Other pullback lemma for effective epimorphisms Consider the following diagram in a given category
$\require{AMScd}$
\begin{CD}
A @>>> B @>>> C\\
@VVV @VVV @VVV\\
A' @>>> B' @>>> C'
\end{CD}
Suppose the outer square and the left square are pullbacks, and suppose $A' \to B'$ is an effective epimorphism. Is then also the right square a pullback? 
I think the above should hold in the stated generality, but I'm not able to prove this directly. 
If it helps, we may further assume that the ambient category is a topos, and I'll already be happy with an argument for the case that all objects are representables, $A \to A'$ also an effective epimorphism, and $C = *$. In this setting I've tried to use the fact that now $B \to B'$ is a local epimorphism, so that a given map $Z \to B'$ locally lifts to a map $Z \to B$. Now we would be done if these local lifts glue together, which is where I'm stuck.
 A: It doesn't hold in the general case you mention, but (assuming your category has pullbacks) it does hold when $A'\to B'$ is stably regular, i.e. when every pullback of $A'\to B'$ is a regular (or just strong) epimorphism. In particular, it holds in every regular category, and thus in every topos.
The proof is not extremely hard if you can draw the right diagrams, but MathJax is not very practical for this. So to give the idea : if you take the pullback $P=B'\times_{C'}C$ you have an induced map $\beta:B\to P$, and it suffices to prove that $\beta$ is an iso. If you take then the pullback $Q=A'\times_{B'}P$, you also have a map $\alpha:A\to Q$ induced by $A\to B\stackrel{\beta}\to P$ and $A\to A'$, which is an iso since
$$Q=A'\times_{B'}P=A'\times_{B'}(B'\times_{C'}C)\cong A'\times_{C'}C=A.$$
Now you have a rectangle
$$\begin{CD}A@>>> B \\ @V{\alpha}VV @VV{\beta}V \\ Q @>>> P \\ @VVV @VVV \\ A'@>>> B' \end{CD}$$
which is a pullback, and such that the bottom square is a pullback, so the top is also a pullback. The stability of $A'\to B'$ implies that the horizontal arrows are strong epimorphisms; since $\alpha$ is an iso, $\beta$ is also a strong epimorphism. Taking the kernel pairs of $\alpha$ and $\beta$ gives you yet another pullback squares, and thus the induced map between them is again a strong epimorphism. Then, since the two projections of the kernel pair of $\alpha$ are equal, those of $\beta$ are equal as well, and thus $\beta$ is also a monomorphism, hence an iso.
See also this MO thread for more on the topic (including some counterexamples for the case where $A'\to B'$ is merely epic, and a proof that pullback stability is necessary).
