Lemma 6.32.1. Let $X$ be a topological space. Let $i : Z \to X$ be the inclusion of a closed subset $Z$ into $X$. Let $\mathcal{F}$ be a sheaf of sets on $Z$. The stalks of $i_*\mathcal{F}$ are described as follows

$i_*\mathcal{F}_ x = \left\{ \begin{matrix} \{ *\} & \text{if} & x \not\in Z \\ \mathcal{F}_ x & \text{if} & x \in Z \end{matrix} \right.$

where $\{ *\}$ denotes a singleton set. Moreover, $i^{-1}i_* = \text{id}$ on the category of sheaves of sets on $Z$. Moreover, the same holds for abelian sheaves on $Z$, resp. sheaves of algebraic structures on $Z$ where $\{ *\}$ has to be replaced by $0$, resp. a final object of the category of algebraic structures.

Proof. If $x \not\in Z$, then there exist arbitrarily small open neighbourhoods $U$ of $x$ which do not meet $Z$. Because $\mathcal{F}$ is a sheaf we have $\mathcal{F}(i^{-1}(U)) = \{ *\}$ for any such $U$, see Remark 6.7.2. This proves the first case. The second case comes from the fact that for $z \in Z$ any open neighbourhood of $z$ is of the form $Z \cap U$ for some open $U$ of $X$. For the statement that $i^{-1}i_* = \text{id}$ consider the canonical map $i^{-1}i_*\mathcal{F} \to \mathcal{F}$. This is an isomorphism on stalks (see above) and hence an isomorphism.

For sheaves of abelian groups, and sheaves of algebraic structures you argue in the same manner. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).