# Why is homology isomorphic to reduced homology plus Z?

1. Jul 5, 2011

### redbowlover

Working through Hatcher...
For any space X, we have an augmented chain complex

$...\rightarrow C_1(X) \rightarrow C_0(X)\rightarrow \mathbb{Z}\stackrel{\epsilon}{\rightarrow}0$
Hathcer says that since $\epsilon$ induces a map $H_0(X)\rightarrow \mathbb{Z}$ with kernel $\tilde{H}_0(X)$, we get an isomorphism $H_0(X)\simeq \tilde{H}_0(X)\oplus \mathbb{Z}$

Where is this isomorphism coming from? I understand where the induced map on $H_0(X)$ comes from...

Thanks

2. Jul 5, 2011

### quasar987

There is a short exact sequence 0-->H(reduced)_0-->H_0-->Z-->0, and Z being free, it splits. That is, H_0=H(reduced)_0 x Z.

3. Jul 7, 2011

thanks!