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Let $f:X\to Y$ be a closed immersion of topological spaces. Let $\mathcal F$ be a sheaf of rings on $X$. Is the canonical morphism of sheaves $\varphi: f^{-1}f_*\mathcal F\to \mathcal F$ an isomorphism?

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  • $\begingroup$ @GeorgesElencwajg Here $f$ is also a closed immersion. $\endgroup$ – Born to be proud Oct 30 '18 at 8:59
  • $\begingroup$ @Born to be proud. You are right, I misread the question. I have deleted my comment and apologize to you and eloiPrime. Thanks for your vigilance. $\endgroup$ – Georges Elencwajg Oct 30 '18 at 9:05
  • $\begingroup$ @GeorgesElencwajg It doesn't matter. $\endgroup$ – Born to be proud Oct 30 '18 at 9:25
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Yes, $\varphi$ is an isomorphism: I showed it here
However if $f$ is not assumed to be a closed immersion (=the inclusion of a closed subspace) , the result is no longer true in general.
Here is a counterexample:

Let $X$ be an arbitrary topological space space, take for $\mathcal F$ the sheaf of continuous functions $\mathcal C$ on $X$ and let $Y=\{y\}$ be a point. Of course $f$ must be the constant map $X\to Y:x\mapsto y$.
Then $f_*\mathcal C= A_Y$, the constant sheaf on $Y$ with (unique!) stalk $A=\mathcal C(X)$.
Hence $f^{-1}f_*\mathcal F=A_X$, the constant sheaf on $X$ with fibre $A$.
Since obviously the sheaf $\mathcal C$ is not constant in general, the sheaf morphism $\varphi: f^{-1}f_*\mathcal C=A_X\to \mathcal C$ cannot be an isomorphism in general.

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$\forall x\in X$, $(f^{-1}f_*\mathcal F)_x\simeq (f_*\mathcal F)_{f(x)}\simeq \mathcal F_x$, so $\varphi$ is an isomorphism. Is it correct?

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    $\begingroup$ Yes it is correct and proved in detail here $\endgroup$ – Georges Elencwajg Oct 30 '18 at 9:29

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