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Let $f: X \to Y$ be a morphism of schemes, when talking about the surjectivity of $f$, there are at least several possibilities.

(1) $f$ is surjective at the level of sets, that is $\forall \ y \in Y$, there exist $x \in X$, such that $f(x)=y$.

(2) $f$ is surjective in the sense of category morphisms. This means for any scheme $Z$, and morphisms $g_1, g_2 : Y \to Z$ such that $g_1 \circ f=g_2 \circ f$ implies $g_1 =g_2$.

(3)$f$ is surjective at the level of schemes, i.e. $\overline{f(X)} = Y$. This seems unlikely to be correct definition of surjectivity, and I do not know how to make sense of the scheme structure of the closure of $f(X)$ (that is $\overline{f(X)}$). But if $Y$ is reduced, then there is no ambiguity.

My question is two folds: (1) Which is "correct" definition of surjective morphism between schemes. (2) Since both definition 1 and 2 seems reasonable in some sense, are they equivalent? If $Y$ is reduced scheme, are three definitions equivalent?

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    $\begingroup$ Just use whatever concept is appropriate for whatever problem you need to address. $\endgroup$ May 23, 2013 at 5:03
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    $\begingroup$ The third one is the definition of $f$ being dominant. The second is usually called being an epimorphism (in the category of schemes). The first is what people usually mean when they say surjective, see e.g. the Stacks project. $\endgroup$
    – user314
    May 23, 2013 at 6:49
  • $\begingroup$ @Adeel Great! This justify my guess from reading the literature. But is there any possibility that the first and the second definition are the same? $\endgroup$
    – Li Yutong
    May 23, 2013 at 7:41
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    $\begingroup$ @Li Yutong: certainly not! $\endgroup$
    – user314
    May 23, 2013 at 7:51
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    $\begingroup$ You may be interested in mathoverflow.net/questions/56564/… $\endgroup$ May 23, 2013 at 14:41

1 Answer 1

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1) The correct definition of "surjective" for a morphism of schemes $f:X\to Y $ is that the underlying map of sets $|f|:|X|\to |Y|$ be surjective.
"Correct" means "as decreed by Grothendieck" : Cf. EGAI, Chap. I, Prop 3.5.2 .

2) The categorical notion mentioned in your point (2) is called epimorphism.

3) If $\overline{f(X)} = Y$, we say that $f$ is dominant.
This is much weaker than surjectivity (even for reduced schemes), as witnessed by the inclusion of any dense open strict subset of a scheme, say $\mathbb A^1\setminus \{0\}\hookrightarrow \mathbb A^1$.

4) Surjective morphisms needn't be epimorphisms:

Let $k$ be a field and let $X=Spec(k)$, $Y=Spec(k[T]/(T^2))=k[\epsilon]$ be respectively the simple point and the double point over $k$.
Let $f:X\hookrightarrow Y$ be the closed immersion of the simple point into the double point corresponding to the $k$-algebra morphism $k[\epsilon]\to k$ sending $\epsilon$ to $0$.
The two $k$-algebra morphisms $k[\epsilon]\to k[\epsilon]$ sending $\epsilon$ to respectively $0$ and $\epsilon $ give rise to two morphisms $g_1, g_2:Y\to Y$ satisfying $g_1 \circ f=g_2 \circ f$ but $g_1 \neq g_2$.
Hence the scheme morphism $f:X\to Y$ is not an epimorphism although it is surjective, as are all maps between singleton sets.

5) Epimorphisms needn't be surjective (?):

Let $Y=\mathbb A^1_k$ (with $k$ an algebraically closed field) and $X$ be the discrete scheme obtained from the disjoint union (=coproduct) of all the closed points of $Y$.
The natural morphism $f:X\to Y$ has image $Y$ minus the generic point of $\mathbb A^1_k$ and is thus not surjective.
However it should be an epimorphism, but I haven't written out a proof of that.

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  • $\begingroup$ @Martin: why not? $\endgroup$ May 23, 2013 at 13:15
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    $\begingroup$ @Martin: I'm baffled by your comment. Not only do you not give any proof nor explanation nor hint for your statement, but in a comment to the OP you refer to a question of yours on MathOverflow where you quote an exercise sheet whose exercise 8 confirms that my example in 5) is an epimorphism! $\endgroup$ May 23, 2013 at 18:07
  • $\begingroup$ @Martin: Dear Martin, Just thinking it through, it seems that (5) is epi. What am I missing? Cheers, $\endgroup$
    – Matt E
    Feb 10, 2014 at 12:28
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    $\begingroup$ Dear Georges, You're welcome! Here was my reasoning (probably the same as yours): Given maps $f, g: X = \mathbb A^1_k \to S$, choose any open affine Spec $A$ of $S$, and pull-back to $X$. Since the pull-backs to $Y$ coincides, these pull-backs contain the same closed points, and so are the same, say equal to Spec $k[x, p(x)^{-1}].$ Thus we are reduced to the considering the situation of $$A \to k[x,p(x)^{-1}] \to \prod_{a \in k, p(a) \neq 0} k,$$ and we use that the second map is an injection. Best wishes, $\endgroup$
    – Matt E
    Feb 10, 2014 at 12:36
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    $\begingroup$ Sorry, I was wrong. $\endgroup$ Feb 11, 2014 at 2:33

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