Monomorphisms preserved in a pullback The question is answered, e.g,  here.
Suppose $(P, p_1, p_2)$ is a pullback for $f:A\to C$ and $g:B\to C$ with $fp_1 = gp_2$.

if $f$ is a monomorphism, show $p_2$ is a monomorphism.

What we do is assume $u,v:Q\to P$ such that $p_2u = p_2v$, then $gp_2u = gp_2v$ i.e $fp_1u = fp_1v$. Then $p_1u = p_1v$ because $f$ is monic. But why do we need this?
We immediately have $f(p_1u) = g(p_2v)$, which implies a unique $m:Q\to P$ with $p_1m = p_1u$ and $p_2m= p_2v$ and $u=m=v$ by uniqueness.
The latter must be wrong, but where am I cheating?

We have
$$p_1u\overset{?} = p_1v \Rightarrow fp_1u = fp_1v = gp_2v  $$
So here we need $f$ to be monic to proceed. But we also assumed
$$p_2u = p_2v \Rightarrow gp_2u = gp_2v = fp_1v $$
So we seemingly don't need $p_1u = p_1v$.
 A: If $f$ isn't monic then it might be the case that $p_1 u \neq p_1 v$. If this is so, then when you introduce $m$ using the property that since $f(p_1u) = g(p_2v)$ there's a unique morphism $m$ with $p_1m = p_1u$ and $p_2m= p_2v$, $v$ isn't a morphism with this property.
Therefore $v$ isn't equal to $m$, so it isn't equal to $u$.
Try it out using some small examples of pullbacks of non-monic morphisms. The easiest cases are probably where $C$ is the terminal object, so the pullback is the product.
Edited: Where the morphisms $u$ and $v$ become uncoupled in your argument is when you made the choice to fill out the diagram using $f(p_1u) = g(p_2v)$. There's a unique morphism $m_u$ coming from this diagram, such that $p_1 m_u = p_1 u$ and $p_2 m_u = p_2 v$. (Hint: $m_u$ is in fact $u$. If $p_1 u \neq p_1 v$, then $m_u$ can't be $v$)
You could have instead chosen to fill out the diagram using $f(p_1v) = g(p_2u)$. If $p_1 u \neq p_1 v$ then this is an entirely different diagram, so it has an entirely different induced morphism $m_v$. $m_v$ is the unique morphism such that $p_1 m_v = p_1 v$ and $p_2 m_v = p_2 u$ (Hint: $m_v$ is in fact $v$. If $p_1 u \neq p_1 v$, then $m_v$ can't be $u$). Since the diagrams they are induced by are different, there's no reason for $m_v$ to be equal to $m_u$.
