How can i calculate $H_i(S^2 \times S^1)$?

Question

How can i calculate $H_i(S^2 \times S^1)$? I'm ultimately trying to show that $S^2 \times S^1$ is not homeomorphic to $\mathbb{R}P^2 - \{p\} \cong \mathbb{R}P$

Answer 1

Building on ThorbenK's suggestion, set $U=D_1\times S^1$, $V=D_2\times S^1$ where each $D_j$ is a (slightly fattened) hemisphere (i.e. a disc, so, contractible). Obviously $U\cap V= (D_1\cap D_2)\times S^1=(-c,c)\times S^1\times S^1$, which retracts to a $2$-torus.

So $$H_*(U)=\mathbb{Z},\mathbb{Z}, 0,0$$ $$H_*(V)=\mathbb{Z},\mathbb{Z}, 0,0$$ $$H_*(U\cap V)=\mathbb{Z},\mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z},0$$

Set $X=S^2\times S^1=U\cup V$. By the Mayer-Vietoris sequence $$0\to H_3(X)\to H_2(U\cap V)\to H_2(U)\oplus H_2(V)\to H_2(X)\to H_1(U\cap V)\to H_1(U)\oplus H_1(V)\to H_1(X)\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$$

Or, as we already know the homology groups of $U$, $V$ and $U\cap V$, $$0\to H_3(X)\to \mathbb{Z}\to 0 \to H_2(X)\to \mathbb{Z}^2\to \mathbb{Z}^2\to H_1(X)\to \mathbb{Z}\to \mathbb{Z}^2\to H_0(X)\to 0$$

We immediately see that $H_3(X)=\mathbb{Z}$.

The maps $\rho_j:H_j(U\cap V)\to H_j(U)\oplus H_j(V)$ are given by $((i_1)_*, (i_2)_*)$ where $i_1:U\cap V\to U$ and $i_2:U\cap V\to V$ are the inclusions.

So, at the level $0$, homology is generated by the class of a point; given $[p]\in H_0(U\cap V)$, then obviously $\rho_0([p])=([i_1(p)], [i_2(p)])$, so $\rho_0:\mathbb{Z}\to\mathbb{Z}^2$ is $\rho_0(x)=(x,x)$, hence the cokernel is $$H_0(X)=\mathbb{Z}$$

As $\rho_0$ is injective, the map $H_1(X)\to H_0(U\cap V)$ has image $0$ and its kernel, which is $H_1(X)$, is the image of $\sigma:H_1(U)\oplus H_1(V)\to H_1(X)$, which is then surjective. Therefore $H_1(X)=(H_1(U)\oplus H_1(V))/\rho_1(H_1(U\cap V))$.

Moreover, consider two generators of $H_1(U\cap V)$: the cycle $[\gamma]$ given by the $S^1$ that is also a factor of $X$ and the cycle $[\gamma']$ given by the intersection between the two disc $D_1$ and $D_2$ (the equator of $S^2$). We have that $(i_j)_*([\gamma])=[i_j\circ\gamma]$, whereas $(i_j)_*[\gamma']=0$. So $\rho_1:\mathbb{Z}^2\to\mathbb{Z}^2$ is $\rho_1(x,y)=(x,x)$.

Its kernel is $\mathbb{Z}$ and it is homomorphic to $H_2(X)$.

On the other hand, its image is $\mathbb{Z}$, therefore $H_1(X)=\mathbb{Z}^2/(1,1)\mathbb{Z}=\mathbb{Z}$.

In conclusion $$H_*(X)=\mathbb{Z}, \mathbb{Z},\mathbb{Z},\mathbb{Z}$$