# Relation between $\mathcal{O}_Y\rightarrow f_*\mathcal{O}_{X}$ and $f^{-1}\mathcal{O}_Y\rightarrow \mathcal{O}_X$ being epimorphism/monomorphism.

Let $f:(X,\mathcal{O}_X)\rightarrow (Y,\mathcal{O}_Y)$ be a morphism of ringed spaces. This is the data of a map $f:X\rightarrow Y$ between the topological spaces and either a morphism of sheaves $$f^{\flat}:\mathcal{O}_Y\rightarrow f_*\mathcal{O}_{X} \ \ \text{ or } \ \ f^{\sharp}:f^{-1}\mathcal{O}_Y\rightarrow \mathcal{O}_X$$

Giving one morphism or the other is the same thanks to the adjuntion $$\text{Hom}_X(f^{-1}\mathcal{O}_Y,\mathcal{O}_X)\cong \text{Hom}_Y(\mathcal{O}_Y,f_*\mathcal{O}_X)$$

Is it true that under this bijection, monomorphisms/epimorphisms correspond to monomorphisms/epimorphisms?

I wonder this because we now that for every $x\in X$ there is an induced morphism between stalks $$\mathcal{O}_{Y,f(x)}\rightarrow\mathcal{O}_{X,x}$$ This map coincide with $f^{\sharp}_x:(f^{-1}\mathcal{O}_{X})_x=\mathcal{O}_{Y,f(x)}\rightarrow \mathcal{O}_{X,x}$ but is not in general equal to $f^{\flat}_{f(x)}:\mathcal{O}_{Y,f(x)}\rightarrow (f_*\mathcal{O}_{X})_{f(x)}$ because in general we don't have $(f_*\mathcal{O}_{X})_{f(x)}=\mathcal{O}_{X,x}$.

Now if we want to show that $f$ is an epimorphism, this (sometimes by definition) means that $f^{\flat}$ is an epimorphism. In the other hand if we prove that $f$ is an epimorphism on stalks we are proving that $f^{\sharp}$ is an epimorphism. So if I want to show that $f$ is an epimorphism iff is an epimorphism on stalks I need the above result (same with monomorphism).

Also, I have another minor question related with this. It is true that if the canonical map $(f_*\mathcal{O}_{X})_{f(x)}\rightarrow\mathcal{O}_{X,x}$ is an isomorphism then the above maps between stalks are equal?

No, this is not true.

Let $X = \operatorname{Spec} R$, $Y = \operatorname{Spec}(R \times S)$ and $f:X \to Y$ be the inclusion, which is an open and closed immmersion.

The map $f^{-1}\mathcal O_Y \to \mathcal O_X$ is the identity map $R \to R$, while the map $\mathcal O_Y \to f_* \mathcal O_X$ is the projection $R \times S \to S$.

And what do you mean by "$f$ is an epimorphism on stalks"? First of all, $f$ is a map of topological spaces. It does not make sense to ask how $f$ "behaves on stalks".

In general adjunctions rarely preserve monomorphisms or epimorphism. Obviously, you can just test this on the unit and co-unit of the adjunction.

• When I said the $f$ is an epimorphism on stalks I mean that the morphism between ringed spaces induced an epimorphism on stalks, i.e, that the maps $f^{\sharp}_x$ are epimorphisms. When I denote a morphism of ringed spaces I use the same notation as for the map between topological spaces, perhaps this is not a good thing to do. – yamete kudasai Nov 17 '17 at 10:35
• But how does this relate to your question? Surjectivity and injectivity of morphism of sheaves can be checked on stalks. This is very basic and is not related to your question about the adjunction. – MooS Nov 17 '17 at 10:42
• Yes, I know. So if you prove that $f^{\sharp}_x$ are epimorphism for every $x$ we have that $f^{\sharp}$ is an epimorphism. But for example, if we want to prove that $f$ is a closed inmersion we need that $f^{\flat}$ is an epimorphism, so I wanted to relate this two things. (Hartshorne use $f^{\sharp}$ for what I'm calling $f^{\flat}$, sorry if this is confusing. I'm using the notation of Görtz, Wedhorn) – yamete kudasai Nov 17 '17 at 10:47
• For the open immersion $f:\operatorname{Spec} A_a \hookrightarrow \operatorname{Spec} A$, the map $f^\sharp$ is just the identity $A_a \to A_a$, i.e. surjective, while the map $f^\flat$ is the localization map $A \to A_a$ i.e. not surjective. – MooS Nov 17 '17 at 11:45
• IMPORTANT note: For ring maps, surjective and epimorphism are not the same notions. For a closed immersion, you need the map to be surjective. – MooS Nov 17 '17 at 11:46