User:TakuyaMurata/Analysis on complex manifolds

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 $$V$$.