Abelian varieties: 3. Abelian schemes are commutative group schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. We also follow Mumford's book.

Convention. When we discuss an abelian scheme over a base scheme $S,$ we will assume that $S$ is nonempty.

We have discussed how an abelian variety $A$ over $\mathbb{C}$ is a commutative group scheme using analytic techniques. More specifically, we have gone through the following two steps:

Step 1. We have studied the conjugation action of each element of $A(\mathbb{C})$ on the tangent space $T_{e}A(\mathbb{C})$ at the identity $e$.

Step 2. We have made use of the exponential map $T_{e}A(\mathbb{C}) \rightarrow A(\mathbb{C}).$

We are going to show that an abelian variety over any field is a commutative group scheme. Step 1 works in algebraic setting, but Step 2 does not.

Remark. In characteristic $p > 0,$ it is known that there are examples of automorphisms that act trivially on the tangent space but not on the variety. (We might add some examples later.)

Let $A$ be an abelian variety over a field $k.$ The strategy we take is that we are going to consider the conjugations $c_{t} : A(k) \rightarrow A(k)$ as a flat family parametrized by $t \in A(k).$ We will show that $c_{t}$ is constant in $t.$

Our specific goal. Given an abelian scheme $A$ over $S,$ for $a \in A(S)$ (i.e., an $S$-scheme map $a : S \rightarrow A$), we define the left-translation by $a$ as the map $$l_{a} := m_{A} \circ (a \circ \pi_{A}, \mathrm{id}_{A}) : A \rightarrow A \times_{S} A \rightarrow A,$$ where $m_{A}$ is the multiplication map of the group $S$-scheme $A,$ and $\pi_{A} : A \rightarrow S$ is the structure map. Even though the above definition for $l_{a}$ is simple, it is not the best definition to work with. Another way to describe it is that given any $S$-scheme $T,$ the map $l_{a} : A \rightarrow A$ gives $l_{a, T} : A(T) \rightarrow A(T)$ defined by $x \mapsto \pi_{T}^{*}(a) \cdot x = (a \circ \pi_{T}) \cdot x,$ where $\pi_{T} : T \rightarrow S$ is the structure map. This allows us to show that $l_{e} = \mathrm{id}_{G}$ and $l_{g \cdot g'} = l_{g} \circ l_{g'}$ for all $g, g' \in G(S).$ In particular, we have $l_{g \cdot g^{-1}} = \mathrm{id}_{G} = l_{g^{-1} \cdot g}.$ This discussion works for any group scheme $G$ over $S.$ Details can be found in this posting.

Theorem. Let $A, B$ be abelian $S$-schemes with identities $e_{A}, e_{B}.$ If $S$ is a Noetherian scheme, given any $S$-scheme map $f : A \rightarrow B,$ there is a factorization $f = l_{f_{S}(e_{A})} \circ h$ for some map $h : A \rightarrow B$ of $S$-groups, where $f_{S}(e_{A}) = f \circ e_{A} : S \rightarrow A \rightarrow B$ as usual.

Corollary. If $S$ is a Noetherian scheme, any abelian $S$-scheme $A$ is a commutative group scheme.

Proof. Denote by $i_{A} : A \rightarrow A$ the inversion map. Since $i_{A}(e_{A}) = e_{A} : S \rightarrow A,$ we have $l_{i_{A}(e_{A})} = l_{e_{A}} = m_{A} \circ (e_{A} \circ \pi, \mathrm{id}_{A}) = \mathrm{id}_{A},$ where the last equality follows from a group scheme axiom for $A.$ Thus, applying Theorem for $f = i_{A},$ it shows that $i_{A}$ is a group scheme map. From here, it follows that $A$ is a commutative group scheme. $\Box$

Hence, it remains to show Theorem. First, we note that it is enough to show that $l_{f_{S}(e_{A})^{-1}} \circ f : A \rightarrow B$ is a map of $S$-groups. Since $$\begin{align*}(l_{f_{S}(e_{A})^{-1}} \circ f)_{S}(e_{A}) &= l_{f_{S}(e_{A})^{-1}} \circ f \circ e_{A} \\ &= l_{f_{S}(e_{A})^{-1}} \circ f_{S}(e_{A}) \\ &= l_{f_{S}(e_{A})^{-1},S}(f_{S}(e_{A})) \\ &= f_{S}(e_{A})^{-1} \cdot f_{S}(e_{A}) \\ &= e_{B}\end{align*}$$ in $B(S)$ because the $S$-scheme structure map for $S$ is $\mathrm{id}_{S}.$ Thus, this reduces the problem to the case where $f_{S}(e_{A}) = e_{B},$ and we desire to prove that $f : A \rightarrow B$ is an $S$-group map. This means that given any $S$-scheme $T$ and $a, a' \in A(T),$ we have $$f_{T}(m_{A,T}(a, a')) = f_{T}(a \cdot a') = f_{T}(a) \cdot f_{T}(a') = m_{B,T}(f_{T}(a), f_{T}(a')).$$ In terms of $S$-scheme map language, this precisely states that $$f \circ m_{A} = m_{B} \circ (f \times_{S} f) : A \times_{S} A\rightarrow B,$$ and thus this is what we need to prove. For each $S$-scheme $T,$ we may instead try to prove $$m_{B,T}(f_{T}(m_{A,T}(a, a')), i_{B}(m_{B,T}(f_{T}(a), f_{T}(a')))) = e_{B,T} \in B(T)$$ for all $a, a' \in A(T).$ In the $S$-scheme map language, this means that we want to prove $$m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) = e_{B} \circ \pi_{A \times_{S} A} : A \times_{S} A \rightarrow B,$$ where again $e_{B} = e_{B,S} : S \rightarrow B$ is the identity element of the group $B(S)$ and $\pi_{A \times_{S} A} : A \times_{S} A \rightarrow S$ is the $S$-scheme structure map. The map on the left-hand side is the composition $$A \times_{S} A \rightarrow B \times_{S} B \rightarrow B$$ of the relevant maps.

Reduction of the problem. We write $$g := m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) : A \times_{S} A \rightarrow B.$$ We want to show that $g = e_{B} \circ \pi_{A \times_{S} A}.$ Note that given any $S$-scheme $T,$ the map $g_{T} : A(T) \times A(T) \rightarrow B(T)$ is given by $(a, a') \mapsto f_{T}(a \cdot a') \cdot (f_{T}(a) \cdot f_{T}(a'))^{-1}.$ Hence, we have $$\begin{align*}g \circ (\mathrm{id}_{A}, e_{A} \circ \pi_{A}) \circ a &= g_{T}(a, e_{A,T}) \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot f_{T}(e_{A,T}))^{-1} \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot e_{B,T})^{-1} \\ &= e_{B,T} \\ &= e_{B} \circ \pi_{T}\end{align*}$$ because, using $f_{S}(e_{A}) = e_{B},$ we have $$\begin{align*}f_{T}(e_{A,T}) &= f \circ e_{A} \circ \pi_{T}\\ &= f_{S}(e_{A}) \circ \pi_{T}\\ &= e_{B} \circ \pi_{T} \\&= e_{B,T}.\end{align*}$$ Hence, if we prove that $g : A \times_{S} A \rightarrow B$ is a constant map, then it will follow that for every $a, a' \in A(T),$ we have $g_{T}(a, a') = e_{B, T}.$ The only $S$-scheme map $A \times_{S} A \rightarrow B$ such that $A(T) \times A(T) \rightarrow B(T)$ is giving the constant value of $e_{B,T}$ is $e_{B} \circ \pi_{A \times_{S} A},$ so this will finish the proof. Thus, it remains to show that $g : A \times_{S} A \rightarrow B$ is constant.

To finish the proof of Theorem, we use the following "rigidity" lemma:

Lemma (Rigidity). Fix any Noetherian scheme $S.$ Let  $\pi_{X} : X \rightarrow S$ be a proper flat $S$-scheme such that $\dim_{\kappa(s)}(H^{0}(X_{s}, \mathscr{O}_{X_{s}})) = 1$ for all $s \in S,$ where $$X_{s} = X \times_{S} \mathrm{Spec}(\kappa(s)),$$ the fiber at $s.$ If $\phi : X \rightarrow Y$ is any $S$-scheme map such that there exists any $s \in S$ such that the restriction $X_{s} \rightarrow Y_{s}$ of $\phi$ is constant, then $\phi$ is constant on the connected component of $s$ in $S.$

Remark. An $S$-scheme map $X \rightarrow Y$ is said to be constant if it factors through the structure map of $X.$ In particular, in such situation, for any $S$-scheme $T,$ the induced map $X(T) \rightarrow Y(T)$ is constant in set-theoretic sense. Given a field $k,$ a constant map of $k$-schemes $X \rightarrow Y$ necessarily give a constant map on the underlying topological spaces, as it factors as $X \rightarrow \mathrm{Spec}(k) \rightarrow Y.$ I don't think the converse is true: there are probably many examples where a $k$-scheme map is topologically constant but not a constant map according to the definition we are using. However, I have not thought deeply about such examples. (I may add them later if there are such though.)

Proof of Theorem. We use the notations given before Lemma. Write $$p_{1}^{-1}(e_{A}) := S \times_{A \times_{S} A} A,$$ with respect to $e_{A} : S \rightarrow A$ and $p_{1} : A \times_{S} A \rightarrow A.$ Given any $S$-scheme $T,$ the $S$-scheme map $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A$$ induces the set map $$(p_{1}^{-1}(e_{A}))(T) \rightarrow A(T) \times A(T),$$ which necessarily maps to elements of the form $(a, e_{A,T})$ where $a \in A(T)$ may vary. We have checked before that $g_{T}(a, e_{A,T}) = e_{B,T}$ constantly regardless of the choice of $a,$ so this implies that the composition $p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{g} B$ is constant. This implies that the following composition is also constant: $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B.$$ Then now note that $(p_{1}, g)$ is an $A$-scheme map that is constant on the fiber at any point in $A$ in the image of $e_{A} : S \rightarrow A.$ Now note that

• $A$ is connected,
• both $A \times_{S} A$ and $A \times_{S} B$ are proper and smooth over $A$ (base change) and hence also flat over $A$ (e.g., 25.2.2. (iii) in Vakil), and
• for $s \in A,$ we have $(A \times_{S} A)_{s} = p_{1}^{-1}(s) \simeq \mathrm{Spec}(\kappa(s)) \times_{A} A \times_{S} A \simeq A_{s},$ which is geometrically connected, so $\dim_{\kappa(s)} (A \times_{S} A)_{s} = 1.$

Thus, we may apply Rigidity Lemma to conclude that that $(p_{1}, g)$ is constant. Since $g$ is the composition $$A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B \rightarrow B,$$ where the latter one is the projection map onto $B,$ the fact that $(p_{1}, g)$ is constant implies that $g$ is constant. This finishes the proof $\Box$

Group schemes: basic definitions

I realized that I am still not quite comfortable with basic properties of group schemes, so I have decided to give myself a self-exercise to organize the material in my own words. I am not directly following but taking a look at Vistoli's notes for this.

Let $S$ be any scheme. An $S$-scheme $G$ is called a group scheme over $S$ (or an $S$-group) if the functor $h_{G} = \mathrm{Hom}_{\textbf{Sch}_{S}}(-, G) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ factors as $$\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Grp} \rightarrow \textbf{Set},$$ where the second functor is the forgetful functor.

This means that for each $S$-scheme $T,$ the set $G(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, G)$ has a group structure. In particular, we have the multiplication (set) map $$m_{T} : G(T) \times G(T) \rightarrow G(T)$$ and the inverse map $$i_{T} : G(T) \rightarrow G(T).$$ Moreover, both $m_{T}$ and $i_{T}$ are functorial in $T.$ To see this, fix any $S$-scheme map $\phi : T' \rightarrow T.$ The induced map $\phi^{*} : G(T) \rightarrow G(T')$ is a group map, so we must have $$\phi^{*} \circ m_{T} = m_{T'} \circ \phi^{*},$$ which shows the functoriality of $m_{-}.$ The functoriality of $i_{-}$ holds for the same reason. Since $i_{-} : h_{G} \rightarrow h_{G}$ is the map of functors, using the Yoneda embedding $\textbf{Sch}_{S} \hookrightarrow \textbf{PSh}_{\textbf{Set}}(\textbf{Sch}_{S})$ given by $T \mapsto h_{T} = T(-),$ there is a unique $S$-scheme map $i : G \rightarrow G$ that maps to $i_{-} : G(-) \rightarrow G(-).$ We remark that this uniqueness is up to an isomorphism of $S$-schemes. Given any $S$-scheme $T,$ we have $$i_{T} : G(T) \rightarrow G(T)$$ in $\textbf{Set},$ given by $f \mapsto i \circ f.$ We need slightly more for the multiplication map. First, given any $S$-scheme $T,$ we have $$G(T) \times G(T) \simeq (G \times_{S} G)(T)$$ given by $(x, y) \mapsto (x, y),$ where the left-hand side means a pair of two $S$-scheme maps $x, y : T \rightarrow G,$ while the right-hand side means the map $$(x, y) : T \rightarrow G \times_{S} G$$ of $S$-schemes induced by the fiber product. That is, if we denote by $\pi_{1}, \pi_{2} : G \times_{S} G \rightarrow G$ two projections, then

• $x = \pi_{1} \circ (x, y)$ and 
• $y = \pi_{2} \circ (x, y).$ 

Given any $S$-scheme map $\phi : T' \rightarrow T,$ we have

• $x \circ \phi = \pi_{1} \circ (x, y) \circ \phi$ and 
• $y \circ \phi = \pi_{2} \circ (x, y) \circ \phi,$

so $(x, y) \circ \phi = (x \circ \phi, y \circ \phi),$ so we see that $G(T) \times G(T) \simeq (G \times_{S} G)(T)$ is functorial in $T.$ Hence, we may realize $m_{-}$ as a map $(G \times_{S} G)(-) \rightarrow G(-),$ and using the Yoneda embedding, we may find a unique $S$-scheme map $$m : G \times_{S} G \rightarrow G$$ mapping to $m_{-}.$ We note that $m_{T} : (G \times_{S} G)(T) \rightarrow G(T)$ is given by $f \mapsto m \circ f,$ and if we look at $m_{T}$ as a map $G(T) \times G(T) \rightarrow G(T),$ the map is given by $(x, y) \mapsto m \circ (x, y),$ where $(x, y)$ on the right-hand side means the map $G \rightarrow G \times_{S} G.$

