Topics in Abstract Algebra/Sheaf theory

Sheaf theory

We say $\mathcal{F}$ is a pre-sheaf on a topological space $X$ if

• (i) $\mathcal{F}(U)$ is an abelian group for every open subset $U \subset X$
• (ii) For each inclusion $U \hookrightarrow V$, we have the group morphism $\rho_{V, U}: \mathcal{F}(V) \to \mathcal{F}(U)$ such that
$\rho_{U,U}$ is the identity and $\rho_{W,U} = \rho_{V,U} \circ \rho_{W,V}$ for any inclusion $U \hookrightarrow V \hookrightarrow W$

A pre-sheaf is called a sheaf if the following "gluing axiom" holds:

For each open subset $U$ and its open cover $U_j$, if $f_j \in \mathcal F(U_j)$ are such that $f_j = f_k$ in $U_j \cap U_k$, then there exists a unique $f \in \mathcal F(U)$ such that $f|_{U_j} = f_j$ for all $j$.

Note that the uniqueness implies that if $f, g \in \mathcal F(U)$ and $f|_{U_j} = g|_{U_j}$ for all $j$, then $f = g$. In particular, $f|_{U_j} = 0$ for all $j$ implies $f = 0$.

4 Example: Let $G$ be a topological group (e.g., $\mathbf R$). Let $\mathcal{F}(U)$ be the set of all continuous maps from open subsets $U \subset X$ to $G$. Then $\mathcal F$ forms a sheaf. In particular, suppose the topology for $G$ is discrete. Then $\mathcal{F}$ is called a constant sheaf.

Given sheaves $\mathcal F$ and $\mathcal G$, a sheaf morphism $\phi: \mathcal F \to \mathcal G$ is a collection of group morphisms $\phi_U: \mathcal F(U) \to \mathcal G(U)$ satisfying: for every open subset $U \subset V$,

$\phi_U \circ \rho_{V, U} = \rho_{V, U} \circ \phi_V$

where the first $\rho_{V, U}$ is one that comes with $\mathcal F$ and the second $\mathcal G$.

Define $(\operatorname{ker}\phi)(U) = \operatorname{ker}\phi_U$ for each open subset $U$. $\operatorname{ker}\phi$ is then a sheaf. In fact, suppose $f_j \in \operatorname{ker}\phi_{U_j}$. Then there is $f \in \mathcal F(U)$ such that $f|_{U_j} = f_j$. But since

$(\phi_U f)|_{U_j} = \phi_{U_j} (f|_{U_j}) = \phi_{U_j} f_j = 0$

for all $j$, we have $\phi_U f = 0$. Unfortunately, $\operatorname{im}\phi$ does not turn out to be a sheaf if it is defined in the same way. We thus define $(\operatorname{im}\phi)(U)$ to be the set of all $f \in \mathcal G(U)$ such that there is an open cover $U_j$ of $U$ such that $f|_{U_j}$ is in the image of $\phi_{U_j}$. This is a sheaf. In fact, as before, let $f \in \mathcal G(U)$ be such that $f|_{U_j} \in \operatorname{im}\phi_{U_j}$. Then we have an open cover of $U$ such that $f$ restricted to each member $V$ of the cover is in the image of $\phi_V$.

Let $\mathcal{F}^0, \mathcal{F}^1, \mathcal{F}^2$ be sheaves on the same topological space.

A sheaf $\mathcal{F}$ on $X$ is said to be flabby if $\rho_{X, U}:\mathcal{F}(X) \to \mathcal{F}(U)$ is surjective. Let $\mathcal{F}_p = \lim_{U \ni p} \mathcal{F}(U)$, and, for each $f \in \mathcal{F}(U)$, define $\operatorname{supp}f = \{ x \in U | f|_p \ne 0 \}$. $\operatorname{supp}f$ is closed since $f|_p = 0$ implies $p$ has a neighborhood of $U$ such that $f|_q = 0$ for every $q \in U$. Define $\operatorname{Supp}\mathcal{F} = \{ x \in X | \mathcal{F}_x \ne 0 \}$. In particular, if $i: Z \hookrightarrow X$ is a closed subset and $\operatorname{Supp}\mathcal{F} \subset Z$, then the natural map $\mathcal{F} \to i_* i^{-1} \mathcal{F}$ is an isomorphism.

