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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:

$\newcommand{\la}[1]{\kern-1.5ex\xleftarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} D & \la{f_1} & 0 \\ \da{g} & & \da{h}\\ 0 & \la{f_2} & 0 \\ \end{array}$

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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    Worldly imperatives. This is a mathematical question, I don't think any academic authority should be in the position to regulate the internets content.2012-12-12

1 Answers 1

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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 $g = 0$.