# Does the direct image functor on sheaves reflect epimorphisms?

Let $f:X\to Y$ be a morphism of schemes, and let $f_*:\mathbf{Sh}(X)\to\mathbf{Sh}(Y)$ be the direct image functor. Does $f_*$ reflect epimorphisms? That is, suppose $\alpha:\mathscr{F}\to\mathscr{G}$ is a map of sheaves on $X$ such that $f_*\alpha:f_*\mathscr{F}\to f_*\mathscr{G}$ is a surjective map of sheaves on $Y$. Does it follow that $\alpha$ is surjective?

I believe the result holds (at least) when $X$ and $Y$ are affine schemes, and $f$ is induced by a surjective ring homomorphism, but I'm even a bit shaky on this case. I would appreciate verification of this fact even if the general case is not true.

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The only general condition I know for a functor to reflect epimorphisms is that it be faithful, which needn't be the case here (e.g. if $Y$ is a point, so $f_{\ast}$ is global sections). Is it true in this special case? – Qiaochu Yuan Apr 10 '13 at 1:06
Its not true for affine schemes etc., perhaps you only want to consider qc sheaves? – Martin Brandenburg Apr 10 '13 at 1:23
Thanks Martin. What about in the case when $f:\text{Spec }A\to\text{Spec }B$ is induced by a surjective ring map $B\to A$? Do you know of a counter-example? – Jared Apr 10 '13 at 1:27

Here is my answer in the case that $X=\text{Spec }A$, $Y=\text{Spec }B$, and $f:X\rightarrow Y$ is induced by a surjective ring map $\varphi:B\to A$. (Can anybody verify whether or not my reasoning is correct?)

To see that $\alpha:\mathscr{F}\to\mathscr{G}$ is a surjective map of sheaves on $X$, let's consider the induced map on stalks at a point $\mathfrak{p}\in X$. This is the map $$\begin{array}{rcl}\alpha_{\mathfrak{p}}:\mathscr{F}_{\mathfrak{p}}&\longrightarrow&\mathscr{G}_{\mathfrak{p}}\\\langle s,U\rangle&\longmapsto&\langle\alpha_U(s),U\rangle\end{array}$$

This map fits into a commutative diagram as follows:

$$\begin{array}{ccc}(f_*\mathscr{F})_{f(\mathfrak{p})}&\xrightarrow{f_*\alpha_{f(\mathfrak{p})}}&(f_*\mathscr{G})_{f(\mathfrak{p})}\\\downarrow&&\downarrow\\\mathscr{F}_{\mathfrak{p}}&\xrightarrow{\alpha_{\mathfrak{p}}}&\mathscr{G}_{\mathfrak{p}}\\\end{array}$$

The left vertical map in this diagram is just inclusion as we're taking the limit over fewer open sets: $$\begin{array}{rcl}\varinjlim_{p\in f^{-1}(V)}\mathscr{F}(f^{-1}(V))&\longrightarrow&\varinjlim_{p\in U}\mathscr{F}(U)\\\langle s,f^{-1}(V)\rangle&\longmapsto&\langle s,f^{-1}(V)\rangle\end{array}$$

And similarly for the right vertical map. It is enough to show that the vertical maps are surjective. This follows from the surjectivity of $\varphi$ as follows. Given any basic open set $U=D(a)\subset X$, find some $b\in B$ such that $\varphi(b)=a$. Then $f^{-1}(D(b))=D(a)$ so that the limit on the left can be taken over basic open sets, which recovers the entire stalk.

Does this make sense? I can't seem to find any references for this result, which is why I'd like some confirmation because it seems basic enough.

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