Since each $G(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, G)$ is a group, we have the identity element $e_{T} \in G(T)$ for any $S$-scheme $T.$ Any $S$-scheme $T$ comes with its structure map $\pi_{T} : T \rightarrow S,$ and $\pi_{T}^{*} : G(S) \rightarrow G(T)$ is a group map, so we must have $e_{S} \mapsto e_{T}.$ This implies that $e_{T} = e_{S} \circ \pi_{T} : T \rightarrow S \rightarrow G.$ We thus denote by $e := e_{S} : S \rightarrow G,$ as we can recover any $e_{T}$ from it for any $S$-scheme $\pi_{T} : T \rightarrow S.$

The fact that $e_{T}$ is the identity of the group $G(T)$ means that for any $x \in G(T),$ we have $$m \circ (e_{T}, x) = x = m \circ (x, e_{T}).$$ In other words, we have $m_{T}(e_{T}, x) =x = m_{T}(x, e_{T}),$ when we look at $m_{T}$ as a map $G(T) \times G(T) \rightarrow G(T).$ We can make this condition present in $\textbf{Sch}_{S}.$ First, note that $S(T) = \{\pi_{T}\}.$ Then we have the set map $S(T) \rightarrow G(T)$ given by $\pi_{T} \mapsto e_{T},$ which gives $$e_{T} \times \mathrm{id}_{G(T)} : S(T) \times G(T) \rightarrow G(T) \times G(T)$$ given by $(\pi_{T}, x) \mapsto (e_{T}, x).$ Given any $S$-scheme map $\phi : T' \rightarrow T$, we have $\pi_{T} \circ \phi = \pi_{T'}$ and $$\phi_{T} \circ \phi = \phi^{*}(e_{T}) = e_{T'},$$ where the last equality uses that $\phi^{*} : G(T) \rightarrow G(T')$ is a group map. We see that $S(T') \times G(T') \rightarrow G(T') \times G(T')$ gives $$(\pi_{T} \circ \phi, x \circ \phi) = (\pi_{T'}, x \circ \phi) \mapsto (e_{T'}, x \circ \phi) = (e_{T} \circ \phi, x \circ \phi).$$ Hence, the map $e_{T} \times \mathrm{id}_{G(T)}$ is functorial in $T.$ Now, the property that $e_{T}$ is the identity of $G(T)$ is equivalent to saying that $$m_{T} \circ (e_{T} \times \mathrm{id}_{G(T)}) \circ \pi_{2,T} = \mathrm{id}_{G(T)} = m_{T} \circ (\mathrm{id}_{G(T)} \times e_{T}) \circ \pi_{2,T}$$ where the left-hand side is given by the following chain of the compositions $$G(T) \simeq S(T) \times G(T) \rightarrow G(T) \times G(T) \rightarrow G(T)$$ with $\pi_{2,T}$ being the first functorial bijection in $T$ given by the projection onto $G(T).$ Note that the map $e_{-} \times \mathrm{id}_{G(-)} : S(-) \times G(-) \rightarrow G(-) \times G(-)$ of functors corresponds to $e \times_{S} \mathrm{id}_{G} : S \times_{S} G \rightarrow G \times_{S} G,$ and likewise $\mathrm{id}_{G(-)} \times e_{-}$ corresponds to $\mathrm{id}_{G} \times_{S} e.$ Therefore, saying that $e_{T}$ is the identity element of $G(T)$ for every $T$ is equivalent to saying $$m \circ (e \times_{S} \mathrm{id}_{G}) \circ \pi_{2} = \mathrm{id}_{G} = m \circ (\mathrm{id}_{G} \times_{S} e) \circ \pi_{2},$$ which is often called the identity axiom for the group scheme $G$ over $S.$

Now, let's think about associativity of $G(T)$ given an $S$-scheme $T.$ Note that the following are equivalent:

• $(x \cdot y) \cdot z = x \cdot (y \cdot z)$ for all $x, y, z \in G(T)$;
• $m_{T} \circ (m_{T} \times \mathrm{id}_{G(T)}) = m_{T} \circ (\mathrm{id}_{G(T)} \times m_{T})$;
• $m \circ (m \times_{S} \mathrm{id}_{G}) = m \circ (\mathrm{id}_{G} \times_{S} m).$

The last condition is called the associative axiom for $G.$

Given $g \in G(T),$ the inverse $g^{-1} = i_{T}(g)$ satisfies $$m_{T}(g, i_{T}(g)) = e_{T} = m_{T}(i_{T}(g), g).$$ Giving such condition for all $g \in G(T)$ is equivalent to saying that $$m_{T}(g, i_{T}(g)) = e_{T} = m_{T}(i_{T}(g), g),$$ which is equivalent to saying that $$m \circ (\mathrm{id}_{G}, i) = e \circ \pi_{T} = m \circ (i, \mathrm{id}_{G}).$$ This is called the inverse axiom for $G.$

Left multiplication map. Let $G$ be an $S$-group. Given any $g \in G(S),$ for any $S$-scheme $\pi_{T} : T \rightarrow S,$ we have $\pi_{T}^{*}(g) = g \circ \pi_{T} \ \in G(T).$ We then have the map $l_{g, T} :G(T) \rightarrow G(T)$ given by $x \mapsto (g \circ \pi_{T}) \cdot x = m_{T}(g \circ \pi_{T}, x),$ which is functorial in $T$ because given any $S$-scheme map $\phi : T' \rightarrow T,$ we have $$\phi^{*}(\pi_{T}^{*}(g) \cdot x) = \phi^{*}(\pi_{T}^{*}(g)) \cdot \phi^{*}(x) = (\pi_{T} \circ \phi)^{*}(g) \cdot \phi^{*}(x) = \pi_{T'}^{*}(g) \cdot \phi^{*}(x)$$ for all $x \in G(T),$ which implies that $\phi^{*} \circ l_{g,T} = l_{g,T'} \circ \phi^{*}.$ Thus, there must be a unique $S$-scheme map $l_{g} : G \rightarrow G$ such that for any $x \in G(T),$ we have $$l_{g} \circ x = l_{g, T}(x) \in G(T).$$ Since $\pi_{T} = \pi_{G} \circ x = \pi_{T}$ and $x = \mathrm{id}_{G} \circ x,$ we have $$(g \circ \pi_{G}, \mathrm{id}_{G}) \circ x : T \rightarrow G \times_{S} G,$$ so $$\begin{align*}l_{g, T}(x) &= m_{T}(g \circ \pi_{T}, x) \\ &= m \circ (g \circ \pi_{T}, x) \\ &= m \circ (g \circ \pi_{G}, \mathrm{id}_{G}) \circ x\end{align*}$$ for every $x \in G(T).$ This implies that $$l_{g} = m \circ (g \circ \pi_{G}, \mathrm{id}_{G}).$$ Of course, it is quite easy to define $l_{g} : G \rightarrow G$ as the $S$-scheme map given by this formula, but without understanding all these functorial descriptions, it seems difficult to prove something like following:

Theorem. Let $G$ be an $S$-group. Given $g, g' \in G(S),$ we have $$l_{g \cdot g'} = l_{g} \circ l_{g'} : G \rightarrow G.$$

Proof. Fix any $S$-scheme $T.$ For each $x \in G(T),$ we have $$\begin{align*}l_{g \cdot g'} \circ x &= l_{g \cdot g', T}(x) \\ &= \pi_{T}^{*}(g \cdot g') \cdot x \\ &= \pi_{T}^{*}(g) \cdot \pi_{T}^{*}(g') \cdot x \\ &= l_{g,T}(l_{g',T}(x)) \\ &= l_{g} \circ (l_{g',T}(x)) \\ &= l_{g} \circ (l_{g'} \circ x) \\ &= (l_{g} \circ l_{g'}) \circ x.\end{align*}$$ Hence, take $T = G$ and $x = \mathrm{id}_{G}$ in the above computation to finish the proof. $\Box$

Abelian varieties: 2. Abelian schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. We also follow Mumford's book.

Abelian schemes. Given a Noetherian scheme $S,$ an abelian scheme $A$ over $S$ is a group scheme $A$ over $S$ such that the structure map $A \rightarrow S$

1. is proper,
2. is smooth, and
3. has geometrically connected fibers.

In particular, when $S = \mathrm{Spec}(k)$ for a field $k,$ an abelian scheme over $k$ (i.e., $\mathrm{Spec}(k)$) is a proper and geometrically connected variety over $k,$ which we call an abelian variety.

Remark 1. Given an abelian scheme over $S$ and a scheme map $T \rightarrow S,$ the base change $A \times_{S} T \rightarrow T$ is proper and smooth, as these properties are preserved under any base change. For any point $x \in T,$ we have $$A \times_{S} T \times_{T} \mathrm{Spec}(\kappa(x)) \simeq A \times_{S} \mathrm{Spec}(\kappa(x)),$$ which is geometrically connected because $A$ is an abelian $S$-scheme. Hence, the base change $A \times_{S} T \rightarrow T$ is an abelian $T$-scheme.

Personal Remark. Property about smooth maps can be reviewed in Chapter 2 of BLR.

Remark 2. An abelian scheme $A$ over $S$ can be seen as a family of abelian varieties. Indeed, if we consider the structure map $A \rightarrow S$ and a point $s \in S,$ the base change $A \times_{S} \mathrm{Spec}(\kappa(s))$ is an abelian variety.

Remark 3. Any abelian variety (over a field) is geometrically integral. More generally, any scheme that is smooth and geometrically connected over a field is geometrically integral. This is because smoothness (or more generally, regularity) ensures that every stalk is a domain. Hence, this comes down to the fact that if there is a point $x$ in the intersection of two irreducible components of a Noetherian scheme $X,$ the stalk $\mathscr{O}_{X,x}$ is not an integral domain (5.3.C, Vakil); this is because a domain cannot have two minimal primes (as $(0)$ is its unique minimal prime).

Analytic example. Let $A$ be an abelian variety over $\mathbb{C}.$ Then by GAGA, we see $A(\mathbb{C})$ is complex manifold that is compact and connected. We shall sketch how to prove that $A(\mathbb{C})$ is an abelian group using analytic technique. For any $x \in A(\mathbb{C}),$ denote by $c_{x} : A(\mathbb{C}) \rightarrow A(\mathbb{C})$ the holomorphic map given by the conjugation: $y \mapsto xyx^{-1}.$ To show that $A(\mathbb{C})$ is an abelian group, we may show that $c_{x} = \mathrm{id}_{A(\mathbb{C})}.$ Consider $$A(\mathbb{C}) \rightarrow \mathrm{Aut}_{\mathrm{LieGrp}}(A(\mathbb{C})) \rightarrow \mathrm{Aut}_{\mathbb{C}}(T_{e}A(\mathbb{C})) \simeq \mathrm{GL}_{g}(\mathbb{C}),$$ given by $x \mapsto c_{x} \mapsto dc_{x},$ where $e \in A(\mathbb{C})$ is the identity with respect to the group structure. One may check that this is a holomorphic map and since $A(\mathbb{C})$ is a compact complex manifold, the map must be constant. This implies that $$(dc_{x})_{e} = (dc_{e})_{e} = \mathrm{id}_{T_{e}A(\mathbb{C})}.$$ Now, the magic happens when we use the exponential map $$\mathrm{exp} : T_{e}A(\mathbb{C}) \rightarrow A(\mathbb{C}),$$ which will allow us to see that $c_{x} = \mathrm{id}_{A(\mathbb{C})},$ showing that $A(\mathbb{C})$ is an abelian group. It follows that $A$ is a commutative group scheme over $\mathbb{C}.$

Exponential map. Given a complex Lie group $G,$ write $\mathfrak{g} := T_{e}G,$ where $e$ is the identity of $G.$ It is known (e.g., p.79 of Hochschild) that for each $v \in \mathfrak{g},$ there is a unique holomorphic group map $\phi_{v} : \mathbb{C} \rightarrow G$ such that $d\phi_{v} : \mathbb{C} \rightarrow \mathfrak{g}$ is given by $1 \mapsto v.$ We then define the exponential map $$\mathrm{exp}_{G} = \mathrm{exp} : \mathfrak{g} \rightarrow G$$ of $G$ by $v \mapsto \phi_{v}(1).$ The map $\mathrm{exp}$ turns out to be holomorphic (e.g., p.80 of Hochschild).

Given any $s \in \mathbb{C},$ we have $$d\phi_{v}(s t) = stv = tsv = d \phi_{sv}(t),$$ so $\phi_{v}(t) = \phi_{tv}(1) = \mathrm{exp}(tv).$ In particular, we have $$\mathrm{exp}(0) = \phi_{0}(1) = e,$$ because the trivial group map $\mathfrak{g} \rightarrow G$ mapping everything to $e$ has the zero derivation. Denoting by $f_{v} : \mathbb{C} \rightarrow \mathfrak{g}$ the map given by $t \mapsto tv$ so that $\phi_{v} = \mathrm{exp} \circ f_{v}.$ We have $$(d\phi_{v})_{0} = (d\mathrm{exp})_{0}(df_{v})_{0}$$ by the Chain Rule, so $$\begin{align*}(d\mathrm{exp})_{0}(v) &= (d\mathrm{exp})_{0}(df_{v})_{0}(1) \\ &= (d\phi_{v})_{0}(1)  \\ &= v,\end{align*}$$ showing that $d\mathrm{exp} = \mathrm{id}_{\mathfrak{g}}.$ By the Implicit Function Theorem, this implies that $\mathrm{exp}$ restricts to a homeomorphism from an open neighborhood of $0 \in \mathfrak{g}$ to an open neighborhood of $e \in G.$ This implies that if $G$ is connected, then $\mathrm{exp}(\mathfrak{g})$ generates the whole $G$ using the fact that in any connected Lie group, a nonempty open subset generates the whole group.

