Sheafs and closed immersion

Question

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?

Answer 1

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).

Answer 2

1. 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.

2. 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 $Y$.