Prove that the following commutative square has a monomorphism Prove that if
$$\require{AMScd}
\begin{CD}
A @>f>> C \\
@VmVV @VVnV \\
B @>g>> D
\end{CD}
$$
is a pullback and $n$ is a monomorphism then $m$ is also a monomorphism.
Attempt
First we use the fact that $n$ is a monomorphism. As such, take $u,v:Z\rightarrow A$. We will show that $u=v$ by showing that all projects are equivalent. Indeed,
$$\begin{align}
n\circ f\circ u&=n\circ f\circ v\\
&=g\circ m\circ u\\
&= g\circ m\circ v
\end{align}.$$
Thus, $u=v$.
Next, using the fact that the given square is a pullback we know there exists a unique factorization from $Z$ to $A$. Take this morphism to be $u$ from above, then we have that $u'=m\circ u$ and $v'=f\circ u$ (i.e both triangles commute).
It remains to show that $u'=v'$ to show that $m$ is a monomorphism. This is where I am having a bit of trouble (that is I hope the above is correct). 
The idea I am relying on is that if I can show all the projections are equivalent then we get the equality. However, I am not certain what follows is accurate.
See that,
$$\begin{align}
g\circ u'&=g\circ m\circ u\\
&=n\circ f\circ u\\
&=n\circ v'
\end{align}$$
which give $u'=v'$.
Cheers, for hints and suggestions
 A: I think you have it backwards. You are given $u,v:Z\to A$ such that $m\circ u=m\circ v$. You want the proof to end with "thus $u=v$", but you have that phrase in the middle. Here is a full proof:
Since $m\circ u=m\circ v$, we have $g\circ m\circ u=g\circ m\circ v$. Then, because the square commutes, $n\circ f\circ u=n\circ f\circ v$. Adding parentheses for clarity, the same equality says that $n\circ (f\circ u)=n\circ (f\circ v)$, and because $n$ is mono, we have that $f\circ u=f\circ v$.
This means that $u$ and $v$ induce the same map $n':Z\to B$, and they induce the same map $f':Z\to C$. Furthermore, $g\circ n'=f'\circ m$, so by definition of pullback square, the maps $Z\to A$ they induce must be equal as well. Thus $u=v$.
A: Suppose $m\circ u=m\circ v$. Then also $g\circ m\circ u=g\circ m\circ v$ and therefore
$$
n\circ f\circ u=n\circ f\circ v
$$
Since $n$ is a monomorphism, we obtain $f\circ u=f\circ v$.
Set $\alpha=f\circ u=f\circ v$ and $\beta=m\circ u=m\circ v$. Then $n\circ\alpha=g\circ\beta$ and, by the pull-back property, there is a unique $\gamma\colon Z\to A$ such that $f\circ\gamma=\alpha$ and $m\circ\gamma=\beta$.
Since both $u$ and $v$ satisfy the same property as $\gamma$, we conclude $u=v$.
