TITLE: Reduced homology
QUESTION [14 upvotes]: I'm trying to understand why $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$. In Hatcher on page 110 he writes
''...Since $\varepsilon \partial_1 = 0$, $\varepsilon$ vanishes on $im \partial_1$ and hence induces a map $H_0(X) \rightarrow \mathbb{Z}$ with kernel $\tilde{H}_0(X)$...''
The bit I don't understand is how exactly it induces this map. I mean, where does $\mathbb{Z}$ come from all of a sudden. It's in the extended diagram for the reduced homology $\tilde{H}_0$ and the $H_0(X)$ is in the regular diagram. So somehow there is a map from normal homology into reduced homology (the exact sequences), but how? Many thanks for your help.

REPLY [2 votes]: How about using first and third isomorphism theorems of groups?
We have $\text{Im} \;\partial_1 \le \text{Ker}\; \epsilon \le \text{Ker}\; \partial_0=C_0(X)$ and both $\text{Im}\; \partial_1$ and $\text{Ker}\;\epsilon$ are normal subgroups of $C_0(X)$. Thus by third isomorphism theorem of groups, $\cfrac {\cfrac{C_0(X)}{\text{Im} \;\partial_1}}{\cfrac{\text{Ker} \;\epsilon}{\text{Im}\; \partial_1}} \approx \cfrac {C_0(X)}{\text{Ker} \;\epsilon}.$
But since $\epsilon$ is surjective, we have $\cfrac {C_0(X)}{\text{Ker} \;\epsilon} \approx \Bbb Z$ by first isomorphism theorem of groups.
By definition of singular homology, we have $H_0(X)=\cfrac {C_0(X)}{\text{Im} \; \partial_1}.$ From here it comes naturally that $\tilde{H}_0(X)$ must be defined as $\cfrac{\text{Ker} \; \epsilon}{\text{Im} \;\partial_1}.$

EDIT: $\text{Im} \;\partial_1 \le \text{Ker}\; \epsilon$ comes from $\epsilon\partial_1=0$.