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Math Help - Homology & covering spaces

  1. #1
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    Homology & covering spaces

    I am looking at a question for a practice class tomorrow (2.2.12 from Hatcher):

    Show that the quotient map S^1\times S^1 \to S^2 collapsing the subspace S^1 \vee S^1 to a point is not nullhomotopic by showing that it induces an isomorphism on H^2. On the other hand, show via covering spaces that any map S^2 \to S^1\vee S^1 is nullhomotopic.

    For the first part, I'm not sure what the induced map looks like, so I can't use it to form an isomorphism; for the second, I'm unsure how to use the covering space to show that the map is nullhomotopic.
    Last edited by harbottle; October 25th 2009 at 04:47 AM. Reason: changed question
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  2. #2
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    Quote Originally Posted by harbottle View Post
    Show that the quotient map S^1\times S^1 \to S^2 collapsing the subspace S^1 \vee S^1 to a point is not nullhomotopic by showing that it induces an isomorphism on H^2.
    Consider this long exact sequence,

    \cdots \rightarrow H_2(S^1 \vee S^1) \rightarrow H_2(S^1 \times S^1) \rightarrow  H_2(S^1 \times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1) \rightarrow H_1(S^1 \times S^1) \cdots.

    By the Hurewicz theorem, the first homology group of a torus H_1(S^1 \times S^1) is isomorphic to the fundamental group of a torus \pi_1(S^1 \times S^1) = \mathbb{Z} \times \mathbb{Z}. By applying the Mayer-Vietoris sequence, we have  H_2(S^1 \vee S^1) = 0 and H_1(S^1 \vee S^1) = \mathbb{Z} \times \mathbb{Z}.

    Now we see that H_2(S^1 \times S^1) \rightarrow H_2(S^1 \times S^1, S^1 \vee S^1) is an isomorphism in the above long exact sequence. Since H_2(S^1 \times S^1) is isomorphic to H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} , the quotient map S^1\times S^1 \to S^2 collapsing S^1 \vee S^1 to a point is not null homotopic.

    On the other hand, show via covering spaces that any map S^2 \to S^1\vee S^1 is nullhomotopic.
    Let f:S^2 \to S^1\vee S^1. The universal cover for S^1\vee S^1 is a \mathbb{Re} \times \mathbb{Re} such that g:\mathbb{Re} \times \mathbb{Re} \rightarrow S^1\vee S^1 is a covering map. By using the lifting criterion (Hatcher p61), we have a lifting map h:S^2 \rightarrow \mathbb{Re} \times \mathbb{Re} such that f=gh. Since \mathbb{Re} \times \mathbb{Re} is contractible, h is nulhomotopic. It follows that f:S^2 \to S^1\vee S^1 is nullhomotopic.
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post
    Consider this long exact sequence,

    \cdots \rightarrow H_2(S^1 \vee S^1) \rightarrow H_2(S^1 \times S^1) \rightarrow  H_2(S^1 \times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1) \rightarrow H_1(S^1 \times S^1) \cdots.

    By the Hurewicz theorem, the first homology group of a torus H_1(S^1 \times S^1) is isomorphic to the fundamental group of a torus \pi_1(S^1 \times S^1) = \mathbb{Z} \times \mathbb{Z}. By applying the Mayer-Vietoris sequence, we have  H_2(S^1 \vee S^1) = 0 and H_1(S^1 \vee S^1) = \mathbb{Z} \times \mathbb{Z}.

    Now we see that H_2(S^1 \times S^1) \rightarrow H_2(S^1 \times S^1, S^1 \vee S^1) is an isomorphism in the above long exact sequence. Since H_2(S^1 \times S^1) is isomorphic to H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} , the quotient map S^1\times S^1 \to S^2 collapsing S^1 \vee S^1 to a point is not null homotopic.

    Let f:S^2 \to S^1\vee S^1. The universal cover for S^1\vee S^1 is a \mathbb{Re} \times \mathbb{Re} such that g:\mathbb{Re} \times \mathbb{Re} \rightarrow S^1\vee S^1 is a covering map. By using the lifting criterion (Hatcher p61), we have a lifting map h:S^2 \rightarrow \mathbb{Re} \times \mathbb{Re} such that f=gh. Since \mathbb{Re} \times \mathbb{Re} is contractible, h is nulhomotopic. It follows that f:S^2 \to S^1\vee S^1 is nullhomotopic.
    I see-- H_1 is the abelianization of pi_1 and of course Z x Z is abelian

    The second part is a bit obivous, now that I can see the idea!

    Thanks so much!
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