Let $T : G \rightarrow H$ be any map of complex Lie groups and denote by $\mathfrak{g}$ and $\mathfrak{h}$ the tangent spaces of $G$ and $H,$ respectively. We have $$T \circ \mathrm{exp}_{G} = \mathrm{exp}_{H} \circ (dT)_{e},$$ where the first map is given by the composition $\mathfrak{g} \rightarrow G \rightarrow H,$ while the second is $\mathfrak{g} \rightarrow \mathfrak{h} \rightarrow H.$ To check this, fix $v \in \mathfrak{g}$ and let $\phi_{v} : \mathbb{C} \rightarrow G$ the unique holomorphic group map satisfying $(d\phi_{v})_{0}(1) = v$ so that $\mathrm{exp}_{G}(v) = \phi_{v}(1).$ Then $T \circ \phi_{v} : \mathbb{C} \rightarrow G \rightarrow H$ is a holomorphic group map such that $$d(T \circ \phi_{v})_{0} = (dT)_{e} \circ (d\phi_{v})_{0} : \mathbb{C} \rightarrow \mathfrak{g} \rightarrow \mathfrak{h}$$ is given by $1 \mapsto v \mapsto (dT)_{e}(v).$ Hence, we must have $$\mathrm{exp}_{H}((dT)_{e}(v)) = T(\phi_{v}(1)) = T(\mathrm{exp}_{G}(v)),$$ as desired.

We want to consider the above identity when $T = c_{g} : G \rightarrow G$ the conjugation by an element $g \in G.$ If $G$ is compact (complex manifold), then $(dc_{x})_{e} = (dc_{e})_{e} = \mathrm{id}_{\mathfrak{g}},$ so the above identity buys us $$c_{g} \circ \mathrm{exp} = \mathrm{exp}.$$ In other words, for any $v \in \mathfrak{g},$ we see that $g$ commutes with $\mathrm{exp}(v)$ in $G,$ so this shows that $\mathrm{exp}(v)$ is in the center of $G$ if $G$ is compact. Thus, we have shown (modulo some facts about exponential maps) the following:

Theorem. Let $G$ be any complex Lie group. If $G$ is connected and compact, then it is an abelian group.

Personal Remark. This document seems to discuss the same fact. The author of the document says that this fact was an exercise in Chapter 8 of Fulton and Harris.

Torus description of complex abelian varieties. Let $A$ be an abelain variety of dimension $g$ over $\mathbb{C}.$ Writing $T_{e}A(\mathbb{C}) = \mathbb{C}^{g},$ we have noted that $\mathrm{exp} : \mathbb{C}^{g} \rightarrow A(\mathbb{C})$ is a surjective holomorphic group map. We also have noted that the exponential map restricts to a homeomorphism near $0,$ so let $U \ni 0$ be an open neighborhood in $\mathbb{C}^{g}$ such that $\mathrm{exp}$ gives $U \simeq \mathrm{exp}(U) =: V \subset A(\mathbb{C}).$ We have $e \in V$ and no other elements to $0$ in $U$ map to $e.$ Thus, for any nonero $v \in \ker(\mathrm{exp}),$ we have $U \cap (v + U) = \emptyset.$ This implies that $\ker(\mathrm{exp})$ is a discrete subgroup of $\mathbb{C}^{g}$ and that $\ker(\mathrm{exp})$ acts on $\mathbb{C}^{g}$ properly discontinuously. The induced group isomorphism $$\mathbb{C}^{g}/\ker(\mathrm{exp}) \simeq A(\mathbb{C})$$ is a holomorphic map, and since the differential at the identity of this map is an isomorphism, by the Inverse Function Theorem, its inverse is holomorphic at $e \in A(\mathbb{C})$ and hence everywhere holomorphic by applying translation at a point we test the holomorphy. As $\ker(\mathrm{exp})$ is a discrete subgroup of $\mathbb{C}^{g} = \mathbb{R}^{2g},$ it is generated by an $\mathbb{R}$-linearly independent subset of $\mathbb{R}^{2g}$ with size $\leq 2g.$ Since the quotient is compact (as $A(\mathbb{C})$ is compact), this non-strict inequality must be an equality, or in other words, we see that $\ker(\mathrm{exp})$ is a lattice in $\mathbb{C}^{g} = \mathbb{R}^{2g}.$ Thus, we have $$A(\mathbb{C}) \simeq \mathbb{C}^{g}/\ker(\mathrm{exp}) = \mathbb{R}^{2g}/\ker(\mathrm{exp}) \simeq (\mathbb{R}/\mathbb{Z})^{2g} = (S^{1})^{2g}.$$ In particular, denoting $\Lambda := \ker(\mathrm{exp}),$ the $n$-torsion subgroup of $A(\mathbb{C})$ can be computed as as $$A(\mathbb{C})[n] = (1/n)\Lambda / \Lambda \simeq (\mathbb{Z}/(n))^{2g}.$$ Note that when $g = \dim(A) = 1,$ we now know that $A(\mathbb{C})$ is (analytic) topologically a torus $S^{1} \times S^{1}$ and $A(\mathbb{C})[n] \simeq (\mathbb{Z}/(n))^{2}.$ An $1$-dimensional abelian variety is called an elliptic curve.

Remark. One can find a lattice $\Gamma \subset \mathbb{C}^{g}$ such that $\mathbb{C}^{g}/\Gamma$ is not an analytification of a complex variety. There is a classification of Riemann on which lattices give a quotient coming from a complex variety.

Personal Remark. I may find and add the references to the remark above.

Abelian varieties: 1. Group schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. Various references available in the notes will not be specified here.

Group schemes. Let $S$ be a scheme. A group scheme over $S$ (also called an $S$-group) is a group object in $\textbf{Sch}_{S},$ the category of $S$-schemes. Concretely, this means an $S$-scheme $G$ such that the functor $$h_{G} = \mathrm{Hom}_{S}(-, G) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ factors through the forgetful functor $\textbf{Grp} \rightarrow \textbf{Set}.$ We say a group scheme $G$ over $S$ is commutative if $G(T) := h_{G}(T)$ is an abelian group for each $S$-scheme $T.$

We now give examples. We will focus on the affine case $S = \mathrm{Spec}(R)$, although it should be possible to use general $S$ by gluing affine pieces.

Example 1. Consider $$\mathbb{G}_{a} = \mathbb{G}_{a, R} := \mathrm{Spec}(R[t]).$$ For any $R$-scheme $T,$ we have $$\mathbb{G}_{a}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathbb{G}_{a}) \simeq \Gamma(T, \mathscr{O}_{T}),$$ where the isomorphism is taken in $\textbf{Set},$ given by $$\phi \mapsto \phi^{*}(t).$$ This map is functorial in $T,$ varying in $\textbf{Sch}_{R},$ and the right-hand side gives the functor into $\textbf{Grp},$ where the group structure is given by the addition. We see that $\mathbb{G}_{a}$ is a (commutative) group scheme.

Example 2. Consider $$\mathbb{G}_{m} = \mathbb{G}_{m, R} := \mathrm{Spec}(R[t, t^{-1}]).$$ For any $R$-scheme $T,$ we have $$\mathbb{G}_{m}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathbb{G}_{m}) \simeq \Gamma(T, \mathscr{O}_{T})^{\times},$$ where the isomorphism is taken in $\textbf{Set},$ given by $$\phi \mapsto \phi^{*}(t).$$ This map is functorial in $T,$ varying in $\textbf{Sch}_{R},$ and the right-hand side gives the functor into $\textbf{Grp},$ where the group structure is given by the multiplication. We see that $\mathbb{G}_{m}$ is a (commutative) group scheme.

Example 3. We define $$\mathrm{GL}_{n} = \mathrm{GL}_{n, R} := \mathrm{Spec}(R[t_{ij}]_{1 \leq i,j \leq n}[1/\mathrm{det}]).$$ For any $R$-scheme $T,$ we have $$\mathrm{GL}_{n}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathrm{Spec}(R[t_{ij}]_{1 \leq i,j \leq n}[1/\mathrm{det}])) \simeq \mathrm{GL}_{n}(\mathscr{O}_{T}(T)),$$ the general linear group of $n \times n$ matrices over the global sections of $T,$ given by $\phi \mapsto [\phi^{*}(t_{ij})]_{1 \leq i,j \leq n}.$ This is functorial in $T,$ and the right hand-side gives the functor into $\textbf{Grp}$ given by the matrix multiplication. This shows that $\mathrm{GL}_{n}$ is a group scheme.

Example 4. We define $$\mu_{n} = \mu_{n, R} := \mathrm{Spec}(R[t]/(t^{n}-1)).$$ For any $R$-scheme $T,$ we have $$\mu_{n}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathrm{Spec}(R[t]/(t^{n}-1))) \simeq \{f \in \mathscr{O}_{T}(T)^{\times} : f^{n} = 1\},$$ given by $\phi \mapsto \phi^{*}(\bar{t}),$ which is functorial in $T,$ and the right hand-side gives the functor into $\textbf{Grp}$ given by the matrix multiplication. This shows that $\mu_{n}$ is a (commutative) group scheme. The group structures are coming from multiplications of $\mathscr{O}_{T}(T)^{\times}.$

Remark. The behavior of $\mu_{n, R}$ depends on the characteristic of $R.$

First, consider $R = \mathbb{C}.$ Then the equation $t^{n} = 1$ has $n$ distinct solutions in $\mathbb{C},$ so $$\mu_{n} = \mathrm{Spec}\left(\frac{\mathbb{C}[t]}{(t^{n} - 1)}\right) \simeq \bigsqcup_{z \in \mu_{n}(\mathbb{C})}\mathrm{Spec}(\mathbb{C}) \simeq \mu_{n}(\mathbb{C}),$$ which is in particular smooth over $\mathbb{C}.$ It turns out that all group schemes in characteristic $0$ are smooth, by a theorem due to Oort. (We will see this later.)

On the other hand, when $R = \mathbb{F}_{p}$ for a prime $p,$ then $$\mu_{p} = \mathrm{Spec}(\mathbb{F}_{p}[t]/(t^{p} - 1)) = \mathrm{Spec}(\mathbb{F}_{p}[t]/(t - 1)^{p}),$$ so $\mu_{n}$ is a non-reduced scheme. Moreover, we note that $\mu_{p}$ is not smooth over $\mathbb{F}_{p}$ as the derivative of $(t - 1)^{p}$ in $t$ vanishes. In this case, note that $\mu_{p} = \mu_{p, \mathbb{F}_{p}}$ is a point. Note that for any $\mathbb{F}_{p}$-algebra $A$ that is a domain, there is only one $\mathbb{F}_{p}$-algebra map $\mathbb{F}_{p}[t]/(t-1)^{p} \rightarrow A$ because the only $a \in A$ such that $(a - 1)^{p} = 0$ must be $1$ as $A$ is a domain. Hence, we have $\mu_{p, \mathbb{F}_{p}}(A)$ is a singleton. However, when $A$ is not a domain, this may not be the case. For instance, take $A = \mathbb{F}_{p}[\epsilon]/(\epsilon^{2}).$ We have $$\begin{align*} \mu_{p,\mathbb{F}_{p}}\left( \frac{\mathbb{F}_{p}[\epsilon]}{(\epsilon^{2})} \right) &= \mathrm{Hom}_{\mathrm{Spec}(\mathbb{F}_{p})}\left( \mathrm{Spec}\left( \frac{\mathbb{F}_{p}[\epsilon]}{(\epsilon^{2})} \right), \mu_{p,\mathbb{F}_{p}} \right) \\ &\simeq \{1 + b\bar{\epsilon} \in \mathbb{F}_{p}[\epsilon]/(\epsilon^{2}) : b \in \mathbb{F}_{p}\}\end{align*},$$ given by $\phi \mapsto \phi^{*}(\bar{t}).$ The last set has $p$ elements and $p > 1.$

More conventions/definitions. For convenience, we shall call a group scheme over a base scheme $S$ an $S$-group. Again, an $S$-group is by definition a certain functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Grp},$ and we define an $S$-group map to be a natural transformation between two $S$-groups. With this definition, we see that the $S$-groups form a category. Note that given an $S$-group map $f : G \rightarrow H$ and an $S$-group $T,$ the corresponding map $f_{T} : G(T) \rightarrow H(T)$ is a group homomorphism.

Comparing Grothendeick topologies

In this posting, we follow Vistoli's notes to discuss how to compare various Grothendieck topologies. Our primary examples will be various small sites of a scheme, and this will let us understand sheaves on one site by understanding them as sheaves on another site.

Sieves. Let $U$ be an object in a category $\mathcal{C}$ and consider a set $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of maps into $U.$ Given any object $T$ in $\mathcal{C},$ define $h_{\mathscr{U}}(T)$ be the set of maps $T \rightarrow U$ such that there is a factorization $T \rightarrow U_{i} \rightarrow U$ by a member of $\mathscr{U}.$ We have $h_{\mathscr{U}}(T) \subset h_{U}(T) = \mathrm{Hom}_{\mathcal{C}}(T, U),$ and this inclusion is functorial in $T.$ That is, we have a subfunctor $h_{\mathscr{U}} \hookrightarrow h_{U},$ and we call $h_{\mathscr{U}}$ the sieve associated with the collection $\mathscr{U}.$

In general, we call any subfunctor of $h_{U}$ a sieve on $U,$ and here is the reason: given a subfunctor $S \hookrightarrow h_{U},$ consider the set $\mathscr{U}_{S} = \bigcup_{T \in \mathrm{Ob}(\mathcal{C})}S(T).$ Then given any object $T$ in $\mathcal{C},$ any map $T \rightarrow U$ that factors through some map $T' \rightarrow U$ in $\mathscr{U}_{S}$ belongs to $S(T).$ That is, we have $h_{\mathscr{U}_{S}}(T) = S(T),$ so $h_{\mathscr{U}_{S}} = S.$ That is, the subfunctor $S$ is the sieve associated with the collection $\mathscr{U}_{S}.$

Sheaf condition via sieves. Now, let $\mathcal{C}$ be equipped with a Grothendieck topology. Given a presheaf $F$ and a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of an object $U,$ denote by $F(\mathscr{U})$ the set of elements $(s_{i})_{i \in I} \in \prod_{i \in I}F(U_{i})$ such that $$s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$$ for all $i, j \in I.$ We have a map $F(U) \rightarrow F(\mathscr{U})$ given by $s \mapsto (s|_{U_{i}})_{i \in I}.$ We can immediately note that

• $F$ is a separated presheaf precisely when $F(U) \rightarrow F(\mathscr{U})$ is injective for every cover $\mathscr{U}$ of $U$ and for every object $U$ in $\mathcal{C},$ and
• $F$ is a sheaf precisely when $F(U) \rightarrow F(\mathscr{U})$ is bijective for every cover $\mathscr{U}$ of $U$ and for every object $U$ in $\mathcal{C}.$

Recall the Yoneda map $$\mathrm{Hom}(h_{U}, F) \simeq F(U)$$ given by $\phi \mapsto \phi_{U}(\mathrm{id}_{U}).$ Given a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of $U,$ we consider the map $$\mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow F(\mathscr{U})$$ given by $\phi \mapsto (\phi_{U_{i}}(U_{i} \rightarrow U))_{i \in I}.$ Since $\phi$ is a natural transformation, we have $$\phi_{U_{i}}(U_{i} \rightarrow U)|_{U_{ij}} = \phi_{U_{ij}}(U_{ij} \rightarrow U) = \phi_{U_{j}}(U_{j} \rightarrow U)|_{U_{ij}},$$ where we wrote $U_{ij} := U_{i} \times_{U} U_{j}.$ Hence, we indeed have $$(\phi_{U_{i}}(U_{i} \rightarrow U))_{i \in I} \in F(\mathscr{U}).$$ We claim that this map is a bijection.

