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I have a naive question on algebraic geometry.

To fix a context we consider $\phi:X\to Y$ a morphism between two affine varieties over an algebraic closed field $k$.

This give under the anti-equivalence of categories a k-algebra morphism $\phi^*$ between coordinate algebras of $Y$ and $X$.

However, $\phi^*$ injective doesn't imply $\phi$ surjective. Think for example to the projection of the hyperbola $XY=1$ to the $X$-axis.

Shouldn't the anti-equivalence of categories force a correspondence between injective and surjective morphisms? It must be that surjective morphism of affine varieties are not the epimorphisms in the categorical sense, but I don't understand why?

I know that for finite morphisms for instance, we have this correspondence. So I am wondering what are the epimorphisms and monomorphisms in the category of affine varieties?

Thanks in advance!

Regards, Moustik

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    $\begingroup$ Monomorphisms in the category of rings give rise to epimorphisms of affine varieties, namely dominant maps. This is not surprising since it is the same in topology: morphisms of varieties are continuous and such maps are uniquely determined on a dense subset. $\endgroup$
    – MooS
    Feb 13, 2017 at 21:22
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    $\begingroup$ Thank @Moos.I think I just figured one thing: if $\phi ^*$ is injective, it is a monomorphism. Then $\phi$ is of course an epimorphism. But epimorphisms are not necessarily surjective... That's just what I was missing... $\endgroup$
    – Moustik
    Feb 13, 2017 at 21:40
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    $\begingroup$ Anyway the question stands. Is there a description of epimorphisms and monomorphisms in the category of affine varieties? $\endgroup$
    – Moustik
    Feb 13, 2017 at 21:46
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    $\begingroup$ For monomorphisms of affine schemes/epimorphisms of commutative rings, see this MathOverflow question. $\endgroup$ Feb 14, 2017 at 7:25
  • $\begingroup$ @TakumiMurayama I think that link qualifies as answer and should be posted as such. $\endgroup$ May 14, 2018 at 17:29

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I am posting my comment as an answer as requested, with some more concrete references.

We start with monomorphisms:

Proposition [Borceux, Ex. 1.7.7.d]. A homomorphism of unital commutative rings is a monomorphism if and only if it is injective.

Epimorphisms are harder to describe; see this MathOverflow question, and especially David Rydh's answer, which states a characterization of monomorphisms among morphisms of schemes which are locally of finite type.

We state a general characterization of epimorphisms of rings:

Theorem [Roby, Thm. 1]. Let $f \colon R \to S$ be a homomorphism of unital commutative rings. The following are equivalent:

  1. $f$ is an epimorphism;
  2. For every $R$-algebra $T$, there exists at most one $R$-algebra homomorphism $S \to T$;
  3. For all $s \in S$, the relation $1 \otimes s = s \otimes 1$ holds in $S \otimes_R S$;
  4. The multiplication map $\mu \colon S \otimes_R S \to S$ is injective (in which case it is an isomorphism of $R$-algebras);
  5. The canonical injection $i_1 \colon S \to S \otimes_R S$ (resp. $i_2 \colon S \to S \otimes_R S$) defined by $s \mapsto s \otimes 1$ (resp. $s \mapsto 1 \otimes s$) is surjective (in which case it is an isomorphism of $R$-algebras);
  6. The tensor algebra $T_R(S)$ is commutative;
  7. $S \otimes_R S/R = 0$.

I've skipped some of the conditions; see [Roby] for more. Roby's article is from a seminar directed by Samuel, which is probably the most thorough source on epimorphisms of rings. Finally, see [Lazard] for results on flat epimorphisms, and [Stacks, Tag 04VM] for a modern reference.

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    $\begingroup$ I edited condition 2 from "there exists an..." to "there exists at most one....", translating from the original "il existe au plus." (Note that Roby's algebra homomorphisms carry units to units). Of course existence is not guaranteed, even for nice surjections, e.g., $\mathbb{Z}$ surjects onto both $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ but there's no $\mathbb{Z}$-algebra homomorphism between the latter two. $\endgroup$ Jul 11, 2019 at 22:47
  • $\begingroup$ @BadamBaplan thank you! $\endgroup$ Jul 12, 2019 at 1:20

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