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I'm trying to prove whether or not this is true:

Let $\mathcal{F}$ be a presheaf and $\mathcal{G}$ a sheaf (of rings, let's say) on a space $X$. If $\{U_i\}_i$ is a basis for the topology of $X$ and $\mathcal{F}\big|_{U_i}\simeq \mathcal{G}\big|_{U_i}$ for all $i$, then the sheafification $\mathcal{F}^{sh}$ is isomorphic to $\mathcal{G}$.

If this is true, I imagine it could be proven by universal property.

I can see how to build the morphism of pre-sheaves $\mathcal{F}\to\mathcal{G}$: if $f\in\mathcal{F}(U)$ we define it's image in $\mathcal{G}(U)$ as the gluing of the images $f\big|_{U\cap U_i}$ in $\mathcal{G}(U\cap U_i)$.

But don't know how to build the morphisms of sheaves $\mathcal{F}^{sh}\to\mathcal{G}$.

Any advice?

rmdmc89
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    If you have a map $\mathcal{F}\to\mathcal{G}$ which induces your isomorphisms $\mathcal{F}|{U_i}\simeq\mathcal{G}|{U_i}$ then this is true since you can check it is an isomorphism locally. If you don't have a map $\mathcal{F}\to\mathcal{G}$ it is false. It already fails if $\mathcal{F}$ is a sheaf as you can take $\mathcal{F},\mathcal{G}$ be the sheaf of sections of two non isomorphic vector bundles of the same rank. Note that your construction of $\mathcal{F}\to\mathcal{G}$ is wrong : you don't know if the $f|_{U\cap U_i}$ can be glued together. – Roland Apr 25 '20 at 11:54
  • @Roland, I realize I didn't understood the universal property correctly, but now I think I do. If we have had a morphism $\mathcal{F}\to\mathcal{G}$, then the by the universal property would not only tell us that $\mathcal{F}^{sh}\simeq \mathcal{G}$, but also that there is an unique isomorphism between them. So your point is that we do need some gluing conditions for the isomorphisms $\mathcal{F}\big|{U_i}\simeq \mathcal{G}\big|{U_i}$, right? – rmdmc89 Apr 25 '20 at 18:49
  • A unique isomorphism such that for any $i$, the morphism $\mathcal{F}|{U_i}\to\mathcal{F}^{sh}|{U_i}\to\mathcal{G}|_{U_i}$ is the given isomorphism. Otherwise there may be plenty of isomorphisms $\mathcal{F}^{sh}\simeq\mathcal{G}$. – Roland Apr 25 '20 at 21:02
  • @Roland, yes, I forgot to mention the commutativity of the diagram. Thank you – rmdmc89 Apr 26 '20 at 16:24

2 Answers2

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I think the assertion is true. Note however that I am learning algebraic geometry at the moment, so the following might be wrong. Please correct me, if this is the case and I will happily edit/delete it...

If you have a morphism of presheaves $\varphi:\mathcal{F} \rightarrow \mathcal{G}$ with the property that $$\varphi \vert_{U_i}: \mathcal{F}\vert_{U_i} \rightarrow \mathcal{G}\vert_{U_i}$$ is an isomorphism for all $U_i$, then functoriality of sheafification (by the universal property) yields a morphism of sheaves $$\varphi^+:\mathcal{F}^+ \rightarrow \mathcal{G}^+$$ with the property that $$\varphi^+\vert_{U_i}:\mathcal{F}^+\vert_{U_i} \cong \mathcal{G}^+\vert_{U_i}.$$ Hence for any $p\in X$ there is some $i$ with $p\in U_i$ and we get that on stalks $$(\varphi^+)_p:(\mathcal F^+)_p \rightarrow (\mathcal G^+)_p = (\varphi^+\vert_{U_i})_p:(\mathcal F^+\vert_{U_i})_p \rightarrow (\mathcal G^+\vert_{U_i})_p$$ is an isomorphism. Thus $\varphi^+$ is an isomorphism and we get $$\mathcal F^+ \cong \mathcal G^+ \cong \mathcal G$$

Jonas Linssen
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  • I think $(\varphi^+)p$ being an isomorphism for all $p$ implies that $\varphi^+$ is both a monomorphism and an epimorphism, but not necessarily an isomorphism. But maybe we can solve this problem by showing that $\varphi^+\big|_V$ is surjective for all $V$, which follows from the fact the $\varphi^+\big|{U_i}$'s are isomorphisms – rmdmc89 Jun 08 '20 at 17:47
  • By the way, you implicitly found the gap in my solution, namely that the morphism of pre-sheaves $\varphi:\mathcal{F}\to\mathcal{G}$ should be given as a hypothesis, after all we cannot say a priori that the isomorphisms $\mathcal{F}\big|{U_i}\simeq \mathcal{G}\big|{U_i}$ can be glued together. Thank you! – rmdmc89 Jun 08 '20 at 18:00
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I actually proved this true on an answer I recently gave here, and coincidently I've just stumbled upon this question asking exactly for this.

It's Corollary 8 here, whose proof relies on Lemma 7. On the statements, what it is meant by a sheaf of sets over a basis is defined in 009J.

Elías Guisado Villalgordo
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