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 have an split exact complex in an abelian category, that is, a chain complex which is exact and maps $s_n : C_n \to C_{n+1}$ st. $dsd = d$. I would like to prove that this implies that $C_n \cong B_n \oplus B_{n-1}$.

I would like to do this by showing that $d_{n+1} s_n + s_{n-1}d_n = id_{C_n}$. Is this even true? If it is how do I prove it?a picture showing the diagram

share|cite|improve this question
Yes, a chain complex $C_\bullet$ is split exact if and only if the identity map of the complex is null-homotopic. Take a look here – Andrea Gagna Feb 13 '13 at 23:39
yes. but can you use $s_n$ for the chain homotopy, rather than defining $t_n$? – krey Feb 14 '13 at 12:13
Oh, sorry! No, it is false in general. If you can wait till the end of the next week, I promise I will write a (class of) easy conterexample(s) and a sufficient condition for that to work, with a good amount of details. – Andrea Gagna Feb 14 '13 at 17:25
up vote 3 down vote accepted

It is not true, in general. You can find a sufficient condition and a conter-example here.

Note: I assume few standard facts about abelian categories, that you can find in Freyd's Abelian Categories, Borceux' Handbook of Categorical Algebra. Vol. 2 or in the last chapter of Aluffi's Algebra: Chapter 0.

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.