Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am just learning about abelian categories and I would like to hear some advice on how to think about the concepts of kernel, cokernel, image etc correctly; I am a bit confused. I am trying to prove the following, probably very easy result, but I am stuck.

Let $0\to A\to B\to C\to 0$ is a short exact sequence of complexes in an abelian category. Show that if two of the three complexes $A$, $B$, $C$ are exact, then so is the third.

Say $A$ and $C$ are exact, and denote the differentials of $B$ by $\beta_{n+1}:B_{n+1}\to B_n$. We need to prove that $\text{im}(\beta_{n+1})=\text{ker}(\beta_n)$. First of all, what does this equality mean? Does it mean we have to prove equivalence as subobjects (one factors through the other and vice versa)? Or do we need to just take the image and the kernel and prove they are equal as arrows (for example by a "pre-compositions with a monic are equal so the arrows are equal" argument)? - but then what is the domain of those arrows and is it the same for both the image and the kernel?

Or am I at a completely wrong path?

Any advice would be highly appreciated, thanks.

share|improve this question
possible duplicate of Short exact sequence of exact chain complexes –  Martin May 26 '13 at 15:17
The $\operatorname{im} = \operatorname{ker}$ thing is understood to mean "isomorphic as subobjects of the relevant object". –  Zhen Lin May 26 '13 at 15:17
Ok, thanks for the link. But, how does the solution they give there work in an arbitrary abelian category? Are they not working with modules over a ring? If the equality required for the exactness means an isomorphism of subobjects, how can you use elements of the objects of the category to prove it? What does, for example, the sentence "Now, consider the element $b-db' \in B_n$" mean? Shouldn't we be dealing with arrows here? Sorry if the question is stupid. –  Terence Maligan May 26 '13 at 16:02
Certain element-chasing methods are valid in general abelian categories. See Ch. VIII in Categories for the working mathematician. –  Zhen Lin May 26 '13 at 16:38
@Zhen Lin: with far more sophisticated methods every element-chasing method is valid in an abelian category (of course every has to be explained, I'm referring to en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem) –  Lano Jun 1 '13 at 17:02

2 Answers 2

up vote 0 down vote accepted

That equality between image and kernel means that if you consider the epi-mono factorization of $\beta_{n+1}$ then "the monic" part is a kernel for $\beta_{n}$, which we know to be unique only up to iso. Now just chase the diagrams as if you were in a category of modules over some ring (if this doesn't bother you) otherwise you simply need to check that the composition is the zero morphism and that you have the universal property of the kernel.

share|improve this answer
"Otherwise you simply need to check that the composition is the zero morphism and that you have the universal property of the kernel." Well, it's not quite as simple as that. You'll need at least some form of the nine lemma if you want to do it diagrammatically, i.e., in an element-free way. Note that the nine lemma is a special case of the result OP asks about. –  Martin May 26 '13 at 16:56
I have not mentioned what you need to prove the result, but what you have to prove; that "simply" was only quantitative and not qualitative so I really can't see what's there to point out. –  Lano May 26 '13 at 22:21

Since the $\beta_n$ form a chain complex, $\beta_{n+1} \circ \beta_n =0$. This means that $\beta_{n+1}$ will map any arguments to things which $\beta_n$ maps to $0$, i.e im $\beta_{n+1}\subset \text{ ker } \beta_{n}$. If there were something which $\beta_{n}$ mapped to $0$ which $\beta_{n+1}$ did not map anything to but $A$ and $C$ were exact then some part of the image in $B$ from $0\to A \to B$ would fail to end up in the kernel of $B\to C$ and $0\to A \to B \to C \to 0$ would fail to be exact.

share|improve this answer
Are you using the embedding theorem for abelian categories or the tricks to chase diagrams in abelian category? If so you should declare it, otherwise you're only dealing the case of a category of modules over some ring ( which is indeed enough but not in an obvious way) –  Lano May 26 '13 at 15:52

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.