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Here is the question from Rotman, verbatim:

A sequence $S'_*\stackrel{f}{\to} S_* \stackrel{g}{\to} S''_*$ is exact in Comp if and only if $S'_{n}*\stackrel{f_n}{\to} S_n \stackrel{g_n}{\to} S''_{n}$ is exact in Ab for every $n\in \mathbb{Z}$

(Note here Comp and Ab are the categories of chain complexes and abelian groups respectively.

I am slightly confused - I thought this is the definition of an exact sequence of chain complexes. Indeed this is what Massey appears to say (Definition 2.6)

What am I missing here?

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Maybe Rotman defines exactness via kernels and images (in the categorical sense). –  Rasmus Mar 1 '11 at 7:48
    
@Rasmus - right, Rotman says that a sequence of complexes and chain maps $\cdots \to A_*^{q+1}\stackrel{f^{q+1}}{\to} A_*^{q} \stackrel {f_q}{\to} A_*^{q-1}\to \cdots $ is exact if $\mathrm{Im} f^{q+1}=\mathrm{Ker} f^q$ for every $q$ –  Juan S Mar 1 '11 at 7:58
    
Then you have to ask yourself: what are the kernel and the image of a chain map? –  Rasmus Mar 1 '11 at 8:09
    
@Rasmus - OK I know what these are - what again almost, trivially implies the result. Is this a simple question, or am I still missing the point? –  Juan S Mar 1 '11 at 8:16
4  
Just an observation: The category of complexes is an abelian category, so there is an intrinsic notion of exactness and it turns out to be what you'd expect it to be. As @Rasmus pointed out, in order to check that the category of complexes is abelian, you have to think about what kernel, cokernel of a chain map are - they are the obvious candidates just check the factorization properties. This also follows from the fact that $\operatorname{Comp}$ can be written as a functor category (with some care), so kernels and cokernels (and all limits and colimits for that matter) can be taken pointwise. –  t.b. Mar 1 '11 at 10:57

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