Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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



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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $h = k$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.