Proving $H^k(M\times S^1)\cong H^k(M)\times H^{k-1}(M)$

Question

Let M be a smooth Manifold. Prove that $$H^k(M\times S^1)\cong H^k(M)\times H^{k-1}(M)$$.

I tried to use the Mayer-Vieotris sequence in the following way:

$U=\{(\alpha,\beta)\in M\times S^1|\alpha\in (0,\pi]\}\\ V=\{(\alpha,\beta)\in M\times S^1|\alpha\in (pi,2\pi)\}$

$$...\to H^{k-1}(U\cap V)\to H^k(M\times S^1)\to H^k(U)\oplus H^k(V)\to...$$

Since $ U\cap V=M$ we have:

$$...\to H^{k-1}(M)\to H^k(M\times S^1)\to H^k(U)\oplus H^k(V)\to...$$

If I had a torus I could use the cylinder cohomology, but with an arbitrary manifold, I am not seeing how to tackle the $H^k(U)\oplus H^k(V)$. Besides, I do not see the isomorphism coming.

Questions:

Can anyone help me finish the proof?

How can I restrict the Mayer Vietoris sequence to maps regarding the degrees k-1 and k?

Thanks in advance!

Answer 1

First of all, you should rather take something like $$U=\{(\alpha,\beta)\in M\times S^1|\beta\in (0,2\pi)\}\\ V=\{(\alpha,\beta)\in M\times S^1|\beta\in (-\pi,\pi)\}$$ or something similar for MV. Now you can notice that $U\simeq V\simeq M$ since $S^1\backslash\{pt\}\simeq\{pt\}$ and that $U\cap V\simeq M\sqcup M$ because $S^1$ minus two distinct points is homotopic to two points. After this, you need to use the form of the maps in the LES of MV plus exactness. Be careful about the end points of the sequence.