# 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?

• I don't believe the first statement, $f_* f^{-1}$ must have support inside (image in $Y$) of $X$, while your $\mathcal{O}$ can definitely have a larger support to start with. The second canonical morphism should be an isomorphism, and that can be checked at the level of sections. – user27126 Nov 4 '12 at 8:39

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).

• I think he's working with topological space with a structure sheaf rather than a scheme. (In particular, he's using inverse image sheaf rather than $O_X$-modules.) – user27126 Nov 4 '12 at 9:27
• Ah, sorry. How do we define closed immersions for topological spaces? – user314 Nov 4 '12 at 9:37
• I think it means that $X$ maps homeomorphically to a closed subset of $Y$ through $f$, and that $O_Y \to f_* O_X$ is surjective. – user27126 Nov 4 '12 at 18:15
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 $$Y$$.

• In the second example, the morphism $\mathbb{P}^1 \rightarrow \{q\}$ is not a closed immersion. – user46336 Nov 4 '12 at 10:53
• Ah yes, you're right: I had forgotten about this condition of yours. I have modified my answer in order that it take this condition into account. – Georges Elencwajg 1 hour ago – Georges Elencwajg Nov 4 '12 at 12:32
• Since in the second part we do not use that $f(X)$ is closed, am I right that $f^{-1} f_* \mathcal F \rightarrow \mathcal F$ is an isomorphism for any immersion $f$? – punctured dusk Oct 6 '17 at 17:39
• @barto: yes, you are right. – Georges Elencwajg Oct 6 '17 at 17:59
• Thanks. Strange, because I always see this with the assumption that $f$ is a closed or open immersion. I guess there's some more general context where that condition becomes necessary. – punctured dusk Oct 6 '17 at 18:01