Are isomorphisms stable under pullbacks? In a category which has pullbacks, do isomorphisms necessarily pull back to isomorphisms?
I can certainly prove that retractions pull back to retractions, but I haven't been able to find a proof that isomorphisms also do.
If not, is there a simple counterexample.
 A: Actually, an isomorphism always has a pull-back (no assumption needed on the category) : if $f: X\to Y$ is an isomorphism, and $g:S\to Y$ is any morphism, then 
$$\require{AMScd}
\begin{CD}
S @>{f^{-1}\circ g}>> X\\
@V{Id}VV @V{f}VV \\
S @>{g}>> Y
\end{CD}$$
is clearly a pull-back diagram (the universal property is easy to check).
So by unicity of fiber product, any pull-back of $f$ along $g$ must be an isomorphism.
A: Lemma: Pullbacks of right-invertible morphisms are right-invertible
Proof: Suppose $(P, \alpha:P{\rightarrow}A, \beta:P{\rightarrow}B)$ is a pullback for the diagram $\{f:A{\rightarrow}C, g:B{\rightarrow}C\}$ where $g$ is right-invertible. Then since $1_A:A{\rightarrow}A$, $g_r^{-1}{\circ}f:A{\rightarrow}B$ and $f{\circ}1_A=f=1_C{\circ}f=g{\circ}g_r^{-1}{\circ}f$, by the definition of pullbacks we find that there is a unique $\theta:A{\rightarrow}P$ such that
$1_A=\alpha{\circ}\theta$ and $g_r^{-1}{\circ}f={\beta}{\circ}\theta$. The first equation shows that $\alpha$ is right-invertible.
Theorem: Pullbacks of isomorphisms are isomorphisms.
Proof: Suppose instead that $g$ is an isomorphism. By the definition of pullbacks we know there is a unique $\rho:P{\rightarrow}P$ such that $\alpha{\circ}\rho=\alpha$ and $\beta{\circ}\rho=\beta$. These equations are clearly satisfied by $1_P$ so we get $\rho=1_P$.
Now by the lemma we know that $\alpha$ is right-invertible and that $g^{-1}{\circ}f={\beta}{\circ}\alpha_r^{-1}$, so we get:
$$\alpha{\circ}(\alpha_r^{-1}{\circ}\alpha)=(\alpha{\circ}\alpha_r^{-1}){\circ}\alpha=1_A{\circ}\alpha=\alpha$$ and
$$\beta{\circ}(\alpha_r^{-1}{\circ}\alpha)=(\beta{\circ}\alpha_r^{-1}){\circ}\alpha=g^{-1}{\circ}(f{\circ}\alpha)=(g^{-1}{\circ}g){\circ}\beta=1_B{\circ}\beta=\beta$$ and so we get that $\rho=\alpha_r^{-1}{\circ}\alpha$, and since $\rho$ is unique $\alpha_r^{-1}{\circ}\alpha=1_P$. This means $\alpha$ is left-invertible and we know it's right-invertible as well, so $\alpha$ is an isomorphism.
