If one vertical arrow in a pullback is an iso, then so is the other Consider the pullback square:
$\require{AMScd}$
\begin{CD}
    A @>f>> B\\
    @Vg VV @VV h V\\
    C @>>j > D
\end{CD}
Suppose $h$ is an isomorphism. I'm trying to show that $g$ is an isomorphism. I considered arrows $1_C,h^{-1}\circ j:C\to C$ and defined $\alpha:C\to A$ as the unique map that makes the appropriate triangles commute. In particular that gave $g\circ \alpha=1_C$. But how to prove that $\alpha\circ g = 1_A$? The best I got is $f\circ \alpha\circ g = f$. I don't know how to get rid of $f$.
 A: You can show that $\alpha \circ g$ is equal to $1_A$ using the uniqueness condition for maps into a pullback.
$f:A \rightarrow B$ and $g: A \rightarrow C$ are maps such that $h \circ f = j \circ g$ so there is a unique $x:A \rightarrow A$ such that $f \circ x = f$ and $g \circ x = g$. Obviously, $1_A$ is one such $x$.
We need to show that $(\alpha \circ g)$ is another such map. You've proved that $f \circ (\alpha \circ g) = f$. You can also show that $g \circ (\alpha \circ g) = g$. So $(\alpha \circ g)$ satisfies the condition.
Since there is a unique map $x$ satisfying $f \circ x = f$ and $g \circ x = g$, and both $1_A$ and $(\alpha \circ g)$ satisfy this, they must be equal. 
A: HINT. Show that $A=C$ with the proper map; that is, you have 
$\require{AMScd}$
\begin{CD}
    C @>h^{-1} \circ j>> B\\
    @V1_C VV @VV h V\\
    C @>>j > D
\end{CD}
This amounts to showing that this diagram has all the desired properties. Note that this shows that isomorphisms always have a pullback (in any category where the pullback exists). 
