Why is homology isomorphic to reduced homology plus Z?

  1. 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...

  2. jcsd
  3. quasar987

    quasar987 4,773
    Science Advisor
    Homework Helper
    Gold Member

    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.
  4. thanks!
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