# Simple question on exact sequences.

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.

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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) – Edoardo Lanari Jun 1 '13 at 17:02

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.
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.