# Hartshorne Ex. II 1.16 b) Flasque sheaves and exact sequences

The exercise states that when we have an exact sequence
$$0\to\mathcal{F}'\to\mathcal{F}\to\mathcal{F}''\to 0$$
of sheaves (say of Abelian groups) over a topological space $$X$$, and when $$\mathcal{F}'$$ is flasque, then for any open set $$U\subset X$$, the sequence

$$0\to\mathcal{F}'(U)\to\mathcal{F}(U)\to\mathcal{F}''(U)\to 0$$
is again exact.

By a previous exercise it is enough to show surjectivity.
I could also figure out that when we have $$s\in\mathcal{F}''(U)$$ and open subsets $$V_1,V_2\subset U$$, such that on both there is a lift of $$s$$ (i.e. there is $$t_i\in\mathcal{F}(V_i)$$, s.t. the image of $$t_i$$ in $$\mathcal{F}''(V_i)$$ is equal to the restriction $$s|_{V_i}$$), then $$s$$ can be lifted on their union.
It is also clear to me that there is an open cover of $$U$$ consisting of sets on which $$s$$ can be lifted. From here on I do not know how to proceed.

I have looked at other solutions and they want to apply Zorns Lemma, but it is not clear to me how this works here. They seem to use that given a chain (w.r.t. inclusion) of open subsets ($$U_\iota$$) on which $$s$$ can be lifted, then there is a lift of $$s$$ on $$\bigcup U_\iota$$ because $$\mathcal{F}$$ is a sheaf. However, I think this does not work, because we have no reason to assume that all the different lifts are compatible. Can anybody help out here?

• Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$. – Georges Elencwajg Nov 17 '14 at 13:08

I organize here facts already discussed.

Let $$0 \rightarrow \mathcal{F}' \stackrel{f}{\rightarrow} \mathcal{F} \stackrel{g}{\rightarrow} \mathcal{F}'' \rightarrow 0$$ be an exact sequence of sheaves on a topological space $$X$$. Let $$U$$ be an open set. By exercise II.1.8 taking sections on $$U$$ is a left exact functor, i.e., we have an exact sequence $$0 \rightarrow \mathcal{F}'(U) \stackrel{f_U}{\rightarrow} \mathcal{F}(U) \stackrel{g_U}{\rightarrow} \mathcal{F}''(U)$$. It thus remains to show that $$g_U$$ is surjective.

Let $$t'' \in \mathcal{F}''(U)$$. By the surjectivity of $$g$$ and exercise II.1.3.(a), there is an open covering $$U = \cup_{i \in I}U_i$$ and sections $$t_i \in \mathcal{F}(U_i)$$ such that $$g_{U_i}(t_i) = t''|_{U_i}$$. Let $$\mathscr{T}$$ be the set of all pairs $$(U_i,t_i)$$ equipped with the following patial order: $$(U_i,t_i) \le (U_j,t_j)$$ if $$U_i \subset U_j$$ and $$t_j|_{U_i} = t_i$$. We show that $$\mathscr{T}$$ has a maximal element.

Let $$\mathscr{C}$$ be a chain of $$\mathscr{T}$$ and set $$C = \{i: \, (U_i,t_i) \in \mathscr{C}\}$$. Set $$V = \cup_{i \in C} U_i$$. Then the sections $$t_i$$ for $$i \in C$$ agree on intersections so that by the sheaf axiom there is a section $$w$$ of $$V$$ that restricted on $$U_i$$ is $$t_i$$ for every $$i \in C$$. This implies that $$g_V(w) = t''_V$$ so that $$(V,w)$$ is an upper bound of $$\mathscr{C}$$ in $$\mathscr{T}$$. Then Zorn's lemma asserts the existence of a maximal element $$(U^*,t^*)$$ of $$\mathscr{T}$$.

Here is where the assumption on $$\mathcal{F}'$$ being flasque comes into play. Suppose $$U^* \subsetneq U$$. Then there exists some $$U_k$$ such that $$U^* \subsetneq U^* \cup U_k$$. Since the images under $$g$$ of $$t^*$$ and $$t_k$$ agree on $$U^* \cap U_k$$, we have that $$t^*|_{U^*\cap U_k} - t_k|_{U^*\cap U_k} \in \ker g_{U^* \cap U_k}$$. Then the flasque property of $$\mathcal{F}'$$ gives a section $$z \in \mathcal{F}(U^*)$$ such that $$g_{U^*}(z) = 0$$ and $$z|_{U^* \cap U_k} = t^*|_{U^*\cap U_k} - t_k|_{U^*\cap U_k}$$. Set $$\bar{t}^* = t^* - z$$. Then $$\bar{t}^*$$ and $$t_k$$ agree on $$U^* \cap U_k$$ and by the sheaf axiom there is some $$t^{**} \in \mathcal{F}(U^{**})$$, $$U^{**} = U^* \cup U_k$$, that restricts to $$\bar{t}^*$$ and $$t_k$$ on $$U^*$$ and $$U_k$$ respectively. Moreover, $$g_{U^{**}} (t^{**}) = t''|_{U^{**}}$$ so that $$(U^{**},t^{**}) \in \mathscr{T}$$. But this contradicts the maximality of $$(U^*,t^*)$$. Hence $$U^* = U$$ and $$g_U(t^*) = t''$$.

yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.

• Would you provide a bit more detail, please? :) – Shaun Mar 9 '15 at 10:56