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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?

user46336
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    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. –  Nov 04 '12 at 08:39

2 Answers2

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  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 $Y$.

Georges Elencwajg
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  • In the second example, the morphism $\mathbb{P}^1 \rightarrow \{q\}$ is not a closed immersion. – user46336 Nov 04 '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 04 '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 $f$? – Bart Michels Oct 06 '17 at 17:39
  • @barto: yes, you are right. – Georges Elencwajg Oct 06 '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. – Bart Michels Oct 06 '17 at 18:01
  • To be precise: the result is correct whenever $f: X\hookrightarrow Y$ is the inclusion of a **topological subspace** $X$ (with induced topology) into **a topological space** $Y$. I don't know what immersions of topological spaces are and I'm not saying anything about those. Let me emphasize that my answer concerns **topological spaces**, and **not** schemes. – Georges Elencwajg Oct 06 '17 at 18:24
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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).

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    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.) –  Nov 04 '12 at 09:27
  • Ah, sorry. How do we define closed immersions for topological spaces? –  Nov 04 '12 at 09:37
  • 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. –  Nov 04 '12 at 18:15