Suppose $A\leq A',B$ and $C' \leq C$ are (finite dimensional) vector spaces. Suppose that $$ 0 \to A \to B \to C \to 0 $$

$$ 0 \to A' \to B \to C' \to 0 $$ are exact. Then using a dimension argument it follows $A'/A \cong C/C'$.

However, I was wondering whether there is a canonical map realising this isomorphism. By that, I mean, that after enhancing the diagram with the inclusion maps $A \to A'$, $B\to B$ and $C' \to C$ and supposing that it commutes, is there a diagram chasing argument providing the isomorphism $A'/A \cong C/C'$?

If this is the case, what is essential about vector spaces for this to work? In other words, to what categories does the argument generalize.

If this is a simple question in homological algebra/diagram chasing I would also be content with a reference for this statement.

EDIT: Following Andrews suggestion I redraw the diagram. Consider $$ 0 \ \ \ \to A \ \ \ \stackrel{f}{\to} \ \ \ B \stackrel{g}{\to} C \to 0 $$ $$ \downarrow \subseteq \ \ \ \ \ \updownarrow \cong $$ $$ 0 \ \ \ \to A' \stackrel{f}{\to} \ \ \ B \stackrel{g'}{\to} C' \to 0 $$ So $f:A \to B$ is just the restriction of $f:A' \to B$.

As in Andrews answer there is of course a map $A' \to C \to C/C'$, and $A$ is in the kernel. From that we get a map $A'/A \to C/C'$. But why has this map to be an isomorphism? I think it does not have to be, it could just be the map sending everything to 0, right?

So is there a map which we can read off the diagram, which realizes the isomorphism $A'/A \to C/C'$?