Morphisms in the Category of Epimorphisms

Consider the category of epimorphisms $\mathcal E$ in a given abelian category, where epimorphisms are objects, and morphisms of this category are pairs of arrows which make its objects commute. That is, let $f_1,f_2\in Ob(\mathcal E)$ be epimorphisms and $\alpha:(g,h):f_1\to f_2$ in $Hom(\mathcal E)$ be such a morphism which makes the following commute:


Is it possible to deduce when $\alpha$ is a epimorphism (or monomorphism)? For example, since $f_1,f_2$ are both epimorphisms, can one show that $\alpha$ is an epimorphism $\iff$ $h$ is an epimorphism?

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This is a question on the Math 253 final at UC Berkeley. I don't think this is appropriate. – Qiaochu Yuan Dec 9 '12 at 6:35
Dear @QiaochuYuan: Can this question be reopened once your final is over? – Rankeya Dec 9 '12 at 7:00
Worldly imperatives. This is a mathematical question, I don't think any academic authority should be in the position to regulate the internets content. – NikolajK Dec 12 '12 at 22:24

Here is the answer to your question (not for you but for anyone else who might be interested). It is true that $\alpha$ is an epimorphism iff $h$ is an epimorphism and also that $\alpha$ is a monomorphism iff $h$ is a monomorphism. There are four implications here, and three of them hold in an arbitrary category.
If $C$ is any category, there is a forgetful functor $U : \text{Epi}(C) \to C$ sending $\alpha$ to $h$. $U$ is faithful, so it reflects epimorphisms and monomorphisms; that's two of the implications. $U$ also has a left adjoint sending an object $c \in C$ to the identity $\text{id}_c : c \to c$, from which it follows that $U$ preserves limits, hence pullbacks, hence monomorphisms; that's the third implication.
The fourth implication is that $U$ preserves epimorphisms. I could not see how to prove this for an arbitrary category (possibly it is false), but assuming now that $C$ is $\text{Ab}$-enriched and has pushouts (which holds in particular if $C$ is abelian), it suffices to show that if $\alpha$ is an epimorphism then $h$ has the property that if $g \circ h = 0$ then $g = 0$. To prove this, construct the pushout of $\alpha$ along $g$. Since pushouts preserves epimorphisms, the result is a morphism in $\text{Epi}(C)$, and you can show that its composition with $\alpha$ is $0$. Since $\alpha$ is an epimorphism, it follows that $g = 0$.