Two pullback diagram Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case. 

Now we seek some addition condition on $X\to Y$ that forces the right square is a pullback too. 
My question: is epic a sufficient condition? (If the category is Sets, then yes.)
Added: Let $P$ be the pullback of the right square, then there exists $B\to P$, and the square $A\to P \to Y$ // $A\to X \to Y$ is a pullback, so we have the following diagram in which the bottom and the whole squares are pullback, so is the upper square. If the category is Sets, $X\to Y$ is surjective then $A\to P $ is also surjective. Since the pullback of $B\to P$ along a surjective map is an bijection, $B\to P$ must be a bijection. This shows the right square of the original diagram is a pullback. We can also see why we consider some nice condition on $X\to Y$.

 A: No, epic is not sufficient; there is a counterexample in Pos, the category of posets.  Take A, B, and X each to be the discrete two-element poset $\{0,1\}$; take C, Y, and Z to be the same elements with their natural ordering. $\{ 0 < 1 \}$; and take all maps to be identities on underlying sets.
On the other hand, split epic is sufficient in any category; this is a reasonably straightforward diagram chase.
Trying to weaken this, I would suspect (though I haven’t checked it) that in regular categories, regular epic may well be a sufficient condition; certainly regular categories (and strengthenings like exact categories) give a natural setting for these sorts of interactions between epis and pullbacks.
A: A sufficient condition in a category with pullbacks is that $X \to Y$ be a pullback stable regular epimorphism: a regular epimorphism all of whose pullbacks are also regular epimorphisms.  This property holds in any regular category.
Stability implies that $A \to B$ and $A \to P$ are (pullback stable) regular epis too.  Thus $B \to P$ is strong epi by 2 out of 3 on the upper square of the lower diagram.  So to show $B \to P$ is an isomorphism you just need to show it is mono.
You need to show that the two maps from the kernel pair $B\times_{P}B$ of $B \to P$ coincide.  To see this we'll use the map between kernel pairs $A=A \times_{A}A \to B \times_{P} B$  induced by the same square as before.  This map is, in fact, the pullback of $A \to P$ along equally the domain or codomain projections for the graph map between kernel pairs, and so a regular epi itself.  Thus the two maps $B\times_{P}B \to B$ coincide because their precomposite with this regular epi does.
(There is a good reason for replacing epis in Set by pullback stable regular epis in a category C with pullbacks.  The reason is that in any such category C the pullback stable regular epis, taken as singleton covers, determine the basis for a Grothendieck topology on C.  Moreover it is subcanonical and, furthermore, the induced functor $C \to Sh(C)$ to the category of sheaves preserves pullbacks and sends pullback stable regular epis to the same.  Thus any exactness property which holds between pullbacks and epis in a Grothendieck topos holds between pullbacks and pullback stable regular epis in an arbitrary category with pullbacks.)
A: Consider the category consisting of that diagram, together with an extra object $P$ with maps $Y\leftarrow P\to C$ that commute with the maps to $Z$.  Then in this category, the whole diagram and the left square are pullbacks, $X\to Y$ is epic, but the right square is not a pullback.
A: I tried to write the explanation of the comment about a dozen of times, but was never satisfied with the result. Finally, I decided to write a full note (I will try to put it on arXiv in a few minutes; here it is) describing the natural setting for such questions:
http://www.mimuw.edu.pl/~mrp/the_other_pullback_lemma.pdf
(the note still needs some improvements, but I am running out of time now...)
I found it easier to characterise your condition by "extremal epimorphisms" rather than "strong epimorphisms" (notice however, that in case of finite connected limits, these concepts coincide). Here is the formal statement:
Let us assume that finite connected limits exist. The following are equivalent:


*

*your condition along $e \colon X \rightarrow Y$ holds 

*$e \colon X \rightarrow Y$ is an extremal morphism stable under pullbacks.

