Square of Three Monos and Identity (Top) is a Pullback

Proposition

Consider the following commutative square in any category $C$. That is, three of the maps are monomorphisms, while the fourth is $\text{id}$.
Then the maps form a pullback square.

Proof

Suppose you draw a cone from an object $Z$ int $X$ and $Z$ like so:

The only map from $u: W \to X$ such that $\text{id}\circ u = q$ is $u = q$. And this choice makes everything commute because since by assumption $m'p = m'' q = m'mq$ (by commutativity) which implies $p = mq$ since $m'$ is a monomorphism.

Now, by symmetry of a pullback diagram along its diagonal, we do not have to check the case when $m$ and $\text{id}$ are swapped.

$\blacksquare$

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