What is the definition of surjective morphism of schemes? 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?
 A: 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.
