# Math Help - Homology & covering spaces

1. ## 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.

2. Originally Posted by harbottle
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.

3. Originally Posted by aliceinwonderland
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!