quotient of subsheaf is a sheaf?

Question

This is the problem 5.49 of Rotman Homoglogical Algebra.

Let $i:F\to G$ be the inclusion map between sheaf of abelian groups over topological space $X$. I want to show $G/F$ is a sheaf as the problem asks to show the map $F\to G\to Coker(i)\to 0$ is exact as sheave exact sequence which is equivalent to say $(G/F)^{\star}$ sheafification of $(G/F)$ is same as $(G/F)$. Since the notation is without $\star$, I guess there is no need to sheafify.

Consider $U\subset X$ open set with cover $U_i$. Let $U_{ij}=U_i\cap U_j$. Denote $F_i=F(U_i), F_{ij}=F(U_{ij})$, $F=F(U)$ and similarly for $G_i,G_{ij},G$ and $G/F$'s. It suffices to consider the following exact sequences.

$0\to F\to G\to G/F\to 0$

$0\to F_i\to G_i\to (G/F)_i\to 0$

$0\to F_{ij}\to G_{ij}\to (G/F)_{ij}\to 0$.

There is induced exact sequence by $F,G$ being sheaf $0\to F\to \prod_i F_i\to \prod_{ij}F_{ij}$ and similarly for $G$'s. So I can induce two other maps $G/F\to \prod_i (G/F)_i\to \prod_{ij}(G/F)_{ij}$. I could show $G/F\to \prod_i(G/F)$ is injective and $G/F\to \prod_{ij}(G/F)_{ij}$ is identically 0.

However, I do not how to glue elements in $G/F(U_i)$ with them agreeing on the overlaps. Take any elements $\sigma_{i,G/F}\in (G/F)_i$ and suppose they agree on overlaps $(G/F)_{ij}$ and surjectivity allows me to construct $\sigma_{i,G}\in G_i$. However, when those elements descends down to $\prod_{ij}G_{ij}$, there is no guarantee that image of $\sigma_{i,G}$ is $0\in \prod_{ij}G_{ij}$.

Any hint will be helpful.

Answer 1

It seems you've misinterpreted something: it's certainly necessary to sheafify the quotient. For instance, consider the topological space $X$ with open sets $\emptyset,A,B,C,X$ such that $A=B\cap C$ and $B, C$ aren't comparable. Then a sheaf on $X$ is essentially a pullback square of abelian groups. We can get a counterexample to the claim that the quotient is always a sheaf by considering the sheaf corresponding to the exact sequence $0\to M\to N\to P$ (Where $X\mapsto M, B\mapsto N,A\mapsto P,C\mapsto 0$) with its subsheaf $0\to 0 \to 0 \to P$. The quotient, $M\to N\to 0$, is not exact if $M \to N$ is not an isomorphism.

The question should, perhaps, be to show that the corresponding sequence is exact after we do sheafify the cokernel. For this, my suggestion would be to use stalks.

Answer 2

It is easy to mix up when a formula is intended to be interpreted in the category of presheaves or the category of sheaves. (and sometimes, source materials aren't entirely clear on the matter)

The formula $\operatorname{coker}(\iota)$, here, is meant to be interpreted in the category of sheaves. In plainer terms, what this question is asking is:

Show that the cokernel of a sheaf morphism can be computed by sheafifying the cokernel of the corresponding morphism of presheaves.

There is, incidentally, a one-line category-theoretic proof:

$$ \operatorname{coker}(\mathcal{F} \xrightarrow{\iota} \mathcal{G}) \cong \operatorname{coker}(\mathbf{a}(\mathbf{i}\mathcal{F} \xrightarrow{\mathbf{i}\iota} \mathbf{i}\mathcal{G})) \cong \mathbf{a} \operatorname{coker}(\mathbf{i}\mathcal{F} \xrightarrow{\mathbf{i}\iota} \mathbf{i}\mathcal{G}) $$

where $\mathbf{a}$ is the sheafification functor and $\mathbf{i}$ is the inclusion functor from sheaves to presheaves. (thus, the three cokernels are computed in sheaves, sheaves, and presheaves respectively)

The main points why this works are:

• $\mathbf{a} \circ \mathbf{i}$ is naturally isomorphic to the identity functor on sheaves
• $\mathbf{a}$ is the left adjoint to $\mathbf{i}$
• Left adjoints preserve all colimits

($\mathbf{a}$ also preserves finite limits, but we don't need that in this calculation)

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The calculation of the "natural map" $\pi$, incidentally, has to take place in the category of presheaves, since it involves the presheaf cokernel $\mathcal{G}/\mathcal{F}$. However, it is a presheaf morphism between sheaves, and thus is also a sheaf morphism. Once we've established what $\pi$ and $\operatorname{coker}(\iota)$ mean, the question "Does $\pi = \operatorname{coker}(\iota)$?" has the same answer whether interpreted in the category of sheaves or of presheaves, so it doesn't matter which context we interpret that formula.