$\newcommand{\cok}{\operatorname{cok}}\newcommand{\ker}{\operatorname{ker}}$This is an exercise from Freyd's introduction to Abelian categories here on page $59$ (w.r.t the manuscript). Everything will be conducted in some Abelian category.
We are given:
$B_{12}\overset{f}{\to}B_{22}, B_{21}\overset{g}{\to}B_{22}$ are given monics and: "their union is $B_{22}$." Show that: $$\frac{B_{12}}{B_{12}\cap B_{21}}\cong\frac{B_{12}\cup B_{21}}{B_{21}}$$
As I understand it, the union will be the pushout of the diagram $B_{11}\overset{i}{\hookrightarrow}B_{12}$, $B_{11}\overset{j}{\hookrightarrow}B_{21}$ where I have set $B_{11}:=B_{12}\cap B_{21}$ the intersection of the subobjects $f,g$. When he says the union "is" $B_{22}$, without specifying arrows, I assume then that the pushout involves $g,f$ as well. I will copy an illustration of Freyd's for visual clarity – we want to show the following commutative diagram to be exact:
Freyd left the diagram blank, but I have labelled all arrows. The diagram is self-explanatorily commutative and exact save for $\alpha,\beta$. I define $\alpha$ to be the unique arrow such that $\cok(f)\circ g=\alpha\circ\cok(j)$, which exists as $\cok(f)\circ g\circ j=\cok(f)\circ f\circ i=0$, and $\beta$ is similarly defined. So the exactness of this diagram is equivalent to the statement that $\alpha,\beta$ are isomorphisms.
I was able to show that $\beta$ is monic, and symmetrically thus also $\alpha$ is monic, via a diagram chase (in Mac Lane's member notation, using the rules described in CWM):
Let $x\in_m\frac{B_{12}}{B_{11}}$ such that $\beta x\equiv0$. $\cok(i)$ is epic, so $\exists_m z_1\in B_{12}$ such that $\cok(i)z_1\equiv x$. Then: $$\cok(g)fz_1\equiv\gamma\cok(i)z_1\equiv\gamma x\equiv0$$Thus, as $g=\ker(\cok(g))$, there exists $z_2\in_m B_{21}$ such that $gz_2\equiv fz_1$, i.e. $gz_2e_2=fz_1e_1$ for suitable epics $e_{1,2}$ that have the same source. By pullback, there then exists a unique $\sigma\in_m B_{11}$ such that $i\sigma\equiv z_1e_1\equiv z_1$ and $j\sigma\equiv z_2e_2\equiv z_2$. Then: $$x\equiv\cok(i)z_1\equiv\cok(i)i\sigma\equiv0$$Which proves that $\beta$ is monic.
But to show that $\beta$ is epic (or that $\alpha$ is epic) I don't think I can just say: "it follows by duality", since the dual diagram has $\beta,\alpha$ in the wrong positioning. I need to directly show $\beta$ is epic, i.e. that, for all $y\in_m\frac{B_{22}}{B_{21}}$ there exists $x\in_m\frac{B_{12}}{B_{11}}$ with $\beta x\equiv y$. Unfortunately the nature of the diagram is such that there is absolutely nothing I can do with this element $y$ – nothing to precompose or postcompose it with, no universal property to exploit – other than use the fact that $\cok(g)$ is epic thus $\exists z\in_m B_{22}$ with $\cok(g)z\equiv y$. My element $z$, in the middle of the diagram, is now stuck as far as I can see. There is nothing useful to compose it with, I don't think, and I can't see any tricks here.
Moreover I need to use the fact that the top left square of my diagram is a pushout, somehow. The diagram doesn't present me with any way to leverage that. I'd really appreciate some hints on this one (or a full answer, if you want to). In familiar set-like categories it is quite clear that $\beta$ surjects but I am lost in the abstract case, here.
Best Answer
where both squares are pushouts by definition of sum and quotient. So the outer rectangle is a pushout too and we conclude.