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Why is homology isomorphic to reduced homology plus Z? 
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#1
Jul511, 03:02 PM

P: 16

Working through Hatcher...
For any space X, we have an augmented chain complex [itex]...\rightarrow C_1(X) \rightarrow C_0(X)\rightarrow \mathbb{Z}\stackrel{\epsilon}{\rightarrow}0[/itex] Hathcer says that since [itex]\epsilon[/itex] induces a map [itex]H_0(X)\rightarrow \mathbb{Z}[/itex] with kernel [itex]\tilde{H}_0(X)[/itex], we get an isomorphism [itex]H_0(X)\simeq \tilde{H}_0(X)\oplus \mathbb{Z}[/itex] Where is this isomorphism coming from? I understand where the induced map on [itex]H_0(X)[/itex] comes from... Thanks 


#2
Jul511, 03:32 PM

Sci Advisor
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P: 4,771

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
Jul711, 03:06 PM

P: 16

thanks!



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