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 am reading the appendix of Charles Weibel's Homological Algebra and have the following question. It is mentioned that every morphism $f: B \to C $ in an abelian category factors as $B \to im(f) \to C$ where $im(f)\equiv ker(coker(f))$ and the morphism $B \to im(f)$ is epi and $im(f) \to C$ is mono. I am able to prove the other parts of the problem but not able to show that the morphism $B \to im(f)$ is epi. Please help. Thanks.

share|cite|improve this question
This appendix is too short to learn the basics properly. Better consult Mac Lane's CWM, the chapter on abelian categories. – Martin Brandenburg Jan 13 '14 at 21:21
What is the definition of Abelian Category in that book? – Berci Jan 13 '14 at 23:44
up vote 3 down vote accepted

A crucial property of Abelian categories (mostly part of the definition) is that the canonically arising morphism ${\rm coim}(f)\to {\rm im}(f)$ is iso for all morphisms $f$.

Using this, you're done, because, by dual argument you have $B\to{\rm coim}(f)$ is epi.

From the assumptions that every mono is kernel and every epi is cokernel, one way to prove $v:B\to{\rm im}(f)$ is epi (or, equivalently, $u:{\rm coim}(f)\to{\rm im}(f)$ is epi) is as follows: Suppose $t\circ v= 0$ and consider the pushout of ${\rm im}(f)\to C$ and $\,t$:

$B \overset{f}\longrightarrow C \to {\rm coker}(f) \\ v\searrow \, \nearrow \ \searrow t_1 \\ \ \ \ \, {\rm im}(f)\quad\ \ \ \cdot \\ \ \ \ \ \ \ t \searrow \ \nearrow i_1 $

Here $t_1\circ f=0$, so $t_1$ goes through ${\rm coker}(f)$, but then $i_1\circ t$ also becomes $0$.

Now the following lemma ensures that $i_1$ is mono, so $t=0$.

Dually one can prove that ${\rm coim}(f)\to{\rm im}(f)$ is also mono, thus, is a kernel and epi, hence iso.

Lemma: In an Abelian category the pushout of a monomorphism is monomorphism, i.e. if $i:A\hookrightarrow B$ and $f:A\to C$ then the arising arrow $i_1:C\to P$ in the pushout diagram is mono.

$\ \phantom{f} A\overset{i}\hookrightarrow\, B \\ f\downarrow\ \ \ \ \ \downarrow \\ \ \phantom{f} C \underset{i_1}\to P $

Hint: Write the pushout as $P={\rm coker}\left(A\overset{[i,-f]}\longrightarrow B\oplus C\right)$ and observe that $[i,-f]:A\to B\oplus C$ is mono (because of $i$), so you can use the condition that $[i,-f]=\ker(B\oplus C\to P)$.

share|cite|improve this answer
Thank you. But I should have clarified the definition of abelian category in use. It is an additive category such that (i) every map has a kernel and a cokernel (ii) a monic is the kernel of its cokernel (iii) every epi is the cokernel of its kernel. Could you please add some comment in the view of this definition ? – user90041 Jan 14 '14 at 3:39
I have updated my answer. – Berci Jan 15 '14 at 1:46
Thanks a lot ! I was stuck on it as I did not know what a push-out was. I read the Wikipedia article on that and can understand your answer now. :-) One more request : Could you please suggest me some good place to learn Abelian categories ? Weibel's appendix is too compact. – user90041 Jan 16 '14 at 5:42
MacLane's Category Theory for the Working Mathematician, for example... And, if you didn't know the pullback, you could have defined $P$ as ${\rm coker}\left([i,-f]\right)$. – Berci Jan 18 '14 at 1:36

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.