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I was going through this proof from Rotman's 'Introduction to homological algebra' (Pages 381-382) and I just can't seem to make sense of it, am not super well-versed in this so I don't know if it's too short and they're assuming that I know things I don't know or it's just simply going over my head. Here is the fragment verbatim:


Let $\mathcal{F}$ be a sheaf over a space $X$.

(i) If $0 \rightarrow \mathcal{F}' \xrightarrow{\iota} \mathcal{F} \xrightarrow{\varphi} \mathcal{F}'' \rightarrow 0$ is an exact sequence of sheaves with $\mathcal{F}'$ flabby, then $0 \rightarrow \Gamma(\mathcal{F}') \rightarrow \Gamma(\mathcal{F}) \rightarrow \Gamma(\mathcal{F}'') \rightarrow 0$ is an exact sequence of abelian groups.

PROOF.

It suffices to prove that $\varphi_X:\Gamma(\mathcal{F}) \rightarrow \Gamma(\mathcal{F}'')$, given by $\varphi_X:s \mapsto \varphi s$, is epic. Let $s'' \in \mathcal{F}''(X) = \Gamma(\mathcal{F}'')$. Define

$\mathcal{X} = \{ (U,s):U \subseteq X$ is open, $s\in \mathcal{F}(U),\varphi s=s''\mid U \} $.

Partially order $\mathcal{X}$ by $(U,s) \preceq (U_1,s_1)$ if $U \subseteq U_1$ and $s_1 \mid U = s$. It is routine to see that chains in $\mathcal{X}$ have upper bounds, and so Zorn's Lemma provides a maximal element $(U_0,s_0)$. If $U_0=X$, then $s_0$ is a global section and $\varphi_X$ is epic. Otherwise, choose $x \in X$ with $x \notin U_0 $. Since $\varphi:\mathcal{F} \rightarrow \mathcal{F}''$ is an epic sheaf map, it is epic on stalks, and so there are an open $V \subseteq X$ with $V \ni x$ and a section $t \in \mathcal{F}(V)$ with $\varphi t = s'' \mid V$. Now $s - t \in \mathcal{F}'(U \cap V)$ (we regard $\iota: \mathcal{F}' \rightarrow \mathcal{F}$ as the inclusion), so that $\mathcal{F}'$ flabby provides $r \in \mathcal{F}'(X)$ extending $s - t$. Hence, $s = t + r \mid (U \cap V)$ in $\mathcal{F}(U \cap V)$. Therefore, these sections may be glued: there is $\tilde{s} \in \mathcal{F}(U \cup V)$ with $\tilde{s} \mid U = s$ and $\tilde{s} \mid V = t + r \mid (U \cap V)$. But $\varphi(\tilde{s}) = s''$, and this contradicts the maximality of $(U_0,s_0)$.


MY QUESTIONS ARE:

1 - Why does it say that 'it is ENOUGH to prove that $\varphi_X$ is epic'?

2 - Why does it say 'If $U_0 = X$, then $s_0$ is a global section and $\varphi_X$ is epic'? Why is $\varphi_X$ epic if $s_0$ is global?

3 - I got stuck there so the rest of the proof I'm basically in La La land too

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For (1), taking sections is automatically left exact. Presumably that's been mentioned before. –  Dylan Moreland Feb 23 '12 at 23:09
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1 Answer 1

1) Because $\Gamma$ is left-exact anyway.

2) The element $s_0\in \Gamma(X,\mathcal F)$ satisfies $\phi_X(s_0)=s''|U_0=s''|X=s''\in \Gamma(X,\mathcal F'')$.
This means that any $s''\in \Gamma(X,\mathcal F)$ is in the image of $\phi_X:\Gamma(X,\mathcal F) \to \Gamma(X,\mathcal F'')$, so that $\phi_X$ is epic(=surjective), which is what we wanted after taking 1. above into account

3) I am a sedentary guy and I don't know LaLaland

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