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Thread: 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 $\displaystyle S^1\times S^1 \to S^2$ collapsing the subspace $\displaystyle S^1 \vee S^1$ to a point is not nullhomotopic by showing that it induces an isomorphism on $\displaystyle H^2$. On the other hand, show via covering spaces that any map $\displaystyle S^2 \to S^1\vee S^1$ 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; Oct 25th 2009 at 05:47 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 $\displaystyle S^1\times S^1 \to S^2$ collapsing the subspace $\displaystyle S^1 \vee S^1$ to a point is not nullhomotopic by showing that it induces an isomorphism on $\displaystyle H^2$.
    Consider this long exact sequence,

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

    Now we see that $\displaystyle 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 $\displaystyle H_2(S^1 \times S^1)$ is isomorphic to $\displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} $, the quotient map $\displaystyle S^1\times S^1 \to S^2$ collapsing $\displaystyle S^1 \vee S^1$ to a point is not null homotopic.

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

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

    Now we see that $\displaystyle 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 $\displaystyle H_2(S^1 \times S^1)$ is isomorphic to $\displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} $, the quotient map $\displaystyle S^1\times S^1 \to S^2$ collapsing $\displaystyle S^1 \vee S^1$ to a point is not null homotopic.

    Let $\displaystyle f:S^2 \to S^1\vee S^1$. The universal cover for $\displaystyle S^1\vee S^1$ is a $\displaystyle \mathbb{Re} \times \mathbb{Re}$ such that $\displaystyle g:\mathbb{Re} \times \mathbb{Re} \rightarrow S^1\vee S^1$ is a covering map. By using the lifting criterion (Hatcher p61), we have a lifting map $\displaystyle h:S^2 \rightarrow \mathbb{Re} \times \mathbb{Re}$ such that $\displaystyle f=gh$. Since $\displaystyle \mathbb{Re} \times \mathbb{Re}$ is contractible, h is nulhomotopic. It follows that $\displaystyle f:S^2 \to S^1\vee S^1$ 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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