# Proving that a morphism of schemes is monic if the underlying continuous map is injective and the map on sheaves is epic.

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?

• 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! Nov 3 '20 at 20:15
• @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. Nov 3 '20 at 20:52
• 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 $f^{-1} O_Y \to O_X$. Nov 3 '20 at 22:39
• @ZhenLin This indeed works. If you would like, you can write this as an answer and I'll gladly accept it. Nov 4 '20 at 19:40

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.