# Homology & covering spaces

• Oct 25th 2009, 04:08 AM
harbottle
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.
• Oct 25th 2009, 02:44 PM
aliceinwonderland
Quote:

Originally Posted by harbottle
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$$\displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1) \rightarrow H_1(S^1 \times S^1) \cdots. By the Hurewicz theorem, the first homology group of a torus \displaystyle H_1(S^1 \times S^1) is isomorphic to the fundamental group of a torus \displaystyle \pi_1(S^1 \times S^1) = \mathbb{Z} \times \mathbb{Z}. By applying the Mayer-Vietoris sequence, we have \displaystyle H_2(S^1 \vee S^1) = 0 and \displaystyle H_1(S^1 \vee S^1) = \mathbb{Z} \times \mathbb{Z}. Now we see that \displaystyle H_2(S^1 \times S^1) \rightarrow H_2(S^1 \times S^1, S^1 \vee S^1) is an isomorphism in the above long exact sequence. Since \displaystyle H_2(S^1 \times S^1) is isomorphic to \displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} , the quotient map \displaystyle S^1\times S^1 \to S^2 collapsing \displaystyle S^1 \vee S^1 to a point is not null homotopic. Quote: On the other hand, show via covering spaces that any map \displaystyle S^2 \to S^1\vee S^1 is nullhomotopic. Let \displaystyle f:S^2 \to S^1\vee S^1. The universal cover for \displaystyle S^1\vee S^1 is a \displaystyle \mathbb{Re} \times \mathbb{Re} such that \displaystyle g:\mathbb{Re} \times \mathbb{Re} \rightarrow S^1\vee S^1 is a covering map. By using the lifting criterion (Hatcher p61), we have a lifting map \displaystyle h:S^2 \rightarrow \mathbb{Re} \times \mathbb{Re} such that \displaystyle f=gh. Since \displaystyle \mathbb{Re} \times \mathbb{Re} is contractible, h is nulhomotopic. It follows that \displaystyle f:S^2 \to S^1\vee S^1 is nullhomotopic. • Oct 26th 2009, 06:11 PM harbottle Quote: Originally Posted by aliceinwonderland Consider this long exact sequence, \displaystyle \cdots \rightarrow H_2(S^1 \vee S^1) \rightarrow H_2(S^1 \times S^1) \rightarrow$$\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!