TITLE: In the category of sets epimorphisms are surjective - Constructive Proof?
QUESTION [20 upvotes]: The statement that surjective maps are epimorphisms in the category of sets can be shown in a constructive way.
What about the inverse?
Is it possible to show that every epimorphism in the category of sets is surjective without reverting to a proof by contradiction / negation?

REPLY [3 votes]: I want to write down a proof that comes naturally, in a way.  This proof assumes that you can form the quotient of a set modulo an equivalence relation, but does not require powersets.  (So it works in any pretopos.)
We have sets $ A $ and $ B $ and a function $ f $ from $ A $ to $ B $.  We know that $ f $ is an epimorphism in the category of sets, and we want to prove that $ f $ is a surjection.
Consider the cokernel pair of $ f $, that is the pushout of $ f $ along itself.  Just as epimorphisms in an abelian category have trivial cokernels, so epimorphisms in more general categories have trivial cokernel pairs.  So this is a natural thing to look at.
$$ \matrix { A & \overset { \textstyle f } \rightarrow & B \\ \llap f \downarrow & & \downarrow \rlap { k _ 1 } \\ B & \underset { \textstyle k _ 0 } \rightarrow & C } $$
The cokernel pair is a set $ C $ defined as a quotient of the disjoint union $ B \uplus B $; if $ y $ is an element of $ B $, then I'll write $ y _ 0 $ and $ y _ 1 $ for elements of $ B \uplus B $.  Then we have an equivalence relation on $ B \uplus B $ according to which $ y _ 0 $ and $ y _ 1 $ are equivalent iff $ y $ belongs to the image of $ f $, and no other equivalences exist besides the reflexive ones.  (More explicitly, $ y _ i $ and $ z _ j $ are equal iff $ y = z $ and $ i = j $ or $ y = z $ and $ y = f ( x ) $ for some $ x $.)  Then $ C $ is the quotient of $ B \uplus B $ under this equivalence relation.  The cokernel pair also comes equipped with two inclusion/quotient maps from $ B $, while I'll call $ k _ 0 $ and $ k _ 1 $; $ k _ i $ maps $ y $ to the equivalence class of $ y _ i $ in $ C $.
Given an element $ x $ in $ A $, $ k _ 0 ( f ( x ) ) = k _ 1 ( f ( x ) ) $ in $ C $ because $ f ( x ) $ is in the image of $ f $.  So since $ f $ is an epimorphism, $ k _ 0 = k _ 1 $.  This means that for every element $ y $ of $ B $, $ k _ 0 ( y ) = k _ 1 ( y ) $, so $ y $ is in the image of $ f $.
So $ f $ is surjective.