Prove that the following commutative square has a monomorphism

Question

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

Answer 1

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 $u=v$.

Answer 2

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 $u=v$.