Is the following proof sound/does anyone have another more elegant (categorical) proof?

The direction $\Rightarrow$ is obvious the "family of stalks"-functor is a functor and functors take isos to isos.

$\Leftarrow$: let us first show we get $\mathcal F \rightarrow \mathcal G$ is mono. We have the diagram

$\require{AMScd}$\begin{CD} \mathcal F(X) @>{\phi}>> \mathcal G(X)\\ @VVV @V{\mu}VV \\ \prod \mathcal F_p @>\bar\phi>> \prod \mathcal G_p \end{CD}

where the vertical and the bottom maps are mono. It follows that $\mu\phi$ is mono; hence $\phi$ is mono.

Now we show that $\phi$ is epi. An element in $\mathcal G(X)$ consists of a colection of compatible germs in $\prod \mathcal G_p$. Since the map on the bottom is an isomorphism, we have a corresponding collection in $\prod\mathcal F_p$; if it is also compatible it will lift up to $\mathcal F$. But it is compatible because it the isomorphism is induced by a morphism upstairs. More explicitly:

Take two germ representatives $(U^q,g^q), (U^t,g^t)$ of a compatible collection in $\prod \mathcal G_p$ such that their preimages via $\bar\phi$ are germs defined over the same open sets $(U^q,f^q), (U^t,f^q)$. We want to show that $$f^q - f^t = 0\qquad \text{in}\, U^q\cap U^t.$$ Since $\bar\phi$ is induced by a morphism of sheaves $\phi$, we get

$$\bar\phi:\prod_{p\in U^q \cap U^t} (f^q-f^t)_p\mapsto\prod_{p\in U^q \cap U^t}(g^q-g^t)_p$$

by looking at appropriate representatives $(V_p,\text{res}$$\vert_{Vp}(f^q-f^t))$ of $(f_q-f_t)_p$, where $V_p\subset U_q \cap U_t$ is small enough that one would be able to find a representative of the germ $(g^q-g^t)_p$ over it. By assumption, the right-hand side is zero and since $\bar\phi$ is an iso so is the left-hand side. We have what we wanted.

  • $\begingroup$ "...seems to suggest that an isomorphism on the family of stalks induces a surjection on the level of sheaves." I think you mean a surjection on the level of (global?) sections? You are correct that this modified statement is not always true. It is surjective on the level of sheaves. Your proof should somehow involve the fact that the image sheaf requires a sheafification. $\endgroup$
    – Matt
    Oct 26, 2013 at 20:01
  • $\begingroup$ @Matt: I meant a surjection on the level of sheaves because on my diagram you could pick any open set $U\subset X$ and argue that a section in $U$ is just a collection of compatible germs in $U$. You pull that collection back to $\prod_{p\in U} \mathcal F_p$ and those germs should still be compatible because we are pulling back by an isomorphism. I didn't understand what "modified statement" you meant since you followed that line by what I thought you were denying "that a morphism induced by the morphism on stalks is surjective on the level of sheaves." $\endgroup$
    – Rodrigo
    Oct 26, 2013 at 20:25
  • $\begingroup$ @Matt now I think I got my proof right and I see where I explicitly made use of the fact that $\bar\phi$ is induced by $\phi$ so my previous worries are settled. I wish I could strike out what comes after Edit:, but that didn't work. Anyway, could you elaborate on how to use the sheafification of the image sheaf? I think that could make for a more elegant proof. $\endgroup$
    – Rodrigo
    Oct 27, 2013 at 0:09
  • 1
    $\begingroup$ I'm just saying the terminology is not standard. You are using the phrase "surjective on the level of sheaves" to mean: for any $U\subset X$ the map $\mathcal{F}(U)\to \mathcal{G}(U)$ is surjective. The phrase ought to mean that the image sheaf is all of $\mathcal{G}$. These are not equivalent. There are surjective maps of sheaves that are not surjective on global sections. $\endgroup$
    – Matt
    Oct 27, 2013 at 0:49
  • 1
    $\begingroup$ @Matt You're right. It's just beginning to down on me that they are not the same, although I'm still forming the picture of what it means for a sheaf map to be surjective (or equivalently the image sheaf be all of $\mathcal G$). This picture should be distinct from just saying that what it means is that the stalks of the image sheaf coincide with the stalks of $\mathcal G$ (although I could take this for a definition) because it is exactly the equivalence of these two statements that I'm trying to prove. $\endgroup$
    – Rodrigo
    Oct 27, 2013 at 1:00


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