Q:

Are these singular chain complexes possible?

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.

A:

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 $B$.

