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I am currently learning singular homology and one question gives these two chain complexes:

$A: 0 \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow 0$

$B: 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$

I am asked to compute a set of maps $A \rightarrow B$ and a set of chain homotopy classes of chain maps. I am puzzled by this question, because with singular complexes, even a single point set has $C_n$ isomorphic to $\mathbb{Z}$ for all $n$, so I don't see how we can have chain complexes that look like the ones above. I feel like I am missing something here.

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1 Answer 1

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The concept of a chain complex can sit independent of topology: it's a sequence of a abelian groups \[ \cdots \xrightarrow{d_{n + 2}} A_{n + 1} \xrightarrow{d_{n - 1}} A_n \xrightarrow{d_n} A_{n - 1} \xrightarrow{d_{n - 1}} \cdots \] such that $d_n \circ d_{n - 1} = 0$ for all $n$. So the question makes sense. But complexes like yours can certainly arise when computing simplicial homology; for example, $S^1$ has a $\Delta$-complex structure (see Hatcher's book) for which the corresponding chain complex would look like your $B$.

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When I first saw this question, I thought the way you explained above, but I guess what really confuses me is the 'chain homotopy classes of chain maps' part. Doesn't a chain map depend on topology? –  Barum Rho Nov 13 '11 at 5:28
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@Barum Chain maps are described later in the article I mentioned. A chain homotopy is again an algebraic definition, although the motivation surely comes from topology. –  Dylan Moreland Nov 13 '11 at 6:03

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