Is this an equivalent formulation of "surjective" resp. "epimorphism"?

Question

An $A$-element $x$ of $X$, written $x\in_A X$, is a map $x:A\to X$. If $f$ is a map with domain $X$, and $x\in_A X$ is an element, we write $f(x)$ to denote the composite of $f$ and $x$.

Now say that a map $f:X\to Y$ is cool if $$\forall A\quad \forall x, x'\in_A X\quad:\quad f(x) = f(x')\implies x = x'.$$

In the category of sets, a map is cool if and only if it is injective! In an arbitrary category, a map is cool if and only it is an monomorphism!

Call a map $f : X\to Y$ fresh if for all $A$, and for all $y\in_A Y$, there is an $x\in_A X$ such that $f(x) = y$. It would now be quite nice if, in the category of sets, a map is fresh if and only if it is surjective. 1. Is this true? 2. Also, in an arbitrary category, is a map fresh if and only if it is an epimorphism?

Answer 1

If $f$ is "fresh", then there is a particularly interesting element you can consider in $Y$ : the $Y$-element $id_Y:Y\to Y$. Now freshness of $f$ tells you that there must be some $s\in_Y X$, thus a map $s:Y\to X$, such that $f(s)=id_Y$, i.e. $f\circ s =id_Y$. In other words, $f$ has a right inverse; it is a split epimorphism.

Conversely, if $f$ is a split epimorphism with right inverse $s$, then for all $y\in_A Y$, $s(y)\in_A X$ is such that $f(s(y))=y$, thus $f$ is fresh. Thus what you call "fresh" is equivalent to being a split epimorphism. Now in the category of sets, being a split epimorphism is equivalent to being surjective (this is one possible way to state the axiom of choice), so the answer to your first question is yes. But in general being a split epimorphism is a stronger property : for example, the quotient map $\Bbb Z\to \Bbb Z/n\Bbb Z$ is an epimorphism but not a split one in the category of groups.

Answer 2

This is saying that the Yoneda embedding, applied to $f$, gives an epimorphic natural transformation, which certainly implies $f$ is an epimorphism, but is much stronger. Indeed, this implies $f$ is a split epi, as we see by taking $y$ to be the identity of $Y$ in your definition.

The difference from the case of monos is that the Yoneda embedding reflects, but does not preserve, epis; this follows from the analogous claim for colimits, whereas Yoneda does preserve limits and thus monos.

In the case that every epi is split, as in Set with the axiom of choice, your condition is indeed equivalent to $f$ being epi; this comes abstractly from the fact that the colimits which Yoneda preserves are precisely the absolute colimits, those preserved by every functor-such as split epis!