When is a pair of morphisms is a pullback of some pair of morphisms?

Question

Given $p$ and $q$ morphisms on a category (preferably arbitrary, but it could be interesting to know on some other class of categories too), are there any useful necessary and/or sufficient conditions for them to be the pullback of some pair of morphisms? That is, are there conditions for there to exist morphisms $f$ and $g$ such that the diagram

is a pullback diagram for $f$ and $g$?

Answer 1

Let $B\times C$ denote a product of $B$ and $C$ with projections $\pi_B$ and $\pi_C$.

Then the unique arrow $e=<p,q>:A\to B\times C$ with $\pi_B\circ e=p$ and $\pi_C\circ e=q$ must be an equalizer.

This of arrows $f\circ\pi_B$ and $g\circ\pi_C$.

So if arrows $p,q$ with common domain can be recognized as pullbacks of arrows $g,f$ then $<p,q>$ is a regular monomorphism (i.e. an equalizer of a pair of arrows).

Unfortunately this condition is not sufficient (see the comment of @Arnaud on this answer) as I first stated wrongly.

Answer 2

To anyone who is interested, I found interesting candidates for $f$ and $g$ as in the question, candidates which are such that $p$ and $q$ are the pullback of some pair of morphisms if, and only if, they are the pullback for $f$ and $g$. This way, all we need to check is whether $p$ and $q$ are the pullback of these morphisms $f$ and $g$ or not in order to figure out whether or not they come from pullbacks. All we need to ask is for a certain coequalizer and the coproduct of $B$ and $C$ to exist.

To be more precise, let $B \coprod C$ be a coproduct for $B$ and $C$, and $u_B$ and $u_C$ be the respective inclusions. So, if we take $d: B \coprod C \rightarrow D$ to be a coequalizer for $u_B \circ p$ and $u_C \circ q$, then the morphisms $f = d \circ u_B$ and $g = d \circ u_C$ are the candidates I spoke of in the previous paragraph.

I won't go into details for the proof, since I am the one who made this question, and I don't know if anyone else is interested in this. If you are interested, however, leave a comment and I may edit this with the proof. (It is not really a difficult proof, all you need to do is to show that any pair of morphisms whose pullback is $p$ and $q$ end up being factored through $f$ and $g$, due to the nature of the coequalizer, and then show how that implies on $f$ and $g$ having $p$ and $q$ as a pullback)