Topics in Abstract Algebra/Sheaf theory

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Sheaf theory[edit]

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.