I have a question about how to yield the reduced homology group. According to the Hatcher's textbook,
A reduced homology group is $H_0(X) \simeq \widetilde{H}_0(X)\oplus \mathbb{Z}$. But I dn not know why it holds when redading the contents. Instead, In my thought, since $\operatorname{Ker} \partial_{0} = C_0(X) = \left \{ \sum n_i \sigma_i : n_i \in \mathbb{Z} \right \} $,
$H_0(X) = {C_0(X)}/\operatorname{Im} \partial _1 = (\left \{ \sum n_i \sigma_i : \sum n_i=0 \right \} \oplus \left \{ \sum n_i \sigma_i : \sum n_i \neq 0 \right \}) / \operatorname{Im} \partial _1 $
$= (\left \{ \sum n_i \sigma_i : \sum n_i=0 \right \}/\operatorname{Im} \partial _1 )\oplus (\left \{ \sum n_i \sigma_i : \sum n_i \neq 0 \right \}/\operatorname{Im} \partial _1 )$
$= \widetilde{H}_0(X) \oplus (\left \{ \sum n_i \sigma_i : \sum n_i \neq 0 \right \}/\operatorname{Im} \partial _1 )$
From here, I expect to get the following claim : $ \left \{ \sum n_i \sigma_i : \sum n_i \neq 0 \right \}/\operatorname{Im} \partial _1 \simeq \mathbb{Z}$, but I am not sure whether or not my claim holds.
Best Answer
Unfortunately your approach does not work.
As Quimey mentions in his comment, it is inadequate to write $$C_0(X) = \left\{ \sum n_i \sigma_i : \sum n_i=0 \right\} \oplus \left\{ \sum n_i \sigma_i : \sum n_i \neq 0 \right\}$$ since $B = \left\{ \sum n_i \sigma_i : \sum n_i \neq 0 \right\}$ is not a subgroup of $C_0(X)$ (note that $0 \notin B$). But even if you had a correct direct sum decomposition $C_0(X) = A \oplus B$, you cannot form both $A/\operatorname{Im} \partial _1$ and $B/\operatorname{Im} \partial _1$ since it is impossible that the subgroup $\operatorname{Im} \partial _1 \subset C_0(X)$ is contained in both direct summands.