I was viewing the solution given here for the following problem:
Verify that $f \simeq g$ implies $f_{\ast} = g_{\ast}$ for induced homomorphisms of reduced homology groups.
I have one question about the proof provided there as the author discusses the case that $n = 0$. The author states the following:
There is a one-to-one correspondence between $H_0(X)/\mathbb{Z}$ and a subgroup $H'(X) \subseteq H_0(X)$, and between $H_0(Y)/\mathbb{Z}$ and a subgroup $H'(Y) \subseteq H_0(Y)$.
How is this deduced? $H_0(X)$ and $H_0(Y)$ are each direct sums of copies of $\mathbb{Z}$, so $H_0(X)/\mathbb{Z}$ collapses one of those summands to a point. It's tough for me to see why this quotient would be isomorphic to a subgroup of the original direct sum.
Thanks!
Consider the tail of your complexes. You have the augmentations $C_0(X)\to C_{-1}(X)$ and $C_0(Y)\to C_{-1}(Y)$. These are induced by the unique maps $X\to*, Y\to*$, so your chain map of non-augmented complexes extends to one of augmented complexes whenever it is induced by a topological map. Now, extend your chain homotopy by defining $s_{-1}\colon C_{-1}(X)\to C_0(Y)$ to be trivial. The rest follows from $f_*, g_*\colon C_*(X)\to C_*(Y)$ being chain maps$ (of augmented complexes).
(that the augmentation works follows from the fact that if $f, g\colon X\to Y$ are any continuous maps, then $X\to Y\to *$ will be the same composite for both of them!)