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.)