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Math Help - Chain maps inducing the suspension isomorphism

  1. #1
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    Chain maps inducing the suspension isomorphism

    Hi,

    Given a space X, it can be shown (either via the Mayer-Vietoris theorem, or by theory of exact sequences and relative homology) that, for all n, we have H~_n(X)=H~_n+1(SX), where SX is the suspension of X, and I'm using H~ to stand for the reduced homology.

    I'm trying to find an explicit chain map from C_n(X) to C_n+1(SX) which induces this isomorphism, but I'm not really sure where to start. Does anyone have any ideas? (This is question 21 in Hatcher, Section 2.1, if that helps at all.)

    Thanks,

    Jonathan.
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  2. #2
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    Quote Originally Posted by jonathan122 View Post
    Hi,

    Given a space X, it can be shown (either via the Mayer-Vietoris theorem, or by theory of exact sequences and relative homology) that, for all n, we have H~_n(X)=H~_n+1(SX), where SX is the suspension of X, and I'm using H~ to stand for the reduced homology.

    I'm trying to find an explicit chain map from C_n(X) to C_n+1(SX) which induces this isomorphism, but I'm not really sure where to start. Does anyone have any ideas? (This is question 21 in Hatcher, Section 2.1, if that helps at all.)

    Thanks,

    Jonathan.
    Choose A and B such that A \cup B = SX and A \cap B = X. The possible choice of A and B are each cone (upper/lower in the below link) of X or contains each cone, respectively, such that A \cap B = X.

    Use the fact that a cone of a space X is contractible, so its reduced homology groups are trivial for all dimensions.

    Now use the Mayer-Vietoris sequence (reduced version) and use the basic property of an exact sequence of groups, i.e., if

    \cdots \rightarrow 0 \rightarrow \tilde{H}_{n+1}(SX) \rightarrow \tilde{H}_{n}(X) \rightarrow 0 \rightarrow \cdots \rightarrow 0,

    then \tilde{H}_{n+1}(SX) \cong \tilde{H}_{n}(X)
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