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Given two subsheaves $\mathcal{F}$ and $\mathcal{G}$ of a sheaf $\mathcal{O}$ on a topological space X, prove that $\mathcal{F}_p = \mathcal{G}_p$ for every p $\in$ X $\Rightarrow$ $\mathcal{F}(U) = \mathcal{G}(U)$ for every open $U \subseteq$ X

I've been trying to prove this in context of showing that if $\mathcal{F}_p \xrightarrow{\phi_p} \mathcal{F'}_p \xrightarrow{\psi_p} \mathcal{F''_p}$ is exact (as a sequence of stalks) at $\mathcal{F'_p}$ then $\mathcal{F} \xrightarrow{\phi} \mathcal{F'} \xrightarrow{\psi} \mathcal{F''}$ is exact (as a sequence of sheaves) at $\mathcal{F'}$.

I got till the point where we have that $(Im(\phi))_p = (Ker(\psi))_p$ for every p in X and I want to use the above proposition to get that $Im(\phi) = Ker(\psi)$ as subsheaves of $\mathcal{F'}$ but I have no idea how to prove the proposition rigorously.

Any help would be appreciated!

nl08
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  • You may want to look at https://math.stackexchange.com/questions/3361681/equality-of-subsheaves-on-stalks-implies-equality-of-subsheaves#comment6918670_3361681 which was posted yesterday and deals with the same thing. – KReiser Sep 20 '19 at 04:34
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    It doesn't seem like the question has been answered though. And the part of the post that seemed to be answered in the comments seems to just deal with the idea of 'equality' of stalks. But thank you very much for pointing this out! – nl08 Sep 20 '19 at 04:38
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    isn't it true that if $x\in O(U)$ then ($x_p \in F_p $ for all p in U iff $x\in F(U)$)? – Tim kinsella Sep 20 '19 at 04:46
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    I can see why it would be true that given x $\in O(U)$, x $\in F(U) \Rightarrow <U,x> \in F_p$ (or $x_p \in F_p$) for every p. This is true for any sheaf F (I think?). But why is it true that $x_p = <U,x> \in F_p$ for every p $\Rightarrow x \in F(U)$? – nl08 Sep 20 '19 at 04:50
  • for each point p, x restricts to an element of F(W) for some nbhd W of p. they patch bc they all come from x. does that work? – Tim kinsella Sep 20 '19 at 05:01
  • Could you explain that in a little more detail possibly? I don’t see where you use the assumption that x is in the larger sheaf O(U). And it doesn’t seem like the original proposition should hold for arbitrary sheaves. – User20354 Sep 20 '19 at 05:13
  • that assumption is used when i say "they all come from x" in my last comment. i.e. you use that assumption to glue together the elements of the various F(W) by showing they agree on overlaps – Tim kinsella Sep 20 '19 at 05:29
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    If I follow correctly, what you’re saying is $x_p \in F_p \Rightarrow x|{V_p} \in F(V_p)$ for every p in U. The $V_p$s for an open cover of U and - because - x is in O(U), the $x|{V_p}$s agree on the intersections $V_p \cap V_q$. Then since F is a sheaf, by the gluing condition, there exists an $x’$ in F(U) such that $x’|{V_p} = x|{V_p}$ for every p. But then by the separated condition on O, we get that $x = x’ \in F(U)$. Is that correct? Apologies if I’m being pedantic about symbolic reasoning. I’m still trying to understand stalks and I find it helpful to write it explicitly. – User20354 Sep 20 '19 at 05:52
  • Edit: where for each p in U, $V_p$ is some open subset of U containing p – User20354 Sep 20 '19 at 06:01
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    Yes exactly. (You're not being pedantic I was being overly terse because I have a wrist injury and typing tex is difficult) – Tim kinsella Sep 20 '19 at 06:17

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