Algebraic Analysis notes Lecture 9 (30 Jan 2019)

Last time we showed that Sh(X;k) is an abelian category. So we’ll get:

a notion of simple objects
complexes, exactness, cohomology of complexes
5-lemma
snake lemma
Jordan-Holder theorem for abelian categories of finite length

For $\phi \colon F \to G$ , we saw $ker (\phi)_x \cong ker (\phi_x)$ and $cok (\phi)_x \cong cok (\phi_x)$ .

Corollary: A sequence $0 \to F \to G \to H \to 0$ in Sh(X;k) is exact iff $0 \to F_x \to G_x \to H_x \to 0$ is exact for any point x in X.

In particular, we can check if $\phi$ is injective/surjective by checking if $\phi_x$ is injective/surjective for all points x in X.

Lemma: $\phi \colon F \to G$ is injective iff $\phi(U) \colon F(U) \to G(U)$ is injective.

Surjectivity is more complicated!

Lemma: $\phi : F \to G$ is surjective iff for any U open in X and $t \in G(U)$ , there is an open cover $(U_\alpha)_{\alpha \in I}$ of U such that for any $\alpha \in I$ , $t|_{U_\alpha}$ is in the image of $\phi_{U_\alpha}$ .

In particular, if $\phi_U$ is surjective for any U open in X, then $\phi$ is surjective, but the converse is not necessarily true.

Lemma: A sequence of sheaves $0 \to F \to G \to H$ is exact iff $0 \to F(U) \to G(U) \to H(U)$ is exact.

proof: If $0 \to F \to G \to H$ is exact, then $F(U) \to G(U)$ is injective. If $F(U) \to G(U) \to H(U)$ failed to be exact for some open U in X, then $F(U) \to ker(G(U) \to H(U))$ would be an isomorphism. Then $F \to ker \psi$ would not be an isomorphism. Contradiction.

Functors between abelian categories

Definition: If A and B are additive categories, and $F : A \to B$ is a functor, we say F is additive if for any objects x,y in A, the map $\mathrm{Hom}_A (x, y) \to \mathrm{Hom}_B (Fx, Fy)$ is a group homomorphism.

Definition: If A and B are abelian categories, we say F is exact if for any short exact sequence $0 \to x \xrightarrow{\phi} y \xrightarrow{\psi} z \to 0$ in A, then $0 \to Fx \xrightarrow{F\phi} Fy \xrightarrow{F\psi} Fz \to 0$ is exact in B. F is called left exact if $0 \to x \to y \to z$ exact implies $0 \to Fx \to Fy \to Fz$ exact. F is called right exact if $x \to y \to z \to 0$ exact implies $0 \to Fx \to Fy \to Fz \to 0$ exact.

Exact functors can be computed inductively. But to compute left/right exact functors, we need homological algebra.

Example/Exercise: Let C be an abelian category, and Y and object in C. We get two functors: $\mathrm{Hom}_C (Y, -) : C \to Ab$ and $\mathrm{Hom}_C (-, Y) : C^{op} \to Ab$ . These are both left exact.

When $\mathrm{Hom}_C (Y, -)$ is exact, then we say that Y is projective, and if $\mathrm{Hom}_C (-, Y)$ is exact, we say that Y is injective.

Definition: Let $F \in Sh(X; k)$ . Then module of global sections of F, denoted $\Gamma(F)$ , is the k-module $\Gamma(F) = F(X)$ . The module of global sections with compact support, $\Gamma_c(F)$ , is the k-module $\Gamma_c(F) = \{ s \in F(X) \mid supp(s) \text{ is compact} \}$ .

Proposition: The functors $\Gamma : Sh(X; k) \to k-mod$ and $\Gamma_c : Sh(X; k) \to k-mod$ are both left exact.

proof: $\Gamma$ is left exact by the previous lemma. Consider $0 \to F \xrightarrow{\phi} G \xrightarrow{\psi} H$ exact.

Since this diagram commutes:

then $\Gamma_c \phi$ is injective. If $s \in \Gamma_c G$ is in the kernel of $\Gamma_c G \to \Gamma_c H$ , then there is a $t \in \Gamma F$ such that $\phi(x)(t) = s$ . Then for any point x in X, $\phi_x (t_x) = s_x$ , but each $\phi_x$ is injective. Then supp(t) = supp(s), which is compact, so t is compactly supported, $t \in \Gamma_c F$ .

Let $f : X \to Y$ be a continuous map of topological spaces. Consider the induced map $f_* : Sh(X; k) \to Sh(Y; k)$ given by $f_*F(U) = F(f^{1}U)$ . Check that this is indeed a sheaf on Y. Also, given a map of sheaves on X $\phi : F \to G$ , For any U open in Y, $\phi$ induces a map $f_* F (U) \to F_* G(U)$ , and $f_*$ is a functor.

Examples:

Let Y={o}, and $f : X \to \{o\}$ . Then $f_* F(Y) = F(X) = \Gamma F$ (this is the simplest example of sheaf pushforward)
if $i : M \hookrightarrow N$ is a closed embedding, then $i_*(F)(U) = F(U \cap M)$ , called the “extension of F by zero”