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why is homology isomorphic to reduced homology plus Z? |
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| Jul5-11, 03:02 PM | #1 |
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why is homology isomorphic to reduced homology plus Z?
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 |
| Jul5-11, 03:32 PM | #2 |
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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.
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| Jul7-11, 03:06 PM | #3 |
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thanks!
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