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Failure of isomorphisms on stalks to arise from an isomorphism of sheaves

Let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over a topological space X.

If we have that $\mathcal{F}_{x} \cong \mathcal{G}_{x}$ for all $x \in X$. (Without having a morphism between the sheaves)

Can we conclude that $\mathcal{F} \cong \mathcal{G}$ ?. How can we show the isomorphism explicitly?

Since $\mathcal{F} \cong \mathcal{F}^{+}$ (the sheafification), I was thinking about defining the morphism stalk by stalk. But the isomorphism between the stalks could be pretty different for $x, y\in X$ and the local condition in the sections in $\mathcal{G}$ would not be satisfied.

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    I didnt see it! Thanks!2012-07-07
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    Should I delete it? How?2012-07-07
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    Don't worry about it – it will be closed as a duplicate in due order.2012-07-07
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    As an unregistered user you can't delete questions yourself (you'd need to register your account to do it). If you *really* want to remove the question you can ping me with `@t.b.` and request deletion and I'll inform the moderators, but there really is no reason to do it. This question might help others to find an answer, for example.2012-07-08

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