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Given a presheaf $\mathcal{F}$on a topological space $X$, one can construct the etale space $\pi_1 : Y_1\to X$. Let us now look at the associated sheaf $\mathcal{F}^+$ as a presheaf and construct the etale space $\pi_2 :Y_2\to X$ corresponding to the presheaf $\mathcal{F}^+$. My question is : what is the relation of $Y_1$ to $Y_2$ ? There is a natural map $\tau : \mathcal{F} \to \mathcal{F}^+ $ which gives rise to maps at the stalk level (by taking direct limits) and hence a map $\tau_{ES} : Y_1 \to Y_2 $. Is this map a homeomorphism ? (If $\mathcal{F}$ were a sheaf to begin with, $\tau$ would have been an isomorphism and hence $\tau_{ES}$ would have been a homeomorphism. However I am not sure if this is the case even if we begin with a presheaf which is not a sheaf.)

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The canonical map $\tau_{ES}:Y_1\to Y_2$ is always a homeomorphism, independently of whether the presheaf $\mathcal F$ is a sheaf or not.
The basic reason is that the presheaf $\mathcal F$ and its sheafification $\mathcal F^+$ have the exact same stalks.
The result is more or less evident if you take for $\mathcal F^+$ the sheaf of sections of the étalé space $Et(\mathcal F)$ associated to the presheaf $\mathcal F$.
The details can be found in Chapter 2,Theorem 3.10 of this extremely clear book by Tennison .

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  • $\begingroup$ It is easy to find a PDF of the book on the Internet. It is probably legal but, since I'm not sure, I'll let users look for the link. $\endgroup$ – Georges Elencwajg Aug 24 '13 at 21:38
  • $\begingroup$ Thanks. I looked at the book and this clears my doubt. This is indeed very clearly explained in the book you mentioned. If I understand correctly, the individual maps taking sections of the presheaf to sections of the associated sheaf need not be bijective (if we start with a presheaf that is not a sheaf) but the induced maps on the stalks are. $\endgroup$ – user90041 Aug 25 '13 at 19:03

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