To show injectivity, fix $\phi, \psi \in \mathrm{Hom}(h_{\mathscr{U}}, F)$ such that $$\phi_{U_{i}}(U_{i} \rightarrow U) = \psi_{U_{i}}(U_{i} \rightarrow U)$$ for all $i \in I.$ Given any object $T$ in $\mathcal{C},$ we know any map $T \rightarrow U$ in $h_{\mathscr{U}}(T)$ is factored as $T \rightarrow U_{i} \rightarrow U$ for some $i \in I,$ which let us see that $\phi_{T}(T \rightarrow U)$ is the image of $\phi_{U_{i}}(U_{i} \rightarrow U)$ under $F(U_{i}) \rightarrow F(T).$ Similarly, we note that $\psi_{T}(T \rightarrow U)$ is the image of $\psi_{U_{i}}(U_{i} \rightarrow U)$ under $F(U_{i}) \rightarrow F(T).$ Thus, we must have $$\phi_{T}(T \rightarrow U) = \psi_{T}(T \rightarrow U),$$ showing $\phi = \psi.$ This establishes injectivity.

To show surjectivity, fix any $(s_{i})_{i \in I} \in F(\mathscr{U}).$ Let $T$ be any object in $\mathcal{C},$ and we want to define a map $h_{\mathscr{U}}(T) \rightarrow F(T).$ Fix any $\eta : T \rightarrow U$ in $h_{\mathscr{U}}(T).$ By definition of $h_{\mathscr{U}}(T),$ we have some factorization $$\eta : T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ for some $i \in I.$ Then consider $(F\eta_{i})(s_{i}) \in F(T),$ where $F\eta_{i} : F(U_{i}) \rightarrow F(T)$ is induced by $\eta_{i} : T \rightarrow U_{i}.$

Claim. The element $(F\eta_{i})(s_{i}) \in F(T)$ is independent to the choice of the factorization of $T \rightarrow U.$

Proof of Claim. Consider any other factorization $$\eta : T \xrightarrow{\eta_{j}} U_{j} \rightarrow U$$ for some $j \in I.$ Consider the map $(\eta_{i}, \eta_{j}) : T \rightarrow U_{i} \times_{U} U_{j}$ induced by the fiber product. Denoting by $p_{i} : U_{i} \times_{U} U_{j} \rightarrow U_{i}$ and $p_{j} : U_{j} \times_{U} U_{j} \rightarrow U_{j}$ projections, we have $\eta_{i} = p_{i} \circ (\eta_{i}, \eta_{j})$ and $\eta_{j} = p_{j} \circ (\eta_{i}, \eta_{j})$ so that

• $F\eta_{i} = F(\eta_{i}, \eta_{j}) \circ Fp_{i}$ and
• $F\eta_{j} = F(\eta_{i}, \eta_{j}) \circ Fp_{j}$

Since $$(Fp_{i})(s_{i}) = s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}} = (Fp_{j})(s_{j}),$$ this finishes the proof. $\Box$

Hence, given $(s_{i})_{i \in I} \in F(\mathscr{U}),$ we get a well-defined map $$h_{\mathscr{U}}(T) \rightarrow F(T)$$ given by $\eta \mapsto (F\eta_{i})(s_{i}),$ where $\eta_{i} : T \rightarrow U_{i}$ is any map such that $$T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ is a factorization of $\eta,$ which always exists by definition. If $\phi : S \rightarrow T$ is any map in $\mathcal{C},$ then for $\eta \in h_{\mathscr{U}}(T),$ we get a factorization $$\eta : T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ so that we get the factorization $\eta \circ \phi : S \rightarrow T \rightarrow U_{i} \rightarrow U$ so that $\eta \circ \phi \in h_{\mathscr{U}}(S).$ Now we know that $h_{\mathscr{U}}(S) \rightarrow F(S)$ gives $$\eta \circ \phi \mapsto F(\eta_{i} \circ \phi)(s_{i}) = ((F\phi) \circ (F\eta_{i}))(s_{i}) = (F\phi)((F\eta_{i})(s_{i})),$$ so we have constructed $\psi : h_{\mathscr{U}} \rightarrow F$ as a map of functors. Moreover, we have $\psi_{U_{i}} : h_{\mathscr{U}}(U_{i}) \rightarrow F(U_{i})$ such that $$[U_{i} \rightarrow U] \mapsto (F(\mathrm{id}_{U_{i}})(s_{i}))_{i \in I} = (s_{i})_{i \in I},$$ so this proves the surjectivity of the map $$\mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow F(\mathscr{U})$$ given by $\psi \mapsto (\psi_{U_{i}}(U_{i} \rightarrow U))_{i \in I},$ so it is a bijection. Note that this bijection restricts to the Yoneda map $\mathrm{Hom}(h_{U}, F) \rightarrow F(U)$ because $$\phi_{U}(\mathrm{id}_{U})|_{U_{i}} = \phi_{U_{i}}(U_{i} \rightarrow U)$$ for all $i \in I.$ Therefore, it follows that:

Corollary. Let $\mathcal{C}$ be a site. Any presheaf $F : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a separated presheaf if and only if for any cover $\mathscr{U}$ of any object $U$ in $\mathcal{C},$ the map $$\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F)$$ given by $$[h_{U} \rightarrow F] \mapsto [h_{\mathscr{U}} \hookrightarrow h_{U} \rightarrow F]$$ is injective. Moreover, the presheaf $F$ is a sheaf if and only if for any cover $\mathscr{U}$ of any object $U$ in $\mathcal{C},$ the map $\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F)$ is bijective.

Refinement. Let $U$ be an object in a category $\mathcal{C}.$ Given a set $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of maps whose target is $U,$ a refinement of $\mathscr{U}$ is a set $\mathscr{U}' = \{U'_{a} \rightarrow U\}_{a \in A}$ of maps with the same target such that every $U'_{a} \rightarrow U$ factors through some $U_{i} \rightarrow U.$ (That is, there is some map $U'_{a} \rightarrow U_{i}$ such that the composition $U'_{a} \rightarrow U_{i} \rightarrow U$ is equal to the map $U'_{a} \rightarrow U.$)

Proposition. Given any two covers $\mathscr{U}$ and $\mathscr{V}$ of an object $U$ in a category $\mathcal{C},$ the cover $\mathscr{V}$ is a refinement of $\mathscr{U}$ if and only if $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$

Proof. Write $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ and $\mathscr{V} = \{V_{j} \rightarrow U\}_{j \in J}.$ If $\mathscr{V}$ is a refinement of $\mathscr{U},$ then $h_{\mathscr{V}}(T) \subset h_{\mathscr{U}}(T)$ for every object $T$ by definition. Conversely, if $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}},$ then for each $j \in J,$ we have $V_{j} \rightarrow U$ in $h_{\mathscr{V}}(V_{j}) \subset h_{\mathscr{U}}(V_{j}),$ so there must be some $i \in I$ such that this map has a factorization $V_{j} \rightarrow U_{i} \rightarrow U.$ $\Box$

The above proposition is easy to prove, but thanks to this, it is easy to notice that refinement gives a partial order on the collection of sets of maps into $U.$ In particular, if $\mathcal{C}$ is a site, then refinements give a partial order on the collection of all the covers of $U.$

Comparing Grothendeick topologies. Let $\mathcal{C}$ be a category and $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ Grothendieck topologies on it. We write $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and read $\mathcal{T}_{1}$ is subordinate to $\mathcal{T}_{2}$ if every cover of $\mathcal{T}_{1}$ has a refinement in $\mathcal{T}_{2}.$ Note that $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ if and only if for every $\mathscr{U}$ in $\mathcal{T}_{1},$ there is $\mathscr{V}$ in $\mathcal{T}_{2}$ such that $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$

Hence, if $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and $\mathcal{T}_{2} \prec \mathcal{T}_{3},$ then $\mathcal{T}_{1} \prec \mathcal{T}_{3}.$ Namely, the subordination of Grothendieck topologies is transitive. Thus, it makes sense for us to say that $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ are equivalent if $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and $\mathcal{T}_{2} \prec \mathcal{T}_{1}.$ This gives an equivalence relation on the Grothendieck topologies on a fixed category $\mathcal{C}.$

Remark. Even though Vistoli does not use this terminology, when $\mathcal{T}_{1} \prec \mathcal{T}_{2},$ I will say $\mathcal{T}_{1}$ is coarser than $\mathcal{T}_{2},$ and $\mathcal{T}_{2}$ is finer than $\mathcal{T}_{2},$ to mean this.

Example. Given any scheme $S,$ on the category $\textbf{Sch}_{S}$ of schemes over $S,$ étale topology is certainly coarser than smooth topology, which has more covers. Vistoli Example 2.51 uses a fact from EGA IV to show that in fact these two Grothendieck topologies are equivalent. I have not checked this proof yet.

Why do we care? Why do we care about comparing Grothendieck topologies? It is because comparison of Grothendieck topologies let us compare the sheaves:

Theorem. Let $\mathcal{C}$ be a category with Grothendieck topologies such that $\mathcal{T}_{1} \prec \mathcal{T}_{2}.$ Then every sheaf on $(\mathcal{C}, \mathcal{T}_{2})$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{1}).$ That is, given two (Grothendieck) topologies, any sheaf on the finer topology is sheaf on the coarser topology.

Proof. Given the hypothesis, let $F$ be a sheaf on $(\mathcal{C}, \mathcal{T}_{2}).$ Given any object $U$ in $\mathcal{C},$ fix any cover $\mathscr{U}$ of $U$ in $(\mathcal{C}, \mathcal{T}_{1}).$ By the hypothesis, there is a cover $\mathscr{V}$ of $U$ in $(\mathcal{C}, \mathcal{T}_{2})$ such that $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$ This gives us $$\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{V}}, F)$$ given by $$\begin{align*}[h_{U} \rightarrow F] \mapsto [h_{\mathscr{U}} \hookrightarrow h_{U} \rightarrow F]  &\mapsto [h_{\mathscr{U}} \hookrightarrow h_{\mathscr{V}} \hookrightarrow h_{U} \rightarrow F] \\ &= [h_{\mathscr{V}} \hookrightarrow h_{U} \rightarrow F].\end{align*}$$ Since $F$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{2}),$ the above composition is bijective, so the first map must be bijective as well, showing that $F$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{1}).$ $\Box$

General sheafification

In this posting, we follow Vistoli's notes to study a sheafification of a presheaf $F$ on a site $\mathcal{C}$ (i.e., a functor $F : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$). We would like to construct a sheaf $F^{\mathrm{a}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ equipped with a map $a : F \rightarrow F^{\mathrm{a}}$ of functors such that for any sheaf $G$ and a map $\phi : F^{\mathrm{a}} \rightarrow G,$ there exists a unique map $\phi : F^{\mathrm{a}} \rightarrow G$ with a factorization $F \xrightarrow{a} F^{\mathrm{a}} \xrightarrow{\phi} G$ of $\phi$. It is immediate that a sheafification is unique up to an isomorphism if it ever exists.

Necessary conditions. Instead, we look for a sheaf $F^{\mathrm{a}}$ together with a map $a : F \rightarrow F^{\mathrm{a}}$ such that given any object $U$ of $\mathcal{C},$

1. if $\xi, \eta \in F(U)$ with $a_{U}(\xi) = a_{U}(\eta),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $\xi|_{U_{i}} = \eta|_{U_{i}}$ for all $i \in I,$ and
2. if $\omega \in F^{a}(U),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a_{U_{i}}(\xi_{i})$ for all $i \in I.$

Remark. In a colloquial term, the sheafification of a presheaf remembers local data of each section of the presheaf.