4 Theorem Suppose

$0 \longrightarrow \mathcal{F}^0 \longrightarrow \mathcal{F}^1 \longrightarrow \mathcal{F}^2 \longrightarrow 0$

is exact. Then, for every open subset $U$

$0 \longrightarrow \Gamma_Z(U, \mathcal{F}^0) \longrightarrow \Gamma_Z(U, \mathcal{F}^1) \longrightarrow \Gamma_Z(U, \mathcal{F}^2)$

is exact. Furthermore, $\Gamma_Z(U, \mathcal{F}^1) \to \Gamma_Z(U, \mathcal{F}^2)$ is surjective if $\mathcal{F}^0$ is flabby.
Proof: That the kernel of $\operatorname{ker}\mathcal{F}^0 \longrightarrow \mathcal{F}^1$ is trivial means that $\operatorname{ker}\mathcal{F}^0(U) \longrightarrow \mathcal{F}^1(U)$ has trivial kernel for any $U$. Thus the first map is clear. Next, denoting $\mathcal{F}^1 \to \mathcal{F}^2$ by $d$, suppose $f \in \mathcal{F}^1(U)$ with $df = 0$. Then there exists an open cover $U_j$ of $U$ and $u_j \in \mathcal{F}(U_j)$ such that $d u_j = f|_{U_j}$. Since $d u_j = f = d u_k$ in $U_j \cap U_k$ and $d_{U_j \cap U_k}$ is injective by the early part of the proof, we have $u_j = u_k$ in $U_j \cap U_k$ and so we get $u \in \mathcal{F}(U)$ such that $du = f$. Finally, to show that the last map is surjective, let $f \in \mathcal{F}^2(U)$, and $\Omega = \{ (U, u) | du = f|_U \}$. If $\{ (U_j, u_j) | j \in J \} \subset \Omega$ is totally ordered, then let $U = \cup_j U_j$. Since $u_j$ agree on overlaps by totally ordered-ness, there is $u \in \mathcal{F}(U)$ with $u|_{U_j} = u_j$. Thus, $(U, u)$ is an upper bound of the collection $(U_j, u_j)$. By Zorn's Lemma, we then find a maximal element $(U_0, u_0)$. We claim $U_0 = U$. Suppose not. Then there exists $(U_1, u_1)$ with $d u_1 = f|_{U_1}$. Since $d(u_0 - u_1) = 0$ in $U_0 \cap U_1$, by the early part of the proof, there exists $a \in \mathcal{F}^0(U_0 \cap U_1)$ with $da = u_0 - u_1$. Then $d(u_1 + da) = du_1 = f|_{U_1}$ (so $(U_1, u_1) \in \Omega$) while $u_1 + da = u_0$ in $U_0 \cap U_1$. This contradicts the maximality of $(U_0, u_0)$. Hence, we conclude $U_0 = U$ and so $du_0 = f$. $\square$

4 Corollary

$0 \longrightarrow \mathcal{F}^0 \longrightarrow \mathcal{F}^1 \longrightarrow \mathcal{F}^2 \longrightarrow 0$

is exact if and only if

$0 \longrightarrow \mathcal{F}_p^0 \longrightarrow \mathcal{F}_p^1 \longrightarrow \mathcal{F}_p^2 \longrightarrow 0$

is exact for every $p \in X$.

Suppose $f: X \to Y$ is a continuous map. The sheaf $f_* \mathcal{F}$ (called the pushforward of $\mathcal{F}$ by $f$) is defined by $f_* \mathcal{F}(U) = \mathcal{F}(f^{-1}(U))$ for an open subset $U \subset Y$. Suppose $f: Y \to X$ is a continuous map. The sheaf $f^{-1} \mathcal{F}$ is then defined by $f^{-1}\mathcal{F}(U) =$ the sheafification of the presheaf $U \mapsto \varinjlim_{V \supset f(U)} \mathcal{F}(V)$ where $V$ is an open subset of $X$. The two are related in the following way. Let $U \subset X$ be an open subset. Then $f^{-1}f_*\mathcal{F}(U)$ consists of elements $f$ in $\mathcal{F}(f^{-1}(V))$ where $V \supset f(U)$. Since $f^{-1}(V) \supset U$, we find a map

$f^{-1}f_*\mathcal{F} \to \mathcal{F}$

by sending $f$ to $f|_U$. The map is well-defined for it doesn't depend on the choice of $V$.