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?

  • 6
    $\begingroup$ 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 $(U_i,s_i)$. $\endgroup$ – 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.

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

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