let $i : Z \hookrightarrow X$ be a closed embedding of topological spaces, and consider a sheaf $\mathscr{F}$ (valued in some concrete category) on $X.$

Given any open $U \subset Z,$ the set $\Gamma(U, i^{-1}\mathscr{F})$ consists of $$s : U \rightarrow \bigsqcup_{x \in Z}\mathscr{F}_{x}$$ such that for each $x \in U,$ we have

1. $s(x) \in \mathscr{F}_{x}$ and
2. there are an open subset $U' \ni x$ in $U$ (i.e., open in $Z$) and an open subset $V \supset U'$ in $X$ with $f \in \mathscr{F}(V)$ such that $\tilde{f}|_{U'} = s|_{U'}.$

Adjointness. If we just look at the continuous map $\phi : X \rightarrow Y,$ we have $$\mathrm{Hom}_{\textbf{Sh}_{X}}(\phi^{-1}\mathscr{G}, \mathscr{F}) \simeq \mathrm{Hom}_{\textbf{Sh}_{Y}}(\mathscr{G}, \phi_{*}\mathscr{F})$$ functorial in $\mathscr{F}$ and $\mathscr{G},$ whose values are in $\textbf{Set}$ or $\textbf{Ab}.$ (See Vakil 2.7.C.)

When we have a ringed space map $\phi : (X, \mathscr{O}_{X}) \rightarrow (Y, \mathscr{O}_{Y}),$ for any $\mathscr{O}_{Y}$-module $\mathscr{G},$ we saw that $\phi^{-1}\mathscr{G}$ has a $\phi^{-1}\mathscr{O}_{Y}$-module structure, so to get an $\mathscr{O}_{X}$-module, we need to consider $$\phi^{*}\mathscr{G} := \phi^{-1}\mathscr{G} \otimes_{\phi^{-1}\mathscr{O}_{Y}} \mathscr{O}_{X},$$ which is called the pullback of $\mathscr{G}$ under $\phi,$ and this is an $\mathscr{O}_{X}$-module. This ad-hoc definition turns out to be canonical, that is, we have $$\mathrm{Hom}_{\mathscr{O}_{X}}(\phi^{*}\mathscr{G}, \mathscr{F}) \simeq \mathrm{Hom}_{\mathscr{O}_{Y}}(\mathscr{G}, \phi_{*}\mathscr{F})$$ functorial in $\mathscr{F}$ and $\mathscr{G}.$ (See Vakil 16.3.4.)