TITLE: Pullbacks of monic morphisms.
QUESTION [5 upvotes]: I'm trying to prove that pullbacks of monics are monic.
Let
$\require{AMScd}$
\begin{CD}
X_1 @>f>> X_2\\
@V m' V  V  @VV m V\\
X_3 @>>g > X_4
\end{CD}
be a pullback square with $m'$ monic.
Let $h, k$ be parallel such that $m\circ h=m\circ k$. Let $x_{0}$ be the domain of $h, k$.
Suppose there exists $\phi : x_{0}\to x_{3}$ such that $g\circ\phi =m\circ h=m\circ k$. Then, since the above is a pullback square, there exist unique $u, v: x_{0}\to x_{1}$ such that
$$h=f\circ u, \phi =m'\circ u$$ and
$$k=f\circ v, \phi =m'\circ v.$$
Since $u, v$ are parallel and $m'$ is monic, $u=v$. Thus $h=f\circ u=f\circ v=k$. But what if $\phi$ doesn't exist?

REPLY [4 votes]: What you're trying to prove is wrong, if I read you correctly. You're trying to prove that if $m'$ is monic then $m$ is if the diagram 

is a pullback diagram. This is false. Consider the category of sets where $x_3 = x_1 = \emptyset$, $x_2$ is an arbitrary set with at least two elements and $x_4 = 1$. Then the diagram is clearly a pullback diagram but $m$ is not injective (not a mono).
On the other hand, if $m$ is monic then $m'$ will always be monic and this is the statement that is usually meant by "pullbacks of monics are monics".
To show that suppose you're given two parallel arrows $h, k$ such that $m' \circ k = m' \circ h$. Then consider the diagram

It should be easy to show that it commutes and that $f \circ h = f \circ k$ by using the fact that $m$ is a mono and that $m' \circ h = m' \circ k$. This should allow you to conclude that $h = k$.