Why are these necessary conditions? Given the condition above, let $\psi : F \rightarrow G$ any map where $G$ is a sheaf. Fix any object $U$ of $\mathcal{C}.$ Given $\omega \in F^{a}(U),$ we may choose a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a(\xi_{i})$ for all $i \in I.$ We have $$\begin{align*}a_{U_{ij}}(\xi_{i}|_{U_{ij}}) &= a_{U_{i}}(\xi_{i})|_{U_{ij}} \\ &= \omega|_{U_{ij}} \\ &= a_{U_{j}}(\xi_{j})|_{U_{ij}} \\ &= a_{U_{ij}}(\xi_{j}|_{U_{ij}})\end{align*}$$ in $F^{a}(U_{ij}).$ Thus, there is a cover $\{V_{s} \rightarrow U_{ij}\}_{s \in J}$ such that $$\xi_{i}|_{V_{s}} = \xi_{j}|_{V_{s}}$$ for all $s \in J.$ This implies that $$\psi_{U_{i}}(\xi_{i})|_{V_{s}} = \psi_{V_{s}}(\xi_{i}|_{V_{s}}) = \psi_{V_{s}}(\xi_{j}|_{V_{s}}) = \psi_{U_{j}}(\xi_{j})|_{V_{s}}$$ for all $s \in J,$ so we must have $$\psi_{U_{i}}(\xi_{i})|_{U_{ij}} = \psi_{U_{j}}(\xi_{j})|_{U_{ij}}$$ in $G(U_{ij})$ for $i, j \in I.$ Since $G$ is a sheaf, this implies that there is a unique $\tilde{\psi}_{\mathscr{U}}(\omega) \in G(U)$ such that $\tilde{\psi}_{\mathscr{U}}(\omega)|_{U_{i}} = \psi_{U_{i}}(\xi_{i})$ for all $i \in I.$ If we have two covers of $U,$ we may take their refinements, and since $G$ is a sheaf, this implies that $\tilde{\psi}_{\mathscr{U}}(\omega)$ does not depend on the choice of the cover $\mathscr{U},$ so we may write $\tilde{\psi}_{U}(\omega) := \tilde{\psi}_{\mathscr{U}}(\omega)$ instead. Hence, we have defined a map $\tilde{\psi}_{U} : F^{a}(U) \rightarrow G(U).$ If $V \rightarrow U$ is a map in $\mathcal{C}$ and $s \in F^{a}(U),$ then taking a cover $\{U_{i} \rightarrow U\}_{i \in I}$ for $U$ to get a cover $\{V_{i} \rightarrow V\}_{i \in I}$ for $V$ where $V_{i} = V \times_{U} U_{i},$ we have $$(\tilde{\psi}_{U}(s)|_{V})|_{V_{i}} = \tilde{\psi}_{U}(s)|_{V_{i}} = \tilde{\psi}_{V_{i}}(s) = \tilde{\psi}_{V}(s)|_{V_{i}}$$ for all $i \in I,$ so $\tilde{\psi}_{U}(s)|_{V} = \tilde{\psi}_{V}(s).$ Thus, our definition gives a map $\tilde{\psi} : F^{a} \rightarrow G$ of sheaves. The definition of $\tilde{\psi}$ immediately tells us that $\tilde{\psi} \circ a = \psi.$ To show uniqueness, say we have another $\psi' : F^{a} \rightarrow G$ such that $\psi' \circ a = \psi.$ Then given any $\omega \in F^{a}(U),$ we choose a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $a_{U_{i}}(\xi_{i}) = \omega|_{U_{i}}$ for $i \in I.$ This implies that $$\begin{align*}\psi'_{U}(\omega)|_{U_{i}} &= \psi'_{U_{i}}(\omega|_{U_{i}}) \\ &= \psi'_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \psi_{U_{i}}(\xi_{i}) \\ &= \tilde{\psi}_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \tilde{\psi}_{U_{i}}(\omega|_{U_{i}}) \\ &= \tilde{\psi}_{U}(\omega)|_{U_{i}}\end{align*}$$ for all $i \in I,$ so $\psi'_{U}(\omega) = \tilde{\psi}_{U}(\omega).$ Hence, we have $\psi' = \tilde{\psi},$ showing the uniqueness.

Sheafifications valued in different categories. The discussion above applies even when we replace $\textbf{Set}$ with $\textbf{Ab}$ or the category of modules over a ring.

Construction. For any $s, t \in F(U),$ we write $s \sim t$ if there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ This is an equivalence relation where we can check transitivity by taking the joint cover of two given covers.

Define $F^{\mathrm{s}}(U)$ be the set of equivalence classes of elements in $F(U)$ with respect to this equivalence relation. Given any morphism $V \rightarrow U$ in $\mathcal{C},$ if $s \sim t$ in $F(U),$ we have a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I,$ so $(s|_{V})|_{V_{i}} = (t|_{V})|_{V_{i}}$ for all $i \in I,$ where $\{V \times_{U} U_{i} = V_{i} \rightarrow V\}_{i \in I}$ is a cover for $V.$ Hence, we have $s|_{V} \sim t|_{V}$ in $F(V).$ This lets us define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[s] \mapsto [s|_{V}].$ This immediately makes $F^{\mathrm{s}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ a functor (i.e., a presheaf). If $\{U_{i} \rightarrow U\}_{i \in I}$ is a cover and $s, t \in F(U)$ with $s|_{U_{i}} \sim t|_{U_{i}} \in F(U_{i})$ for all $i \in I,$ then we may take a cover $\{V_{ij} \rightarrow U_{i}\}_{j \in J_{i}}$ for each $i \in I$ such that $s|_{V_{ij}} = t|_{V_{ij}}$ for all $j \in J_{i}.$ Since $\{V_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}}$ is a cover for $U,$ we have $s \sim t$ in $F(U).$ This implies that $F^{\mathrm{s}}$ is a separated presheaf.

Given an object $U$ of $\mathcal{C},$ we consider the set of pairs $(\mathscr{U}, \boldsymbol{s}),$ where $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ is a cover of $U$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ such that $[s_{i}|_{U_{ij}}] = [s_{j}|_{U_{ij}}] \in F^{\mathrm{s}}(U_{ij})$ for all $i, j \in I$ where $U_{ij} = U_{i} \times_{U} U_{j}.$ For two pairs $(\mathscr{U}, \boldsymbol{s})$ and $(\mathscr{V}, \boldsymbol{t}),$ we write $$(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$$ if $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j})$ for all $i \in I$ and $j \in J,$ where we wrote 

• $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I},$
• $\mathscr{V} = \{V_{j} \rightarrow U\}_{j \in J},$
• $\boldsymbol{s} = ([s_{i}])_{i \in I},$
• $\boldsymbol{t} = ([t_{j}])_{j \in J}.$

We claim that this gives an equivalence relation among such pairs. Reflexivity is evident, and for symmetry, we just need to use the isomorphisms $V_{j} \times_{U} U_{i} \simeq U_{i} \times_{U} V_{j}$ because the covers always include isomorphisms and we can compose covers to get another. For transitivity, suppose that $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$ and $(\mathscr{V}, \boldsymbol{t}) \sim (\mathscr{W}, \boldsymbol{u}),$ where we write  

• $\mathscr{W} = \{W_{k} \rightarrow U\}_{k \in K},$
• $\boldsymbol{u} = ([u_{k}])_{k \in K}.$

We have $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ for all $i \in I$ and $j \in J,$ while $$[t_{j}|_{V_{j} \times_{U} W_{k}}] = [u_{k}|_{V_{j} \times_{U} W_{k}}]$$ for all $j \in J$ and $k \in K.$ This gives us $$[s_{i}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}].$$ Running over $j \in V,$ since $F^{\mathrm{s}}$ is a separated presheaf, we necessarily have $$[s_{i}|_{U_{i} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} W_{k}}].$$ This shows the transitivity, so we indeed get an equivalence relation.

We define $F^{\mathrm{a}}(U)$ to be the set of equivalence classes $[(\mathscr{U}, \boldsymbol{s})]$ of the pairs we have discussed above. Given a morphism $V \rightarrow U$ in $\mathcal{C},$ we define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[(\mathscr{U}, \boldsymbol{s})] \mapsto [(V \times_{U} \mathscr{U}, \boldsymbol{s}^{*})],$ where $V \times_{U} \mathscr{U} = \{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ and $\boldsymbol{s}^{*} = (s_{i}|_{V \times U_{i}})_{i \in I}.$ Given $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{W}, \boldsymbol{t}),$ we have $$[s_{i}|_{U_{i} \times_{U} W_{j}}] = [t_{j}|_{U_{i} \times_{U} W_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} W_{j}),$ for all $i \in I$ and $j \in J.$ Since $$(V \times_{U} U_{i}) \times_{V} (V \times_{U} W_{j}) \simeq V \times_{U} U_{i} \times_{U} W_{j},$$ having $$[s_{i}|_{V \times_{U} U_{i} \times_{U} W_{j}}] = [t_{j}|_{V \times_{U} U_{i} \times_{U} W_{j}}]$$ is enough to check that the restrictions $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ are well-defined.

Given a cover $\{U_{i} \rightarrow U\}_{i \in I},$ fix any $[(\mathscr{V}, \boldsymbol{s})], [(\mathscr{W}, \boldsymbol{t})] \in F^{a}(U)$ such that $$[(U_{i} \times_{U} \mathscr{V}, \boldsymbol{s}^{*})] = [(U_{i} \times_{U}\mathscr{W}, \boldsymbol{t}^{*})]$$ for all $i \in I.$ Since $$(U_{i} \times_{U} V_{j}) \times_{U_{i}} (U_{i} \times_{U} W_{k}) \simeq U_{i} \times_{U} V_{j} \times_{U} W_{k},$$ we necessarily have $$[s_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j} \times_{U} W_{k})$. Ranging over all $i \in I,$ using the fact that $F^{\mathrm{s}}$ is a separated presheaf, we have $$[s_{j}|_{V_{j} \times_{U} W_{k}}] = [t_{k}|_{V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(V_{j} \times_{U} W_{k})$. This implies that $$[(\mathscr{V}, \boldsymbol{s})] = [(\mathscr{W}, \boldsymbol{t})].$$ This shows that $F^{\mathrm{a}}$ is a separated presheaf.

To show that $F^{\mathrm{a}}$ is a sheaf, fix $[(\mathscr{V}_{i}, \boldsymbol{s}_{i})] \in F^{\mathrm{a}}(U_{i})$ for $i \in I$ such that $$[(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{i} , \boldsymbol{s}_{i}^{*})] = [(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{j}, \boldsymbol{s}_{j}^{*})].$$ We have $$(U_{i} \times_{U} U_{j} \times_{U} V_{i,l}) \times_{U_{i} \times_{U} U_{j}} (U_{i} \times_{U} U_{j} \times_{U} V_{j,m}) \simeq U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm},$$ so the above identity implies that we have $$s_{i,l} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}} = s_{j,m} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}},$$ but then this makes $[(\mathscr{V}, \boldsymbol{s})] \in F^{\mathrm{a}}(U)$ given by $\mathscr{V} = \{V_{il} \rightarrow U_{i} \rightarrow U\}_{i,l}$ and $\boldsymbol{s} = ([s_{i,l}])_{i,l} \in \prod_{i,l} F^{\mathrm{s}}(V_{il}).$ This shows that $F^{\mathrm{a}}$ is a sheaf.

We define $a_{U} : F(U) \rightarrow F^{\mathrm{a}}(U)$ by $s \mapsto [(\{\mathrm{id}_{U}\}, (s))].$ It is immediate that this gives rise to a map $a : F \rightarrow F^{\mathrm{a}}$ of functors. Given $s, t \in F(U)$ with $$[(\{\mathrm{id}_{U}\}, (s))] = [(\{\mathrm{id}_{U}\}, (t))],$$ we have $[s] = [t] \in F^{\mathrm{s}}(U),$ so by our definition, there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ Any given $[(\mathscr{U}, \textbf{s})] \in F^{\mathrm{a}}(U)$ comes with the cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ so that $$a_{U_{i}}(s_{i}) = [(\{\mathrm{id}_{U_{i}}\}, (s_{i}))] = [(U_{i} \times_{U} \mathscr{U}, \boldsymbol{s}^{*})]| = [(\mathscr{U}, \boldsymbol{s})]|_{U_{i}},$$ where we get the second identity because $$U_{i} \times_{U_{i}} (U_{i} \times_{U} U_{j}) \simeq U_{i} \times_{U} U_{j}$$ and $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$ for all $j \in J.$ Thus, we are done checking the necessary conditions for $a : F \rightarrow F^{\mathrm{a}}$ to be a sheafification of $F.$

Grassmannians - Part 2

We continue a previous posting following Vakil's notes. We work over a fixed base scheme $S$ and fixed integers $0 \leq r \leq n,$ where $n \geq 1.$ Recall that the Grassmanian is a functor $\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ that sends an $S$-scheme $Y$ to the isomorphism class of an epimorphism $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q},$ where $\mathscr{Q}$ is a locally free $\mathscr{O}_{Y}$-module of rank $r.$ When $r = 1,$ we saw in the previous posting that $\mathrm{Gr}_{S}(1, n)$ is represented by $\mathbb{P}^{n-1}_{S}.$

Special case: rational points over a field. Let $S = \mathrm{Spec}(k)$ for some field $k.$ Then the elements of $\mathrm{Gr}(r, n)(k) := \mathrm{Gr}_{\mathrm{Spec}(k)}(r, n)(\mathrm{Spec}(k))$ are given by the $k$-linear surjections $k^{n} \twoheadrightarrow k^{r},$ which can be viewed as $r \times n$ matrices with at least one $r \times r$ submatrix having nonzero determinant. (In other words, the matrices have rank $r,$ which is the full rank.) We are these maps up to isomorphisms, which amounts to the action of $\mathrm{GL}_{r}(k)$ on the left. Thus, we have established: $$\mathrm{Gr}(r, n)(k) \leftrightarrow \mathrm{GL}_{r}(k) \backslash \mathrm{Mat}_{r \times n}(k).$$ The elements of the right-hand side can be seen as picking $r$-linearly independent vectors up to $k$-linear isomorphisms of the vector spaces they generate in $k^{n}.$ Thus, we see that $\mathrm{Gr}(r, n)(k)$ parametrizes $r$-dimensional subspaces of $k^{n}.$ The following lemma explains this a bit more clearly:

Lemma. Let $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ two sets of $r$ $k$-linearly independent vectors in $k^{n}.$ Denote by $A$ the $r \times n$ matrix whose $i$-th row is given by $v_{i}$ and define $B$ similarly using $w_{i}.$ Then the following are equivalent:

• $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$;
• $B = gA$ for some $g \in \mathrm{GL}_{r}(k).$

Proof. Write $v_{i} = (v_{i,1}, \dots, v_{i,n})$ and $w_{i} = (w_{i,1}, \dots, w_{i,n})$ with $v_{i,j}, w_{i,j} \in k.$ Then for any $g = (g_{i,j}) \in \mathrm{Mat}_{r}(k),$ saying that $B = gA$ is equivalent to saying $$w_{ij} = g_{i,1}v_{1,j} + \cdots + g_{i,r}v_{r,j}$$ for $1 \leq i \leq r$ and $1 \leq j \leq n.$ This condition can be rewritten as $$w_{i} = g_{i,1}v_{1} + \cdots + g_{i,r}v_{r}$$ for $1 \leq i \leq r.$ Now, note that $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$ if and only if the last condition we mentioned is true for some $g_{i,j} \in k.$ When this happens, it is immediate that $(g_{ij})$ is necessarily invertible, so this finishes the proof. $\Box$

Back to functorial nonsense. We are going to show that $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ is representable. Since all representable functor is a Zariski sheaf (i.e., a sheaf on the big Zariski site of $S$), the strategy is to show that $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf and then check a certain property on top of that so that we can guarantee that it is representable by some $S$-scheme (which will be necessarily a unique one). That is, after showing $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf, we will use the following theorem we studied in a previous posting:

Theorem. A functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ is representable if and only if the following two conditions are satisfied:

• $F$ is a Zariski sheaf;
• $F$ can be covered by representable open subfunctors.

