why is homology isomorphic to reduced homology plus Z?


by redbowlover
Tags: homology, isomorphic, reduced
redbowlover
redbowlover is offline
#1
Jul5-11, 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
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quasar987
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#2
Jul5-11, 03:32 PM
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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
redbowlover is offline
#3
Jul7-11, 03:06 PM
P: 16
thanks!


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