# A sheaf is flasque if all restriction maps are surjective

A sheaf $\mathcal{F}$ on $X$ is called a flasque sheaf if every restriction map $\mathcal{F}(U)\rightarrow \mathcal{F}(V)$ is a surjective map.

Question is as follows :

If $0\rightarrow \mathcal{F}'\xrightarrow{\psi} \mathcal{F}\xrightarrow{\varphi} \mathcal{F}''\rightarrow 0$ is an exact sequence of sheaves, and $\mathcal{F}'$ is a flasque sheaf, then, the sequence of abelian groups $0\rightarrow \mathcal{F}'(U)\xrightarrow{\psi(U)}\mathcal{F}(U)\xrightarrow{\varphi(U)} \mathcal{F}''(U)\rightarrow 0$ is exact.

Fix an open subset $U\subseteq X$. It suffices to prove that $\mathcal{F}(U)\xrightarrow{\varphi(U)} \mathcal{F}''(U)$ is surjective. Given $s\in \mathcal{F}''(U)$ there exists an open cover $\{V_i\}$ and $t_i\in \mathcal{F}(V_i)$ such that $\varphi(V_i)(t_i)=s|_{V_i}$ in the following commutative diagram I am trying to prove that these sections $t_i\in \mathcal{F}(V_i)$ glue together to give a section $t\in \mathcal{F}(U)$ i.e., $t|_{V_i}=t_i$. Then, $$\varphi(U)(t)|_{V_i}=\varphi(V_i)(t|_{V_i})=\varphi(V_i)(t_i)=s|_{V_i}.$$ By identity axiom, we then conclude that $\varphi(U)(t)=s$.

I still have not used that $\mathcal{F}'$ is a flasque sheaf. I am confused what open sets should I consider whose restriction map is a surjective map. As we are talking about restrcition $t_i|_{V_i\cap V_j}$ of an element in $\mathcal{F}(V_i)$ I think surjectivity of restriction maps of these open sets $V_i,V_i\cap V_j, V_j$ gives some result.

We have To use that restriction maps corresponding to $\mathcal{F}'$ are surjective, we need to get an element in $\mathcal{F}'(V_i\cap V_j)$. Suppose $a\in \mathcal{F}(V_i\cap V_j)$ is such that $\varphi(V_i\cap V_j)(a)=0$ then $a=\psi(b)$ for some $b\in \mathcal{F}'(V_i\cap V_j)$.

We have \begin{align}\varphi(t_i|_{V_i\cap V_j}-t_j|_{V_i\cap V_j}) &=\varphi(t_i|_{V_i\cap V_j})-\varphi(t_j|_{V_i\cap V_j})\\ &=\varphi(V_i)(t_i)|_{V_i\cap V_j}-\varphi(V_j)(t_j)|_{V_i\cap V_j}\\ &=(s|_{V_i})|_{V_i\cap V_j}-(s|_{V_j})|_{V_i\cap V_j}\\ &=s|_{V_i\cap V_j}-s|_{V_i\cap V_j}=0\end{align} So, $$t_i|_{V_i\cap V_j}-t_j|_{V_i\cap V_j}=\psi(a_{ij})$$ for some $a_{ij}\in \mathcal{F}'(V_i\cap V_j)$. As restriction maps are surjective, there exists $b_i\in \mathcal{F}'(V_i)$ such that $b_i|_{V_i\cap V_j}=a_{ij}$ for all $i,j$.

Then $\psi (V_i)(b_i)\in \mathcal{F}(V_i)$. I do not think this way would lead me anywhere near the final step.

Any suggestions regarding this are welcome.

I think you're totally on right track. First, let's solve the special case where your open cover $\{V_i \}$ consists only of two sets, $V_1$ and $V_2$.

In this special case, there exists a $t_1 \in \mathcal F(V_1)$ and a $t_2 \in \mathcal F(V_2)$ such that $$\varphi(t_1) = s|_{V_1}, \ \ \ \ \ \ \varphi(t_2) = s|_{V_2},$$ and your difficulty is that $t_1$ and $t_2$ do not glue together.