Checking Grassmanianns are Zariski sheaves. Saying that $\mathrm{Gr}_{S}(r, n)$ is a sheaf means that for any $S$-scheme $U$ and an open cover $U = \bigcup_{i \in I}U_{i},$ the following two conditions are satisfied:

• if $\phi, \psi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \psi|_{U_{i}}$ for all $i \in I,$ then $\phi = \psi$;
• if $(\phi_{i})_{i \in I} \in \prod_{i \in I}\mathrm{Gr}_{S}(r, n)(U_{i})$ satisfies that $\phi_{i}|_{U_{i} \cap U_{j}} = \phi_{j}|_{U_{i} \cap U_{j}}$ for all $i, j \in I,$ then there is $\phi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \phi_{i}$ for all $i \in I.$

Proof for the first condition. To check the first among the two conditions above, consider $\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ and $\psi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}'.$ Since we are working with representatives of isomorphism classes, instead of checking needed equalities, we aim to establish corresponding isomorphisms. Denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(U)$ and $t_{1}, \dots, t_{n} \in \mathscr{Q}'(U)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}(U)^{\oplus n},$ writing $\mathscr{O} = \mathscr{O}_{U}$ for convenience.

Suppose that $\phi|_{U_{i}} \simeq \psi|_{U_{i}}$ for all $i \in I.$ Our goal is to show that $\phi \simeq \psi.$ For this, we may assume that $U_{i}$'s are refined enough so that both $\mathscr{Q}|_{U_{i}}$ and $\mathscr{Q}'|_{U_{i}}$ are free of rank $r$ over $U_{i}.$ The restrictions $\phi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}|_{U_{i}}$ and $\psi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}'|_{U_{i}}$ are isomorphic so that we have $\eta_{i} : \mathscr{Q}|_{U_{i}} \overset{\sim}{\longrightarrow} \mathscr{Q}'|_{U_{i}}$ with $\psi|_{U_{i}} = \eta_{i} \circ \phi|_{U_{i}}$ for $i \in I.$ Given $i \in I,$ the commutativity condition can be written as $$t_{l}|_{U_{i}} = \eta_{i, U_{i}}(s_{l}|_{U_{i}})$$ for $1 \leq l \leq n.$ An important observation is the following:

Claim. We have $\eta_{i, U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}.$

Proof of Claim. Note that $$\eta_{i, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}}) = t_{l}|_{U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}})$$ for all $1 \leq l \leq n.$ Since $s_{1}|_{U_{i} \cap U_{j}}, \dots, s_{l}|_{U_{i} \cap U_{j}}$ generate $\mathscr{Q}(U_{i} \cap U_{j}),$ this establishes the identity. $\Box$ (Claim)

From now on let us write $U_{ij} = U_{i} \cap U_{j},$ as the notations get messier. Let $V \subset U$ be any open subset. We write $V_{i} = V \cap U_{i}$ and $V_{ij} = V \cap U_{ij}.$ Given any $s \in \mathscr{Q}(V),$ we have $$\eta_{i, V_{i}}(s|_{V_{ij}})|_{V_{ij}} = \eta_{i, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{j}}(s|_{V_{j}})|_{V_{ij}}$$ for $i, j \in I.$ Since $\mathscr{Q}'$ is a sheaf, there is unique $\eta_{V}(s) \in \mathscr{Q}'(V)$ such that $\eta_{V}(s)|_{V_{i}} = \eta_{i, V_{i}}(s|_{V_{i}})$ for $i \in I.$ This defines a map $\eta_{V} : \mathscr{Q}(V) \rightarrow \mathscr{Q}'(V),$ which can be immediately recognized as a sheaf map $\eta : \mathscr{Q} \rightarrow \mathscr{Q}'$ by intersection with $U_{i}$'s to check some commutative diagrams. $\Box$

Proof for the second condition. Consider an $\mathscr{O}_{U_{i}}$-linear epimorphism $$\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}_{i}$$ for each $i \in I$ such that $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ This means that, for $i, j \in I,$ we have $\eta_{ij} : \mathscr{Q}_{i}|_{U_{ij}} \overset{\sim}{\longrightarrow} \mathscr{Q}_{j}|_{U_{ij}}$ such that $\phi_{j}|_{U_{ij}} = \eta_{ij} \circ \phi_{i}|_{U_{ij}}.$ Our goal is to find $$\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$$ such that $\phi|_{U_{i}} \simeq \phi_{i}$ for $i \in I.$ Writing $U_{ijk} = U_{i} \cap U_{j} \cap U_{k},$ we note that we have $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}} = \eta_{jk}|_{U_{ijk}} \circ \phi_{j}|_{U_{ijk}} = \phi_{k}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}}.$$ This implies that we have the cocycle condition $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}},$$ so there must be a unique sheaf (that is necessarily locally free of rank $r$) $\mathscr{Q}$ gluing $\mathscr{Q}_{i}$'s. (For example, see SP:00AK.) Now, we are given $\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \rightarrow \mathscr{Q}|_{U_{i}}$ with $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ Working similarly to the last paragraph of the last proof, we are done. $\Box$ 

Hence, we have shown that $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf.

What remains to show Grassmannians are representable. Now that we know $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf, to show it is representable, it remains to show that $\mathrm{Gr}_{S}(r, n)$ is covered by reprsentable open subfunctors.

Given any subset $I \subset \{1, \dots, n\}$ of size $r,$ and an $S$-scheme $Y,$ denote by $\mathrm{Gr}_{S}(r, n)_{I}(Y)$ the subset of $\mathrm{Gr}_{S}(r, n)$ consisting of $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ such that on every trivializing open subset $U \subset Y$ of $\mathscr{Q}$ (which in particular gives $\mathscr{Q}|_{U} \simeq \mathscr{O}_{U}^{\oplus r}$), the $I$-th minor of the $r \times n$ matrix given by $\mathscr{O}_{U}^{\oplus r} \twoheadrightarrow \mathscr{Q}|_{U}$ is nonzero at every point of $U.$

More concretely, given $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q},$ let $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ be the images of $$e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$$ Then $s_{1}|_{U}, \dots, s_{n}|_{U}$ are the columns of the $r \times n$ matrix in discussion. Fix any $S$-scheme map $\pi : X \rightarrow Y.$ We know this induces $$\mathrm{Gr}_{S}(r, n)(Y) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$$ by the pullback $\pi^{*}.$ Moreover, we note that $\pi^{*}s_{1}, \dots, \pi^{*}s_{n} \in \mathscr{O}_{X}(X)^{\oplus n}$ are the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{X}(X)^{\oplus n},$ so considering trivializing open subsets, we may observe that the above map restricts to $$\mathrm{Gr}_{S}(r, n)_{I}(Y) \rightarrow \mathrm{Gr}_{S}(r, n)_{I}(X)$$ for each $I \subset \{1, \dots, n\}$ of size $r.$ Thus, we have seen that $\mathrm{Gr}_{S}(r, n)_{I}$ is a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ We have a map $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors given by the relevant inclusions of sets.

Claim. The functor $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ is an open subfunctor. That is, given any $S$-scheme $Y$ and a map $h_{Y} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors, the base chage $$\mathrm{Gr}_{S}(r, n)_{I} \times_{\mathrm{Gr}_{S}(r, n)} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V_{I} \hookrightarrow Y$ (i.e., $h_{V_{I}} \rightarrow h_{Y}$).

Construction of open subsets $V_{I}$. We are given be an $S$-scheme $Y$ and an element of $$\mathrm{Hom}(h_{Y}, \mathrm{Gr}_{S}(r, n)) \simeq \mathrm{Gr}_{S}(r, n)(Y),$$ where the hom-set is taken over the functors $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ The correspondence is given by the Yoneda lemma, and explicitly, it is given as $\xi \mapsto \xi_{Y}(\mathrm{id}_{Y}).$ We denote by $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ (a reperesentative of) the given element in $\mathrm{Gr}_{S}(r, n)(Y).$ It is convenient to denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}$ under $\phi_{Y}.$

It necessarily follows that $h_{Y}(X) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$ is given by $$[\pi : X \rightarrow Y] \mapsto [\mathscr{O}_{X}^{\oplus n} \simeq \pi^{*}\mathscr{O}_{Y}^{\oplus n} \xrightarrow{\pi^{*}\phi} \pi^{*}\mathscr{Q}.]$$ Thus, the set $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X)$$ consists of the elements of the form $(\pi^{*}\phi, \pi)$ such that $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$ (i.e., the $I$-th minor of $[(\pi^{*}s_{1})|_{W} | \cdots | (\pi^{*}s_{1})|_{W}]$ is nowhere zero in any trivializing open $W \subset X$ for $\pi^{*}\mathscr{Q}$). The map $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \rightarrow h_{Y}(X)$$ is simply given by $(\pi^{*}\phi, \pi) \mapsto \pi.$ Of course, we note that $\phi$ is only given up to an isomorphism, but the above discussion still makes sense even if we just consider isomorphism classes.

Let's construct $V_{I}$'s now. Again, note that $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ is fixed. Let $U \subset Y$ be any trivializing open subset for $\mathscr{Q}.$ Denote by $f_{I} \in \mathscr{O}_{Y}(U)$ the $I$-th minor (i.e., the determinant of the $I$-th $r \times r$ submatrix of $[s_{1}|_{U} | \cdots | s_{n}|_{U}]$). We define $V_{I}$ to be the union of $D_{U}(f_{I})$ where we vary trivializing open subsets $U \subset Y$ for $\mathscr{Q}.$

Proof of Claim. For any $\pi : X \rightarrow Y$ in $h_{Y}(X) = \mathrm{Hom}_{S}(X, Y),$ we have $\pi^{*}\phi : \mathscr{O}_{X}^{\oplus n} \twoheadrightarrow \pi^{*}\mathscr{Q}$ in $\mathrm{Gr}_{S}(r, n)(X).$ The upshot is that the following are equivalent:

• we have $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$;
• $\pi(X) \subset V_{I}.$

This immediately gives an explicit bijection $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \simeq h_{V_{I}}(X) = \mathrm{Hom}_{S}(V_{I}, X),$$ which does the job. $\Box$

Our $V_{I}$'s form an open cover of $Y$. For each $y \in U,$ we have a surjective $\kappa(y)$-linear map $\mathscr{O}_{Y}^{n}|_{y} \twoheadrightarrow \mathscr{Q}|_{y},$ so there exists at least one $I \subset \{1, \dots, n\}$ with $|I| = r$ such that $y \in D_{U}(f_{I}).$ This implies that we have an open cover $$U = \bigcup_{I \subset \{1, \dots, n\}}D_{U}(f_{I}).$$ Denote by $V_{I}$ the union of all $D_{U}(f_{I}) \subset Y$ where $U$ varies among all the trivializing open subsets of $Y$ for $\mathscr{Q}.$ This gives an open cover $$Y = \bigcup_{I \subset [n] \text{ with } |I| = r} V_{I},$$ where we started to write $[n] = \{1, \dots, n\}$ since the notations are getting messy.

Each $\mathrm{Gr}_{S}(n, k)_{I}$ is represented by $\mathbb{A}^{r(n-r)}_{S}$. Let $Y$ be an $S$-scheme. Given $[\phi] \in \mathrm{Gr}_{S}(r, n),$ the map $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ can be described by $n$ sections $\phi_{j} : \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q}.$ That is, we have $\phi = \phi_{1} \oplus \cdots \oplus \phi_{n}.$ 

Fix $I = \{i_{1}, \dots, i_{r}\}$ with $1 \leq i_{1} < \cdots < i_{r} \leq n.$ If $[\phi] \in \mathrm{Gr}_{S}(r, n)_{I},$ then $\phi_{i_{1}} \oplus \cdots \oplus \phi_{i_{r}} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{Q}$ is an isomorphism. 

