Showing epimorphism without using the Freyd-Mitchell Embedding Theorem

In an Abelian category $\mathscr{C}$ consider a commutative diagram as follows:

$$\require{AMScd}\begin{CD} 0@>>>\ker f@>\theta>>W @>{f}>> Y\\ @. @. @V{\phi}VV @|{id} \\ @. @. A @>{\beta}>> Y. \end{CD}$$ Since $\beta\circ(\phi\circ\theta)=0$, there is a morphism $\gamma\colon \ker f\rightarrow \ker\beta$ that makes a commutative diagram as follows:

$$\require{AMScd}\begin{CD} 0@>>>\ker f@>\theta>>W @>{f}>> Y\\ @. @V\gamma VV @V{\phi}VV @|{id} \\ 0@>>>\ker\beta@>\alpha>>A @>{\beta}>> Y. \end{CD}$$

Now suppose $\phi$ is an epimorphism. Then using the Freyd-Mitchell Embedding Theorem, by taking "elements", it is easy to show that $\gamma$ is also an epimorphism. My question is how can I show this without using the Freyd-Mitchell Embedding Theorem, directly using universal properties of various representable functors such as kernel, cokernel, pull-backs, push-outs, etc.

• What about using MacLane's technique of "elements"? – Pedro Tamaroff May 1 '16 at 18:38
• I want to see a proof avoiding taking "elements". The statement is very simple but I have not succeeded in showing it using universal properties of functors. – Must May 1 '16 at 18:40

Consider the maps $$\ker f\stackrel{\begin{pmatrix}\gamma\\\theta\end{pmatrix}}{\longrightarrow}\ker\beta\oplus W\stackrel{\begin{pmatrix}\alpha&-\phi\end{pmatrix}}{\longrightarrow}A.$$
The first map is the kernel of the second, since if $$\begin{pmatrix}\lambda\\\mu\end{pmatrix}:U\to\ker\beta\oplus W$$ is a map with $\alpha\lambda-\phi\mu=0$ then $f\mu=\beta\phi\mu=\beta\alpha\lambda=0$, so $\mu$ factors uniquely through $\ker f$ as $\mu=\theta\mu'$, and $\alpha\gamma\mu'=\phi\theta\mu'=\phi\mu=\alpha\lambda$, so $\gamma\mu'=\lambda$ by the universal property of $\ker\beta$, and so $\begin{pmatrix}\lambda\\\mu\end{pmatrix}$ factors uniquely through $\ker f$.
The second map $\begin{pmatrix}\alpha&-\phi\end{pmatrix}$ is an epimorphism, since its composition with the inclusion $W\to\ker\beta\oplus W$ is an epimorphism. It follows that the second map is the cokernel of the first, since the natural map $\operatorname{cok}\begin{pmatrix}\alpha&-\phi\end{pmatrix}\to A$ given by the universal property of the cokernel is just the natural map $$\operatorname{cok}\ker\begin{pmatrix}\alpha&-\phi\end{pmatrix}\to\ker\operatorname{cok}\begin{pmatrix}\alpha&-\phi\end{pmatrix},$$ which must be an isomorphism in an abelian category.
Suppose $\delta:\ker\beta\to V$ is a map with $\delta\gamma=0$. Then $\begin{pmatrix}\delta&0\end{pmatrix}:\ker\beta\oplus W\to V$ composes with $\begin{pmatrix}\gamma\\\theta\end{pmatrix}$ to give zero, and so factors through $A$. I.e., we get a unique map $:\xi:A\to V$ with $\xi\alpha=\delta$ and $\xi\phi=0$. But since $\phi$ is an epimorphism, $\xi=0$, and so $\delta=0$. Thus $\gamma$ is an epimorphism.