# Epic implies Surjective

I'm starting to work through Algebra: Chapter 0, and I'm fumbling a bit. One of the problems in the first chapter asks me to formulate a definition for epimorphism based on having seen the definition for monomorphism, and then to tie it to surjectivity as monomorphism is related to injectivity. This is still early in the book, so I'm only hanging out in the world of sets and functions.

Proposed definition: A function $f:A\rightarrow B$ is epic if for all $Z$ and all $\alpha, \alpha ':B\rightarrow Z$, $\alpha \circ f=\alpha' \circ f\implies\alpha=\alpha'$.

With this definition in mind I want the statement that $f$ is epic iff $f$ is surjective. Showing surjectivity implies epic wasn't hard so long as I know surjective means $f$ has a right inverse. But I'm getting stuck with the other direction. I know I'm probably supposed to find some special $\alpha$'s to help my cause but I'm coming up short...

-
As an aside, keep in mind that in general concrete categories, epic doesn't always imply surjective. For example, in the category $\mathbf{Rng}$ of rings and ring homomorphisms, $\mathbb{Z} \to \mathbb{Q}$ is epic. Monic can fail to imply injective as well, although this comes up less often. – Hurkyl Dec 25 '12 at 21:33

Let $Z = \{ 0,1 \}$; then let $\alpha = \chi_{f(A)}$ be the characteristic function of the image $f(A)$ of $f$ and let $\alpha'$ be the constant function with value $1$. If $f$ is epic, what does this tell you?
Well then $\alpha = 1$ whenever its argument is in the image of $f$, so the composites are equal. So then $\alpha=\alpha'$ since $f$ is epic. So since $\alpha$ is defined on all of $B$ and equals 1, then $B=f(A)$. Very clever, thank ya! – AsinglePANCAKE Dec 25 '12 at 20:48