Proof. To see this, it is enough to show that its localization at a point $y \in Y$ is an isomorphism. Surjectivity is immediate from Nakayama. To show injectivity, consider the map $\mathscr{O}_{Y,y}^{\oplus r} \rightarrow \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}$ in question as an $r \times r$ matrix whose columns are given by $$\phi_{i_{1},y}(1), \dots, \phi_{i_{r},y}(1) \in \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}.$$ The determinant of this matrix is not in the maximal ideal $\mathfrak{m}_{Y,y}$ of $\mathscr{O}_{Y,y},$ so it must be invertible in the local ring. By Cramer's rule, this implies that the matrix is invertible over $\mathscr{O}_{Y,y},$ which implies the injectivity we wanted. $\Box$

Hence, we now may write $$\mathscr{Q}(Y) = \mathscr{O}_{Y}(Y)\phi_{i_{1},Y}(1) \oplus \cdots \oplus \mathscr{O}_{Y}(Y)\phi_{i_{r},Y}(1).$$ This implies that given $j \in [n] \setminus I,$ we may write $$\phi_{j,Y}(1) = a_{1,j}(\phi)\phi_{i_{1},Y}(1) + \cdots +a_{r,j}(\phi)\phi_{i_{r},Y}(1) \in \mathscr{Q}(Y),$$ where $a_{i,j}(\phi) \in \mathscr{O}_{Y}(Y) = \Gamma(Y, \mathscr{O}_{Y}),$ which can be viewed as an $S$-scheme map $a_{i,j}(\phi) : Y \rightarrow \mathbb{A}^{1}_{S},$ by gluing $R$-algebra maps $R[t] \rightarrow \Gamma(Y, \mathscr{O}_{Y})$ given by $t \mapsto a_{ij}(\phi)$ for affine open $\mathrm{Spec}(R) \subset S.$

Thus, we have $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S}.$$ This defines a map $$\mathrm{Gr}_{S}(r, n)(Y)_{I} \rightarrow \mathrm{Hom}_{S}(Y, \mathbb{A}^{r(n-r)}_{S}),$$ as we may note that $a_{i,j}(\phi)$ only depends on the isomorphism class of $\phi.$

Given a map $\pi : X \rightarrow Y$ of $S$-schemes, we have $$(\pi^{*}\mathscr{Q})(X) = \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{1},Y})(1) \oplus \cdots \oplus \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{r},Y})(1),$$ so we may see that $a_{i,j}(\pi^{*}\phi) = \pi^{*}a_{i,j}(\phi) \in \mathscr{O}_{X}(X)$ for $j \in [n] \setminus I.$ Since $\pi : X \rightarrow Y$ comes with the map $\Gamma(Y, \mathscr{O}_{Y}) \rightarrow \Gamma(X, \mathscr{O}_{X})$ such that $a_{i,j}(\phi) \mapsto \pi^{*}a_{i,j}(\phi),$ we see that $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} \circ \pi = (\pi^{*}a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} = (a_{i,j}(\pi^{*}\phi))_{1 \leq i \leq r, j \in [n] \setminus I}.$$ This shows that our map is actually a map $$\mathrm{Gr}_{S}(r, n)(-)_{I} \rightarrow \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S})$$ of functors.

Conversely, given $$(a_{i,j})_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S},$$ which we interpret as $a_{ij} \in \Gamma(Y, \mathscr{O}_{Y}) = \mathscr{O}_{Y}(Y).$ We define $\phi : \mathscr{O}_{Y}^{n} \rightarrow \mathscr{O}_{Y}^{\oplus r}$ as follows. We declare $\phi_{i_{1},Y}(1) := e_{1}, \dots, \phi_{i_{r},Y}(1) := e_{r} \in \mathscr{O}_{Y}(Y)^{\oplus r}$ so that $$\phi_{i_{1},Y} \oplus \cdots \oplus \phi_{i_{r},Y} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{O}_{Y}^{\oplus r}$$ is the identity map. For $j \in [n] \setminus I,$ we may define $$\phi_{j,Y}(1) := a_{1,j}\phi_{i_{1},Y}(1) + \cdots + a_{r,j}\phi_{i_{r},Y}(1) \in \mathscr{O}_{Y}(Y)^{\oplus r}.$$ One may check this gives a desired inverse so that we establish $$\mathrm{Gr}_{S}(r, n)(-)_{I} \simeq \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S}).$$ That is, the $S$-scheme $\mathbb{A}^{r(n-r)}_{S}$ respresents the functor $\mathrm{Gr}_{S}(r, n)(-)_{I}.$

Upshot. Since these are open subfunctors that cover $\mathrm{Gr}_{S}(r, n),$ we see that $\mathrm{Gr}_{S}(r, n)$ is representable by an $S$-scheme.

Grassmannians - Part 1

In this posting, we follow Vakil's notes to start discussing about Grassmannians. Let us fix the base scheme $S$ once and for all. Given an $S$-scheme $Y$ and integers $0 \leq r \leq n,$ we denote by $\mathrm{Gr}_{S}(r, n)(Y)$ the set of isomorphism classes exact sequences of the form $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{Q}$ is a locally free sheaf (or more precisely $\mathscr{O}_{X}$-module) of rank $r$ on $Y.$

Unimportant remark. The latter $\mathscr{Q}$ is supposed to be "Q", for "quotient".

Given any epimorphism $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ of $S$-schemes where $\mathscr{Q}$ is a locally free sheaf of rank $r$ on $Y,$ its kernel $\ker(\phi)$ is locally free of rank $n - r.$ (See Vakil 13.5.B.(a) for details.) Thus, we see that $\mathrm{Gr}_{S}(r, n)(Y)$ can be described as the set of isomorphism classes of short exact sequences of the form $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{S}$ has rank $n - r$ and $\mathscr{Q}$ has rank $r.$

Remark. Given an exact sequence $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n}$$ with a locally free sheaf $\mathscr{S}$ of rank $r,$ its cokernel is not necessarily locally free. An example (which we learn from Vakil 13.4.1) is as follows: take $$Y = \mathbb{A}^{1}_{k} = \mathrm{Spec}(k[t]),$$ where $k$ is a field. If we take $n = 1$ and $\mathscr{S}$ the locally free sheaf of rank $r = 1$ associated to the $k[t]$-submodule $k[t]t$ of $k[t]$ generated by $t.$ Then the quotient $\mathscr{O}_{Y}/\mathscr{S}$ is the coherent sheaf associated to $k$ by giving $t$-action as multiplication by $0$, which is not locally free.

We want to make $\mathrm{Gr}_{S}(r, n)$ into a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$ so let's describe where a morphism $\pi : X \rightarrow Y$ of $S$-schemes is mapped to. Given an exact sequence $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{Y}$-modules, since $\pi^{*}\mathscr{O}_{Y} = \mathscr{O}_{X}$ and $\pi^{*}$ is right-exact, we have an exact sequence $$\mathscr{O}_{X}^{\oplus n} = \pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \rightarrow \pi^{*}\mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{X}$-modules. Since $\mathscr{Q}$ is a locally free $\mathscr{O}_{Y}$-module of rank $r,$ it follows that $\pi^{*}\mathscr{Q}$ is a locally free $\mathscr{O}_{X}$-module of rank $r$ (e.g., Vakil 16.3.7.(3)). If there is another map $\rho : Y \rightarrow Z$ of $S$-schemes, applying $\rho^{*}$ to the last sequence gives $$\mathscr{O}_{Z} = (\pi \circ \rho)^{*}(\mathscr{O}_{Y}^{\oplus n}) = \rho^{*}\pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \twoheadrightarrow \rho^{*}\pi^{*}\mathscr{Q} = (\pi \circ \rho)^{*}\mathscr{Q},$$ which is the same as applying $(\pi \circ \rho)^{*}$ to $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ we began with. It is not hard to see that our discussion is compatible with isomorphisms of sequences, so we have constructed a functor $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$$ which we call the Grassmanian of $r$-subspaces in $n$-spaces over $S.$

Special case: rank $1$. Consider the case $r = 1.$ Locally free sheaves of rank $1$ are line bundles by definition. Given $1 \leq i \leq n,$ denote by $$e_{i} = (0, \dots, 0, 1, 0, \dots, 0) \in \mathscr{O}_{Y}(Y)^{\oplus n},$$ where $1$ only appears in the $i$-th spot. For any open subset $U \subset Y,$ this element restricts to an element $\mathscr{O}_{Y}(U)^{\oplus n}$ with the same description. Fix a line bundle $\mathscr{L}.$ Then a map $\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{L}$ of sheaves is an epimorphism if and only if its localization $\mathscr{O}_{Y, y}^{\oplus n} \rightarrow \mathscr{L}_{y}$ is surjective for all $y \in Y.$ If we denote by $s_{i} \in \mathscr{L}(Y)$ the image of $e_{i}$ under the map $\mathscr{O}_{Y}(Y)^{\oplus n} \rightarrow \mathscr{L}(Y),$ the global section of the given sheaf map, then the last statement is equivalent to saying that one of $s_{1}, \dots, s_{n}$ must have a nonzero image in $\mathscr{L}_{y}$ for each $y \in Y$ (because $\mathscr{L}_{y} \simeq \mathscr{O}_{Y,y},$ so any nonzero element would generate the whole module), or in other words, the global sections $s_{1}, \dots, s_{n}$ do not share a common zero in $Y.$

This gives rise to an $S$-scheme map $[s_{1} : \cdots : s_{n}] : Y \rightarrow \mathbb{P}^{n-1}_{S},$ which can be painlessly obtained by first working over $\mathrm{Spec}(\mathbb{Z})$ and then base change over $S.$ Given another line bundle $\mathscr{L}'$ on $Y,$ an isomorphism of two maps $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}$ and $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}'$ is an isomorphism $\mathscr{L} \rightarrow \mathscr{L}'$ of $\mathscr{O}_{Y}$-modules such that $s_{i} \mapsto s'_{i}$ for $1 \leq i \leq n,$ where $s'_{i} \in \mathscr{L}'(Y)$ is the image of $e_{i} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$ This implies that for each $y \in Y,$ there is an affine open $U \ni y$ in $Y$ such that $\mathscr{L}|_{U} \simeq \mathscr{O}_{Y}|_{U}$ and for $1 \leq i \leq n,$ the image of $s'_{i}$ is the product of the image of $s_{i}$ and an invertible element in $\mathscr{O}_{Y}(U).$ This implies that $[s_{1} : \cdots : s_{n}] = [s'_{1} : \cdots : s'_{n}].$

On the other hand, any $S$-scheme map $\pi : Y \rightarrow \mathbb{P}^{n-1}_{S}$ can be obtained as $\pi = [\pi^{*}x_{0} : \cdots : \pi^{*}x_{n-1}],$ where $x_{0}, \dots, x_{n-1}$ are given by the global sections of $\mathscr{O}_{\mathbb{P}^{n-1}_{S}}(1).$ Explicitly, if $\mathrm{Spec}(R) \subset S$ is an affine open subset, then $$x_{0}, \dots, x_{n-1} \in R[x_{0}, \dots, x_{n-1}].$$ Let's review the meaning of the pullback of a section of a line bundle:

What do we mean by pullback of a global section of a line bundle? Given a global section $s \in \mathscr{L}(Y)$ of a line bundle $\mathscr{L}$ on $Y,$ we can consider the map $\mathscr{O}_{Y}(Y) \rightarrow \mathscr{L}(Y)$ given by $1 \mapsto s.$ This extends to a unique $\mathscr{O}_{Y}$-module map $\mathscr{O}_{Y} \rightarrow \mathscr{L}$ per each $s.$ This way, the section $s$ can be viewed as a map $\mathscr{O}_{Y} \rightarrow \mathscr{L}.$ Now, if $\pi : X \rightarrow Y$ is a map of $S$-schemes, then we may consider the pullback $\pi^{*}s : \mathscr{O}_{X} = \pi^{*}\mathscr{O}_{Y} \rightarrow \pi^{*}\mathscr{L}.$ The image of $1 \in \mathscr{O}_{X}(X)$ under the map $$\mathscr{O}_{X}(X) = (\pi^{*}\mathscr{O}_{Y})(X) \rightarrow (\pi^{*}\mathscr{L})(X)$$ given by $\pi^{*}s$ is also denoted as $\pi^{*}s.$

Going back to the discussion, open sets of the form $\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R))$ cover $Y,$ where $U_{i} = D_{\mathbb{P}^{n-1}}(x_{i})$ and $\mathrm{Spec}(R) \subset S$ is affine open. The restriction $$\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)) \rightarrow U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R) = \mathrm{Spec}(R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}])$$ of $\pi$ corresponds to the ring map $$R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}] \rightarrow \Gamma(\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)), \mathscr{O}_{Y})$$ such that $x_{j}/x_{i} \mapsto \pi^{*}x_{j}/\pi^{*}x_{i}.$ Some more details can be found in (the proof of) Vakil 16.4.1. Hence, we have established an explicit bijection $$\mathrm{Gr}_{S}(1, n)(Y) \leftrightarrow \mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}),$$ which we can briefly write as $(s_{1}, \dots, s_{n}) \mapsto [s_{1} : \cdots : s_{n}].$ Given any map $\rho : X \rightarrow Y,$ the induced map $\mathrm{Gr}_{S}(1, n)(Y) \rightarrow \mathrm{Gr}_{S}(1, n)(X)$ can be described as $(s_{1}, \dots, s_{n}) \mapsto (\rho^{*}s_{1}, \dots, \rho^{*}s_{n}),$ while $\mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}) \rightarrow \mathrm{Hom}_{S}(X, \mathbb{P}^{n-1}_{S})$ is given by $\phi \mapsto \rho \circ \phi.$ We can write $$\phi = [s_{1} : \cdots : s_{n}] = [\phi^{*}x_{1} : \cdots : \phi^{*}x_{n}]$$ and $$\begin{align*}\rho \circ [s_{1} : \cdots : s_{n}] &= \phi \circ \rho \\ &= [(\phi \circ \rho)^{*}x_{1} : \cdots : (\phi \circ \rho)^{*}x_{n}] \\ &= [\rho^{*}\phi^{*}x_{1} : \cdots : \rho^{*}\phi^{*}x_{n}] \\ &= [\rho^{*}s_{n} : \cdots : \rho^{*}s_{n}].\end{align*}.$$ This means that our bijection is now checked to be a natural trasformation so that it is an isomorphism of functors: $$\mathrm{Gr}_{S}(1, n)(-) \simeq \mathrm{Hom}_{S}(-, \mathbb{P}^{n-1}_{S}).$$ This, by definition, means that the functor $\mathrm{Gr}_{S}(1, n)(-)$ is representable by the projective space $\mathbb{P}^{n-1}_{S}$ over $S.$

Example. Take $S = \mathrm{Spec}(k),$ where $k$ is a field. Then we have $$\mathrm{Gr}(1, n)(k) \simeq \mathbb{P}^{n-1}(k),$$ given by $$(a_{1}, \dots, a_{n}) \mapsto [a_{1} : \cdots : a_{n}],$$ where $a_{i} \in k,$ at least one of which is nonzero. We understand the right-hand side quite well, so let's consider the left-hand side. An element is given by (the isomorphism class of) a map $\mathscr{O}_{\mathrm{Spec}(k)}^{\oplus n} \twoheadrightarrow \mathscr{L}.$ This is nothing more than a surjective $k$-linear map $k^{n} \twoheadrightarrow k,$ so the image of each $e_{i}$ is $a_{i} \in k$ we are considering.