However, you showed that there exists a $a_{12} \in \mathcal F'(V_1 \cap V_2)$ such that $$\psi(a_{12}) = t_1 - t_2$$ and using the fact that $\mathcal F'$ is flasque, you also argued that there exists a $b_2 \in \mathcal F'(V_2)$ such that $$b_2 |_{V_1 \cap V_2} = a_{12}.$$ So let's define $$\widetilde t_1=t_1 \in \mathcal F(V_1), \ \ \ \ \ \ \ \ \ \widetilde t_2 = t_2 + \psi(b_2)\in \mathcal F(V_2),$$

What's nice about $\widetilde t_1$ and $\widetilde t_2$ is that:

• Just like the original sections $t_1$ and $t_2$, they map onto $s|_{V_1}$ and $s|_{V_2}$ under $\varphi$: $$\varphi(\widetilde t_1) = s|_{V_1}, \ \ \ \ \ \varphi(\widetilde t_2) = s|_{V_2} + \varphi \circ \psi(b_2) = s|_{V_2} + 0 = s|_{V_2}.$$

• And moreover, $\widetilde t_1$ and $\widetilde t_2$ agree on $V_1 \cap V_2$: $$\widetilde t_2|_{V_1 \cap V_2} = t_2|_{V_1 \cap V_2} + \psi(a_{12}) = t_2|_{V_1 \cap V_2} + (t_1 - t_2)|_{V_1 \cap V_2} = t_1|_{V_1 \cap V_2}.$$ So $\widetilde t_1$ and $\widetilde t_2$ glue together to form a section $\widetilde t \in \mathcal F(U)$ that maps onto $s$ under $\varphi$.

Now let's discuss the general case. We're going to deduce this from the special case, using Zorn's lemma.

Consider the set of tuples $(V, t)$ where $t \in \mathcal F(V)$ and $\varphi(t) = s|_V$, and define an ordering on this set of tuples by declaring that $(V, t) \leq (V', t')$ iff $V \subseteq V'$ and $t'|_{V} = t$. It is not hard to see that every chain $\{ (V_\alpha, t_\alpha) \}$ in our set of tuples has an upper bound: to construct this upper bound, we take $V = \bigcup_\alpha V_\alpha$, and we use the sheaf axioms to glue the $t_\alpha$ together to form a section $t \in \mathcal F(V)$. So we may apply Zorn's lemma to conclude that our set of tuples has a maximal element, $(V_{\rm max}, t_{\rm max})$.

The only thing left to check is that $V_{\rm max} = U$. Suppose that $x$ is a point in $U \ \backslash \ V_{\rm max}$. Then there exists an open neighbourhood $V_{\rm extra}$ of $x$, and a section $t_{\rm extra} \in \mathcal F(V_{\rm extra})$ such that $\varphi(t_{\rm extra}) = s|_{V_{\rm extra}}$. Using our proof of the special case of the theorem from earlier, with $V_{\rm max}$ and $V_{\rm extra}$ in place of $V_1$ and $V_2$, we deduce that there exists a section $t \in \mathcal F(V_{\rm max} \cup V_{\rm extra})$ extending $t_{\rm max}$ such that $\varphi(t) = s|_{V_{\rm max} \cup V_{\rm extra}}$. By maximality, it must be the case that $V_{\rm max} = V_{\rm max} \cup V_{\rm extra}$, contradicting our assumption that $x \notin V_{\rm max}$.

(Note that $t$ does not extend $t_{\rm extra}$, because we have to modify $t_{\rm extra}$ by adding on $\psi(b_{\rm extra})$ so that it glues with $t_{\rm max}$ on $V_{\rm max} \cap V_{\rm extra}$. But this doesn't matter!)

• Ah, So, I gave up in the last but one step. :D... How much necessary is it to use Zorn's lemma? +1 – user312648 Jul 30 '17 at 14:20
• @cello I'm not aware of any proof that doesn''t use Zorn's lemma. I would be interested to know under what conditions we can avoid Zorn's lemma. Unfortunately I'm not an expert on these things... – Kenny Wong Jul 30 '17 at 14:21
• Fine fine. If I do not get any answer that does not use Zorn's lemma, I will accept this.. Nice answer it is. – user312648 Jul 30 '17 at 14:31
• @cello Trevor Wilson in this answer seems to show that you do need choice/Zorn's lemma to prove your statement in full generality. I believe you can get away with less if you assume some things about your set $U$, for example, if $U$ is (quasi-)compact (you can glue finitely many things by hand) or just Lindelöf (use countable choice). – Takumi Murayama Aug 1 '17 at 5:17