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Let $\mathcal G$ be a sheaf on a topological space and $X$ (say a sheaf of sets) and suppose its support is contained in a subset $Z$ of $X$, i.e. $\operatorname{Supp} \mathcal G \subset Z \hookrightarrow X$. Let $i$ denote this inclusion, $i_*$ the direct image sheaf and $i^{-1}$ the inverse image sheaf.

I was able to show that $i_*i^{-1}\mathcal G \cong \mathcal G$ for $Z$ is closed by proving that the stalks are isomorphic. A key step in my proof, which would not have worked so easily if $Z$ were arbitrary, was in proving that $(i_*i^{-1}\mathcal G)_p$ is the terminal object $T$ for $p\notin Z$ (obviously, $\mathcal G_p = T$ for $p\notin Z$). I needed to use that $Z$ was closed to pick a neighborhood $V$ of $p$ disjoint from $Z$ (and hence disjoint from $\operatorname{Supp} \mathcal G$) so that $i_*i^{-1}\mathcal G(V) = i^{-1}\mathcal G(\{\}) = T$.

My question is: can we get a counterexample to the isomorphism $i_*i^{-1}\mathcal G \cong \mathcal G$ when $Z$ is open and not closed (or when $Z$ is arbitrary)?

I think that a counterexample, if it exists, would have to involve a non-Hausdorff space, because if $Z$ is open, the only stalks that could be a "problem" would be the ones at the boundary $p\in \partial Z\cap \partial \operatorname{Supp} \mathcal G$. For such points, $(i_*i^{-1} \mathcal G)_p$ is the colimit over a directed system $\{\tilde V_i\} := \{Z \cap V_i\}$ where $V_i$ are neighborhoods around $p$. In a Hausdorff space, $\cap \tilde V_i = (\cap V_i) \cap Z = \{p\} \cap Z = \{\}$ the empty set. So I'm tempted to say that this colimit is the terminal object, for the any sheaf takes the terminal object on the empty set (but that's not true since a sheaf functor does not have to commute with colimits). On the other hand, that's what we expect since $\mathcal G_p = T$, for $p \notin \operatorname{Supp} \mathcal G$.

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  • $\begingroup$ @GeorgesElencwajg thank you for your comment. I have been noticing that for some reason I stopped receiving up-votes for my questions the way that I used to get. I couldn't think that the quality of my questions had deteriorated, but maybe you made the point: I stopped defining all the my variables because for some misled reason I thought their meaning would be obvious. $\endgroup$ – Rodrigo Nov 1 '13 at 21:20
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Let $Z$ be an open set and define $\mathcal G$ to be the direct image under the inclusion map of the constant sheaf $\mathbb Z$ on $Z$. Then $Z=\operatorname{Supp} \mathcal G$ and for some $p \in \partial Z$ we have that the stalk $(i_*i^{-1} \mathcal G)_p$ is the colimit over $\mathbb Z$. This is $\mathbb Z$.

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    $\begingroup$ Dear Rodrigo, your assertion $Z=\operatorname{Supp} \mathcal G$ is not true: in general the support of $\mathcal G$ will be larger than $Z$. For example if $X=\mathbb R$ and $Z=(0,1)$, can you see that $\operatorname{Supp} \mathcal G=\bar Z$ ? $\endgroup$ – Georges Elencwajg Nov 1 '13 at 21:58
  • $\begingroup$ @GeorgesElencwajg Oops, you're right on the spot. Thank you. $\endgroup$ – Rodrigo Nov 1 '13 at 22:53
  • $\begingroup$ I have deleted my answer, since it didn't take into account your hypothesis on the support of $\mathcal G$. Sorry about that. $\endgroup$ – Georges Elencwajg Nov 1 '13 at 23:11
  • $\begingroup$ @GeorgesElencwajg no problem. I actually didn't understand what you were trying to prove, but I'm glad that you corrected my mistake anyway :) $\endgroup$ – Rodrigo Nov 2 '13 at 1:38

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