Next time. We describe what happens to $\mathrm{Gr}_{S}(r, n),$ when $r$ may be larger than $1.$

$l$-adic cohomology: Lecture 4

References. The following are the references I use for writing this posting:

• Freitag and Kiehl,
• Milne,
• Hochster,
• Vakil,
• Stacks Project (which will be referred as "SP").

Of course, I may cite more references as I go.

Goal. The goal is to follow the first chapter of the first reference.

Presheaves/sheaves on a site (valued in abelian groups). Let $\mathcal{C}$ be any category. A presheaf (valued in the category $\textbf{Ab}$ of abelian groups) on $\mathcal{C}$ is a functor $\mathscr{F} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Ab}.$

Now, suppose that $\mathcal{C}$ is equipped with a Grothendieck topology (i.e., $\mathcal{C}$ is a site). A presheaf $\mathscr{F}$ on $\mathcal{C}$ is called a sheaf if for any object $U$ of $\mathcal{C}$ and a cover $\{U_{i} \rightarrow U\}_{i \in I}$ of $U,$ the induced diagram $$\mathscr{F}(U) \rightarrow \prod_{i \in I}\mathscr{F}(U_{i}) \rightrightarrows \prod_{(i, j) \in I^{2}} \mathscr{F}(U_{i} \times_{U} U_{j})$$ is an equilizer diagram in $\textbf{Ab}.$

How is the above diagram defined? The first map is obtined by applying $\mathscr{F}$ to maps $U_{i} \rightarrow U$ and then using the universal property of the product. There are two maps from the second term to the third. The first one comes from the maps of the form $U_{i} \times_{U} U_{j} \rightarrow U_{i},$ and the second one comes from the maps of the form $U_{i} \times_{U} U_{j} \rightarrow U_{j}.$

Notation. If $\mathcal{C}$ is a site, for any map $U \rightarrow V$ in $\mathcal{C},$ we describe the induced map $\mathscr{F}(V) \rightarrow \mathscr{F}(U)$ as $s \mapsto s|_{U}.$ This notation does not mean that there is a unique map from $U$ to $V,$ but we will use it as long as it does not create any confusion.

Remark. Let $\mathcal{C}$ be a site. Note that a presheaf $\mathscr{F}$ is a sheaf if and only if for every cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of each object $U$ in $\mathcal{C},$ the following conditions are satisfied :

• for each $s, t \in \mathscr{F}(U),$ if $s|_{U_{i}} = t|_{U_{i}}$ for every $U_{i} \rightarrow U$ in $\mathscr{U},$ then $s = t$.
• given any $(s_{i})_{i \in I} \in \prod_{i \in I}\mathscr{F}(U_{i}),$ if $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{i}|_{U_{i} \times_{U} U_{j}}$ for every $U_{i} \rightarrow U$ and $U_{j} \rightarrow U$ in $\mathscr{U}$ (including the case $i = j$), then there is $s \in \mathscr{F}(U)$ such that $s|_{U_{i}} = s_{i}$ for every $U_{i} \rightarrow U$ in $\mathscr{U}.$

We want to define cohomology of a sheaf on the étale site $X_{ét}$ of a scheme $X.$ The following facts let us do this:

Fact 1 (SP03NT). The category of sheaves $\textbf{Sh}(X_{ét})$ is an abelian category that has enough injectives.

A part of showing the above fact also yields the following:

Theorem. The global section functor functor $\Gamma : \textbf{Sh}(X_{ét}) \rightarrow \textbf{Ab}$ given by $\mathscr{F} \mapsto \Gamma(X, \mathscr{F}) := \mathscr{F}(X)$ (with morphisms are mapped to their restrictions to the global sections) is left-exact.

Proof. (cf. SP03CN). Let $$0 \rightarrow \mathscr{F} \xrightarrow{\phi} \mathscr{G} \xrightarrow{\psi} \mathscr{H}$$ be an exact sequence in $\textbf{Sh}(X_{ét}).$ We want to show that the induced sequence $$0 \rightarrow \mathscr{F}(X) \xrightarrow{\phi} \mathscr{G}(X) \xrightarrow{\psi} \mathscr{H}(X)$$ is exact.

To show the exactness at $\mathscr{F}(X),$ it is enough to show the following:

Lemma 1. Let $\mathscr{F} \xrightarrow{\phi} \mathscr{G}$ be any morphism of sheaves in $\textbf{Sh}(X_{ét}).$ Then $\ker(\phi_{U}) = \ker(\phi)(U)$ for every $U$ in $\textbf{Sh}(X_{ét}).$

Proof of Lemma 1. We construct a presheaf $\mathscr{K}$ on $X_{ét}$ by defining $$\mathscr{K}(U) := \ker(\phi_{U}) \subset \mathscr{F}(U).$$ Given any map $U \rightarrow V$ of $X_{ét},$ the restriction map $\mathscr{K}(V) \rightarrow \mathscr{K}(U)$ is given by restricting the map $\mathscr{F}(V) \rightarrow \mathscr{F}(U).$ This defines a functor $\mathscr{K} : X_{ét}^{\mathrm{op}} \rightarrow \textbf{Ab},$ because $\mathscr{F}$ is a functor.

We first show that $\mathscr{K}$ is a sheaf. Let $s \in \mathscr{K}(U)$ and say $\{U_{i} \rightarrow U\}_{i \in I}$ is a cover of $U.$ If $s|_{U_{i}} = 0 \in \mathscr{K}(U) \subset \mathscr{F}(U)$ for each $i,$ then $s = 0$ in $\mathscr{F}(U),$ so $s = 0$ in $\mathscr{K}(U).$ Next, suppose that we have $s_{i} \in \mathscr{K}(U_{i})$ for each $i \in I$ such that $$s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}} \in \mathscr{K}(U_{i} \times_{U} U_{j}) \subset \mathscr{F}(U_{i} \times_{U} U_{j})$$ for all $i, j \in I.$ Since $\mathscr{F}$ is a sheaf, we may find $s \in \mathscr{F}(U)$ such that $$s|_{U_{i}} = s_{i} \in \mathscr{F}(U_{i}).$$ For all $i,$ we have $$\phi_{U}(s)|_{U_{i}} = \phi_{U_{i}}(s_{i}) = 0$$ because $s_{i} \in \mathscr{K}(U_{i}) = \ker(\phi_{U_{i}}).$ This implies that $\phi_{U}(s) = 0,$ so $s \in \mathscr{K}(U).$ This establishes the fact that $\mathscr{K}$ is a sheaf.

We now claim that $\mathscr{K} = \ker(\phi).$ The inclusions $\mathscr{K}(U) \hookrightarrow \mathscr{F}(U)$ build into a map $\iota : \mathscr{K} \rightarrow \mathscr{F}$ of presheaves. It is immediate that the composition $\mathscr{K} \xrightarrow{\iota} \mathscr{F} \xrightarrow{\phi} \mathscr{G}$ is the zero map (i.e., $\phi \circ \iota = 0$), so to finish the check, we fix any other sheaf map $j : \mathscr{K}' \rightarrow \mathscr{F}$ such that $\phi \circ j = 0.$ We want to show that there exists a unique map $\eta : \mathscr{K}' \rightarrow \mathscr{K}$ such that $j = \iota \circ \eta.$ This requirement forces $j_{U} = \iota_{U} \circ \eta_{U},$ so given any object $U$ of $X_{ét},$ the map $\eta_{U} : \mathscr{K}'(U) \rightarrow \mathscr{K}(U)$ has to be given by the usual property of kernels in $\textbf{Ab}.$ More explicitly, since the image of $j_{U}$ lies in $\ker(\phi_{U}) = \mathscr{K}(U),$ it factors as $$\mathscr{K}'(U) \rightarrow \mathscr{K}(U) \hookrightarrow \mathscr{F}(U).$$ It is immediate that these map build a map $\eta : \mathscr{K}' \rightarrow \mathscr{K}$ of sheaves. Hence, we have proved that $\mathscr{K} = \ker(\phi).$ This finishes the proof (of Lemma 1). $\Box$

Going back to the proof of Theorem, it remains to show the exactness at $\mathscr{G}(X).$ That is, we need to show that $\mathrm{im}(\phi_{X}) = \ker(\psi_{X}).$ We have $$\ker(\mathrm{coker}(\phi)) = \mathrm{im}(\phi) = \ker(\psi),$$ the second of which uses the exactness at $\mathscr{G}.$ Thus, by Lemma 1, we have $$\ker(\mathrm{coker}(\phi)(X)) = \ker(\psi_{X}).$$ The inclusion $$\mathrm{im}(\phi_{X}) = \ker(\mathrm{coker}(\phi_{X})) \subset \ker(\mathrm{coker}(\phi)(X)) = \ker(\psi_{X})$$ follows immediately from diagram chasing.

To show the reverse inclusion, we actually need to understand what $\mathrm{coker}(\phi)$ is. It is the "sheafification" of the presheaf on $X_{ét},$ which we briefly discuss now:

Sheafification. Given a presheaf $\mathscr{F}$ on $X_{ét},$ a sheafification $\tilde{\mathscr{F}}$ of $\mathscr{F}$ is a sheaf on $X_{ét}$ together with a presheaf map $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ satisfying the following property: for any sheaf $\mathscr{G}$ on $X_{ét}$ and a presheaf map $\phi : \mathscr{F} \rightarrow \mathscr{G},$ there is a unique sheaf map $\tilde{\mathscr{F}} \rightarrow \mathscr{G}$ that $\phi$ factors through the sheafification.

Remark. Note that if a sheafification of $\mathscr{F}$ exists, it is unique up to a unique isomorphism. It is also immediate that taking sheafification defines a functor $\textbf{Psh}(X_{ét}) \rightarrow \textbf{Sh}(X_{ét})$ from the category of presheaves over $X_{ét}$ to that of sheaves.

The sheafification always exists, but we will not discuss the proof of it but cite a reference instead. At this point, if we denote by $\mathrm{coker}(\phi_{-})$ the presheaf given by $U \mapsto \mathrm{coker}(\phi_{U}),$ one can check that the cokernel $\mathscr{G} \rightarrow \mathrm{coker}(\phi)$ of the sheaf map $\phi : \mathscr{F} \rightarrow \mathscr{G}$ can be constructed as the following composition: $$\mathscr{G} \rightarrow \mathrm{coker}(\phi_{-}) \rightarrow \widetilde{\mathrm{coker}(\phi_{-})} =: \mathrm{coker}(\phi).$$ Note that the usual proof for the small Zariski site of $X$ (Vakil 2.6.1) works here without any change.

We need a little bit more to finish our proof:

Fact 2 (Vistoli Theorem 2.64). The desired sheafification can be constructed so that it satisfies the following two properties:

1. for any $s \in \mathscr{F}(U),$ if its image in $\tilde{\mathscr{F}}(U)$ is zero, then there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = 0$ for each $i.$
2. for any $t \in \tilde{\mathscr{F}}(U),$ there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $t|_{U_{i}}$ is in the image of $\mathscr{F}(U_{i}) \rightarrow \tilde{\mathscr{F}}(U_{i})$ for each $i.$

Remark. Note that if we have any sheaf $\tilde{\mathscr{F}}$ and a presheaf map $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ satisfying the above two properties, then $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ is necessarily a sheafification.

We are now ready to finish our proof of Theorem. Denote by $\eta : \mathscr{G} \rightarrow \mathrm{coker}(\phi)$ the map comes with $\mathrm{coker}(\phi).$ Consider $\eta_{X} : \mathscr{G}(X) \rightarrow \mathrm{coker}(\phi)(X).$ We know $\ker(\psi_{X}) = \ker(\eta_{X}),$ so what remains for us to show is $\ker(\eta_{X}) \subset \mathrm{im}(\phi_{X}).$ 

Fix any $s \in \ker(\eta_{X}) \subset \mathscr{G}(X).$ We want to show that $s \in \mathrm{im}(\phi_{X}).$ Since $\eta_{X}(s) = 0 \in \mathrm{coker}(\phi)(X),$ we have a map $\mathrm{coker}(\phi_{X}) \rightarrow \mathrm{coker}(\phi)(X)$ given by the universal property of the cokernel (source) such that $\bar{s} \mapsto \eta_{X}(s) = 0.$ Note that this is exactly the sheafification map, so by Fact 2-1, we can find a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $\overline{s|_{U_{i}}} = 0$ in $\mathrm{coker}(\phi_{U_{i}})$ (i.e., $s|_{U_{i}} \in \mathrm{im}(\phi_{U_{i}})$) for each $i.$ Hence, we can write $s|_{U_{i}} = \phi_{U_{i}}(t_{i})$ for some (necessarily unique) $t_{i} \in \mathscr{F}(U_{i})$ for each $i.$ Since $\phi_{U_{i} \times_{X} U_{j}} : \mathscr{F}(U_{i} \times_{X} U_{j}) \rightarrow \mathscr{G}(U_{i} \times_{X} U_{j})$ is injective, we have $$t_{i} |_{U_{i} \times_{X} U_{j}} = t_{j} |_{U_{i} \times_{X} U_{j}}$$ for all $i, j \in I.$ As $\mathscr{F}$ is a sheaf, this implies that we have $t \in \mathscr{F}(X)$ such that $t|_{U_{i}} = t_{i}$ for all $i.$ Since $\mathscr{G}$ is a sheaf, it follows that $\phi_{X}(t) = s,$ so $s \in \mathrm{im}(\phi_{X}),$ as desired. This finishes the proof (of Theorem). $\Box$

Definition of étale cohomology group. Using the facts above, by looking at any sheaf $\mathscr{F}$ on $X_{ét}$ as an object of $\textbf{Sh}(X_{ét}),$ one writes $H^{i}(X_{ét}, -)$ to be the $i$-th right derived functor of the global section functor. Given any sheaf $\mathscr{F}$ on $X_{ét},$ the abelian group $H^{i}(X_{ét}, \mathscr{F})$ is called the $i$-th étale cohomology group of the sheaf $\mathscr{F}.$

Remark. It is important to note that our discussion works even if we used the category $\textbf{Mod}_{R}$ of $R$-modules instead of $\textbf{Ab}$ for any given ring $R.$