Let $X$ be a topological space, $i:Z\rightarrow X$ the inclusion and $\mathcal{F}$ a sheaf on $Z$.
Assume that $Z$ is locally connected and that $\mathcal{F}$ is a constant sheaf with value $E$ where $E$ is some set. Show that $i_{*}(\mathcal{F})_{x}=E$ for all $x\in\overline{Z}$.
Question
Because $\mathcal{F}$ is a constant sheaf on $Z$, so naturally for all $x\in Z$, $i_{*}(\mathcal{F})_{x}=\mathcal{F}_{x}=E$.
The question is why would this be true even for $x\in\overline{Z}$? Let's say in some cases where $\overline{Z}-Z$ is not empty, and let $x$ be an element in such a set. Then
$\displaystyle i_{*}(\mathcal{F})_{x}=\lim_{\rightarrow}i_{*}(\mathcal{F})(U)$
where the limit is taken over open sets containing $x$. I tried to reason out using the fact that $\mathcal{F}(U)$ may be seen as a set of continuous functions from $U$ to $E$ where $E$ is given a discrete topology, but I was stuck.
I went to find other sources, but they just say "since every open neighbourhood of $x$ has non-empty intersection with $Z-\{x\}$, so the direct limit lies inside $E$"
I would like to know why this should be true. Or rather, we know that the direct limit should be a set, but why should it be $E$ just because every neighbourhood of $x$ has non-empty intersection with $Z-\{x\}$?
Thanks!
The result you are trying to prove is false:
For a counterxample, take $X=\mathbb R$, $Z=\{\frac 1n|n\in \mathbb N^*\}$.
Then, for $0\in \overline Z=Z\cup \{0\}$, the stalk $(i_*(\mathcal F))_0$ is far from being equal to $E$.
Actually $(i_*(\mathcal F))_0$ consists of "tails" of sequences $(z_1,z_2,z_3,\cdots)\in E^\mathbb N$, i.e. classes of sequences where you identify two sequences differing in finitely many spots.
Edit: the awful terminology "constant sheaf".
Given a topological space $X$ and a set $E$ we can construct a presheaf $\mathcal F$on $X$ defined by $\mathcal F(U)=E$ for all open subsets $U\subset X$.
This presheaf is called the constant presheaf on $X$ associated to $E$.
It is not a sheaf, but has (as all presheaves) an associated sheaf $ \hat {\mathcal F} $ defined by demanding that $ \hat {\mathcal F} (U)$ equal the set of locally constant maps $U\to E$.
The disaster is that this sheaf is called the constant sheaf on $X$ associated to $E$ and this terminology almost forces one to make mistakes!
For example, if $X$ is discrete we have $\hat {\mathcal F} (U)=E^U$ which shows that $\hat {\mathcal F}$ is about as little constant as one can imagine!
This explains the counterexample I gave: the sheaf $i_*(\mathcal F)$ satisfies $i_*(\mathcal F)(U)=\mathcal F(U\cap Z)=E^{U\cap Z}$, a denumerably infinite product of copies of $E$, which explains why the stalk $(i_*(\mathcal F))_0$ is the huge set I said it was .