# Pullbacks of monic morphisms.

I'm trying to prove that pullbacks of monics are monic.

Let

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?

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tikz-cd isn't working on MSE. – Pece Oct 29 '13 at 13:46
@Shaun: The input language here sadly isn't LaTeX, only something similar. I'm afraid you'll have to hack up your diagrams using arrays. – Johannes Kloos Oct 29 '13 at 13:46
@Johannes Kloos I see; thank you. I'm using my phone here so it'll be too fiddly to correct it now. They're not particularly important anyway. Does anyone mind if I just leave it like that (for now)? – Shaun Oct 29 '13 at 13:52
@Shaun: I don't speak for the site, but I don't mind. – Johannes Kloos Oct 29 '13 at 14:04
That should do it . . . I've used the picture in the answer below. – Shaun Oct 29 '13 at 22:37

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 $h = k$.

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Yep: I copied the question down wrong. Thanks. – Shaun Oct 29 '13 at 14:59
going to $f \circ h = f \circ k$ is easy, but I must be missing something as I dont see how that entails $h = k$. could you argue why ? – nicolas Mar 18 '14 at 14:00
@nicolas Then you have that both $h$ and $k$ both make the above diagram commute. Since the square is a pullback this means that $h = k$. – Aleš Bizjak Mar 18 '14 at 16:17
of course, by the universal mapping property on the pullback : that there is a unique $h$ making the diagram commute aka realizing $g m'\circ h = m f\circ h$ – nicolas Mar 19 '14 at 14:26