Sheafs and closed immersion Let $f:X \rightarrow Y$ be a continuous map of topological spaces, such that it is closed immersion. Let $\mathfrak{F}$ and $\mathfrak{G}$ be sheafs on $X$ and $Y$ respectively.
How to show, that canonical morphisms
$$
\mathfrak{G} \rightarrow f_* f^{-1} \mathfrak{G}, \; f^{-1} f_* \mathfrak{F} \rightarrow \mathfrak{F}
$$
are isomorphisms?
Is it true, that taking stalk doesn't commute with direct image, but commutes with inverse image?
 A: *

*Let $Y$ be any topological space, and $ f:X=\lbrace p \rbrace \hookrightarrow Y$, the inclusion of a closed point $p\in X$.
If $\mathcal G$ is a  sheaf on $Y$, the sheaf $f^{-1}\mathcal G$ is the sheaf on the one-point space $X$ with stalk $\mathcal G_p$ and the sheaf $f_*f^{-1}\mathcal G$ restricted to $ Y\setminus \lbrace p \rbrace $ is zero while $\mathcal G$ will not be zero on $ Y\setminus \lbrace p \rbrace $ in general, so that $$\mathcal G \rightarrow f_* f^{-1} \mathcal G \; $$ will certainly not be an isomorphism.  

*Yes, it is true that if $X\subset Y$ is the inclusion of a subspace (closed or not), then for any sheaf $\mathcal F$ on $X$ the canonical map $$ f^{-1} f_* \mathcal F \rightarrow \mathcal F   $$ is an isomorphism of sheaves.
It is enough to check that for each $x\in X$ the map of stalks 
$$ (f^{-1} f_* \mathcal F )_x\rightarrow \mathcal F_x   $$ is bijective.
Since for  any sheaf $\mathcal G$ on $Y$, we have for all $x\in X$ the equality of stalks $$(f^{-1}\mathcal G)_x=\mathcal G_{f(x)}$$ we have to check that the canonical morphism  $(f_*\mathcal F)_{f(x)}\to \mathcal F_x $ is bijective .
But this is immediate from the definition $\Gamma(U,f_*\mathcal F)=\Gamma(U\cap X,\mathcal F)$ (for $U$ open in $Y$) if you remember that all open neighbourhoods of $x$ in $X$  are of the form $U\cap X$ for some ope neighbourhood $U$ of $x$ in $Y$.
A: That $f^\ast(f_\ast(\mathcal{F})) \to \mathcal{F}$ is an isomorphism follows from the fact that for a ring $R$, ideal $I$ and $(R/I)$-module $M$, the canonical homomorphism $M_R \otimes_{R} R/I \to M$ is an isomorphism.  The other one is similar.  See (Stacks project, 24.4.1).
