Let $\mathcal{A}$ be an abeilan category. Consider the following exact, commutative diagram in $\mathcal{A}$ $$\require{AMScd}\begin{CD}\ker a@>{\overline\alpha}>> \ker b@>{\overline \beta}>>\ker c\\ @V{k_a}VV @V{k_b}VV @V{k_c}VV\\ A @>{\alpha}>>B@>{\beta}>>C@>>>0\\ @V{a}VV @V{b}VV @V{c}VV\\ A' @>{\alpha'}>>B'@>{\beta'}>>C'@>>>0\\ @VVV\\ 0 \end{CD}$$, where the first row is kernel of $a,b,c$, respectively. I want to show that $\overline \beta$ is an epimorphism.
If we can use element to do the diagram chasing, then the proof is so elementary.
But, we now live in the world of Abelian category. By definition, let $\ker\mathop\rightarrow\limits^{f} X$ be a morphism such that $f\overline \beta=0$. We want to show $f=0$. My main problem is that $f$ is a morphism that leaves from the diagram at the location $\ker c$. So, I don't know how to use the comutativity of the diagram to argue the epiness of $\overline\beta$.
Does there any command, idea or detailed proof of the proposition on some references? Thanks in advance for your kindly help.
$\newcommand{\coker}{\operatorname{coker}}$We can use the generalised diagram chase (see MacLane's Categories for the Working Mathematician for a proof of the following rules, or have a go yourself):
Here, a statement like "$x\in_m a$" just means ("$m$" for "member"), there is some object $X$ and some arrow $x:X\to a$. If $x,y\in_m a$ are equivalent ($x\equiv y$) then that means to say: "there are epimorphisms $u,v$ with $xu=yv$". Of course, $\equiv$ is an true equivalence relation on "members" of any object.
It will be possibly to prove this directly, but the rules of the generalised diagram chase make it a hellavalot easier!
Ok, so let $z\in_m\ker c$ be arbitrary. I know that $k_cz\in_m C$, and I know by exactness of the middle row that $\beta$ is epic - hence, by rule $\rm(iii)$, there is some $y'\in_m B$ with $\beta y'\equiv k_c z$. Now, $c(k_c z)\equiv 0$ of course and $c(k_c z)\equiv c\beta y'=\beta'by'$ so that by rule $\rm(v)$ and exactness of the last row, there is $x\in_m A'$ with $\alpha'x\equiv by'$.
By rule $\rm(iii)$ and exactness of the leftmost column I know $a$ is epic and that there is $x'\in_m A$ with $ax'\equiv x$, that is, there is $x'\in_m A$ with $b\alpha x'=\alpha'ax'\equiv by'$
By rule $\rm(vi)$ there is $z'\in_m B$ with $bz'\equiv 0$ and the property $f\alpha x'\equiv0\implies fy'\equiv fz'$ for any arrow $f$ with domain $B$.
By rule $\rm(v)$ (or by very definition of "kernel") there is $y\in_m\ker b$ with $k_by\equiv z'$.
Then, $k_c\overline{\beta}y=\beta k_by\equiv\beta z'$. But $\beta\alpha x'=(\beta\alpha)x'\equiv0$ and so $\beta y'\equiv\beta z'$ by definition of $z'$. But, $\beta y'\equiv k_c z$! Therefore $k_c\overline{\beta}y\equiv\beta z'\equiv\beta y'\equiv k_c z$ and by rule $\rm(ii)$, $k_c$ being monic as all kernel arrows are, we have $\overline{\beta}y\equiv z$.
Since $z\in_m\ker c$ was arbitrary and such $y\in_m\ker b$ always exists with $\overline{\beta}y\equiv z$, by rule $\rm(iii)$ we have that $\overline{\beta}$ is an epimorphism.