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 looking for a proof of the fact that in a regular category a commutative square of regular epimorphisms which is a pullback is also a pushfoward.

share|cite|improve this question
I've not seen any lemma like that, but there is one involving a pullback–pushout square of monomorphisms. – Zhen Lin Apr 21 '13 at 20:03
I find this statement on at page 10, but there isn't a proof. – Fabio Lucchini Apr 21 '13 at 23:46
What state the lemma involving a pullback–pushout square of monomorphisms? Thank'you – Fabio Lucchini Apr 24 '13 at 12:11
It's just the statement that if $A = B \cup C$, then $A$ is the pushout of $B \leftarrow B \cap C \rightarrow C$ as well. – Zhen Lin Apr 24 '13 at 12:54
See Proposition 1.4.3 in [Johnstone, Sketches of an elephant, Part A]. – Zhen Lin Apr 26 '13 at 22:24
up vote 1 down vote accepted

Let $f' : A' \to B'$, $v : A' \to A$, $w : B' \to B$, $f : A \to B$ form a pullback square, and let $K (f')$, $K (v)$, $K (w)$, $K (f)$ be the respective kernel pairs. It is not hard to show that $K (v)$ is the pullback along $f'$ of $K (w)$, and $K (f')$ is the pullback along $v$ of $K (f)$; thus we have regular epimorphisms $K (v) \to K (w)$ and $K (f') \to K (f)$. Suppose we have morphisms $h : A \to C$ and $u : B' \to C$ making the evident square commute. Then, by considering an appropriate commutative diagram, we see that $h : A \to C$ must factor through $f : A \to B$, and $u : B' \to C$ must factor through $w : B' \to B$. Since $v : A' \to A$ and $f : A \to B$ are epimorphisms, the two morphisms $B \to C$ we obtain are in fact equal. It is clearly the unique one making the obvious diagram commute, and thus the pullback square we started with is also a pushout square.

share|cite|improve this answer

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.