# Is the canonical morphism of sheaves $f^{-1}f_*\mathcal F\to \mathcal F$ an isomorphism? [duplicate]

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?

• @GeorgesElencwajg Here $f$ is also a closed immersion. – Born to be proud Oct 30 '18 at 8:59
• @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. – Georges Elencwajg Oct 30 '18 at 9:05
• @GeorgesElencwajg It doesn't matter. – Born to be proud Oct 30 '18 at 9:25

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