Why is homology isomorphic to reduced homology plus Z?

[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

 

[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.

 

[redbowlover]

thanks!