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I am trying to understand Tennison / Hitchin's Sheaf Theory proof of the following result.

Let $f:X\to Y$ be a morphism of ringed spaces such that the underlying map of topological spaces is a monomorphism and the morphism $f^\sharp:\mathscr{O}_Y\to f_\ast \mathscr{O}_X$ is an epimorphism. Then $f$ is a monomorphism in the category of ringed spaces.

The proof begins like this: (this is not a citation, it's just what I understood)

Let $(g_1,g_1^\sharp),(g_2,g_2^\sharp):(Z,\mathscr{O}_Z)\to (X,\mathscr{O}_X)$ be morphisms of ringed spaces such that the diagram

commutes. Passing to the diagram of underlying continuous maps, we get that $g_1=g_2$. Henceforth, we'll denote both $g_1$ and $g_2$ by $g$. By the adjunction between direct and inverse images, the diagram of sheaves over $Z$

is also commutative. Finally, passing to the stalks we obtain the commutative diagram

The authors then affirm that, by hypothesis $f^\flat_{g(p)}$ is an epimorphism. I can't see why that is true. The answer Relation between $\mathcal{O}_Y\rightarrow f_*\mathcal{O}_{X}$ and $f^{-1}\mathcal{O}_Y\rightarrow \mathcal{O}_X$ being epimorphism/monomorphism. even shows that this exact adjunction does not preserves epimorphisms. What am I missing?

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  • $\begingroup$ The stalk functor at the point $x$, $Sh(X) \rightarrow Ring$ is a left adjoint to the skyscraper sheaf at $x$ functor. Left adjoints preserve epis and thus we're done! $\endgroup$ Nov 3 '20 at 20:15
  • $\begingroup$ @NoelLundström Perhaps I don't understand how you want to use that but I think that this is not the problem here. Surely if $f^\flat:f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$ is an epimorphism, then so is its stalks. My question is about why $f^\flat$ is an epimorphism if $f^\sharp:\mathscr{O}_Y\to f_\ast\mathscr{O}_X$ is. $\endgroup$
    – Gabriel
    Nov 3 '20 at 20:52
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    $\begingroup$ I think it would be better to phrase the condition in terms of $f^{-1} O_Y \to O_X$, but once you know that $f : X \to Y$ is an injective continuous map then you can deduce it. Indeed, if $f : X \to Y$ is injective, then $f_*$ is fully faithful, so the counit $f^{-1} f_* A \to A$ is an isomorphism for every sheaf $A$ on $X$, so if $O_Y \to f_* O_X$ is an epimorphism then so is $f^{-1} O_Y \to O_X$. $\endgroup$
    – Zhen Lin
    Nov 3 '20 at 22:39
  • $\begingroup$ @ZhenLin This indeed works. If you would like, you can write this as an answer and I'll gladly accept it. $\endgroup$
    – Gabriel
    Nov 4 '20 at 19:40
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If $f : X \to Y$ is an injective continuous map, then $f_* : \textbf{Sh} (X) \to \textbf{Sh} (Y)$ is fully faithful, so the adjunction counit $f^{-1} f_* A \to A$ is an isomorphism for every sheaf $A$ on $X$. Hence, if $f : X \to Y$ is an injective continuous map and $f^\sharp : O_Y \to f_* O_X$ is an epimorphism, then $f^\flat : f^{-1} O_Y \to O_X$ is also an epimorphism.

That said, in my view it would have been better to state the condition in terms of $f^\flat$ in the first place, since $f^\sharp$ may not be an epimorphism even if $f^\flat$ is.

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