# What are the epimorphisms in the category of schemes?

Is there a known characterization of epimorphisms in the category of schemes?

It is easy to see that a morphism $$f : X \to Y$$ such that the underlying map $$\lvert f\rvert$$ is surjective and the homomorphism $$f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$$ is injective, is an epimorphism. But there are other examples, too: if $$Y$$ is reduced and locally of finite type over a field, the obvious morphism from $$X=\bigsqcup_{y\in Y_\text{cl}} \operatorname{Spec}(k(y))$$ to $$Y$$ is an epimorphism (see #8(b) in Mark Haiman's Homework Set 9 for Math 256AB).

If this is is not possible, what about regular, extremal, or effective epimorphisms? Here, again, I know only some examples.

My background is that I want to know if there is a categorical characterization of the spectra of fields in the category of schemes. In the full subcategory of affine schemes, they are characterized by the property: $$X$$ is non-initial and every morphism from a non-initial object to $$X$$ is an epimorphism. But I doubt that this characterization takes over to the category of schemes. EDIT: Kevin Ventullo has shown below that the characterization takes over. Thus my original question has been answered (and I wonder if it's appropriate to accept it as an answer). But of course every other hint about the characterization of epimorphisms of schemes is appreciated.

• One thing to keep in mind, although this is not your original purpose, is that epimorphisms in $Sch$ might in general be 'badly behaved'. In $Aff$ the 'real' notion of epimorphism you need (regular, effective or what-have-you) might coincide with ordinary epis. Feb 24, 2011 at 23:09

$$\DeclareMathOperator\Spec{Spec}$$Actually, your suggested categorical characterization of spectra of fields does work.

Edit: (I had written something incorrect here)

By Martin's comment below, we just have to show that maps from affines into $$\Spec(k)$$ are epis in the full category. But if we had two maps $$f,g: \Spec(k) \rightarrow Y$$ which agreed on some affine mapping into $$\Spec(k)$$, then first of all $$f$$ and $$g$$ would have to be the same topological map. Then both would land inside some affine $$\Spec(R)\subset Y$$, and now we're reduced to the affine situation where we know it holds.

Conversely, suppose $$X$$ is not the spectrum of a field. If every point is dense, $$X$$ is affine and we are done by what we know about the affine subcategory. Otherwise, we can find an open subscheme $$U\subsetneq X$$. Then the inclusion of $$U$$ into $$X$$ is not an epi as is witnessed by the two inclusions

$$X \rightrightarrows X\sqcup_U X,$$

where the last object is $$X$$ glued to itself along $$U$$.

• The observation that maps from $Spec(k)$ to schemes are "completely determined" by maps to affine schemes is neat. It also works for $Spec(R)$, where $R$ is a local ring. This answer (mathoverflow.net/questions/63/…) suggests this approach might be able to produce an example of a coequalizer of schemes which is not surjective, but I wasn't ever able to get it to work. Feb 25, 2011 at 2:40
• I don't think that the use of $Spec(O_X(X))$ is correct here. This would only work if $X \to Spec(O_X(X))$ is an epi. Rather, we compose $X \to Spec(k)$ with the inclusion of a nonempty open affine part of $X$ and it suffices to prove that this composition is epi. Feb 25, 2011 at 7:34
• ... but I didn't manage to prove the other direction, which is now easier than I thought. Thank you! @Anton: I also tried to produce an example some time ago, but it is quite hard. It resulted in mathoverflow.net/questions/24066/…. Feb 25, 2011 at 8:09
• @Martin: Good call; affinification need not be an epi. My mistake. Feb 25, 2011 at 9:35