Using Kunneth theorem I would like to compute the following 4th homology:
$ H_4(S^2\times S^3/\mathbb{Z}_k,\mathbb{Z})\simeq H_4(S^2,\mathbb{Z})\otimes H_0(S^3/\mathbb{Z}_k,\mathbb{Z})\oplus H_3(S^2,\mathbb{Z})\otimes H_1(S^3/\mathbb{Z}_k,\mathbb{Z})\oplus H_2(S^2,\mathbb{Z})\otimes H_2(S^3/\mathbb{Z}_k,\mathbb{Z})\oplus H_1(S^2,\mathbb{Z})\otimes H_3(S^3/\mathbb{Z}_k,\mathbb{Z})\oplus H_0(S^2,\mathbb{Z})\otimes H_4(S^3/\mathbb{Z}_k,\mathbb{Z}) \simeq 0\otimes \mathbb{Z} \oplus 0 \otimes \mathbb{Z}_k \oplus \mathbb{Z}\otimes 0\oplus 0 \otimes \mathbb{Z} \oplus \mathbb{Z}\otimes 0 \simeq 0 $
Is it true that $0\otimes \mathbb{Z}\simeq 0$? I was thinking in terms of tensor products of matrices, but maybe I'm confused. What confuses me is that if I use the Poincaré duality and the universal coefficient theorem, it should be $\mathbb{Z}$. I mean
$ H_4(S^2\times S^3/\mathbb{Z}_k,\mathbb{Z})\simeq H^1(S^2\times S^3/\mathbb{Z}_k,\mathbb{Z})$
by Poincaré duality in 5d. While in general it should be true that
$H^q(X_D,\mathbb{Z})\simeq H_q^{\text{free}}(X_D,\mathbb{Z})\oplus \text{Tor}H_{q-1}(X_D,\mathbb{Z})$
Computing the first homology of my space:
$H_1(S^2\times S^3/\mathbb{Z}_k,\mathbb{Z})\simeq H_1(S^2,\mathbb{Z})\otimes H_0(S^3/\mathbb{Z}_k,\mathbb{Z})\oplus H_0(S^2,\mathbb{Z})\otimes H_1(S^3/\mathbb{Z}_k,\mathbb{Z})\simeq 0\otimes \mathbb{Z} \oplus \mathbb{Z}\otimes \mathbb{Z}_k\simeq \mathbb{Z}\oplus \mathbb{Z}_k$
so first of all I think that I don't understand the $\oplus$ neither, but I don't see my mistake. The first homology should be correct because I know it also for other reason, but then, the free part of $H_1$ is $\mathbb{Z}$ so the 4th homology should be $\mathbb{Z}$. Where am I mistaking?
Yes, that is true. Whenever you have a ring $R$ and a $R$-module $M$, then $M \otimes_R R \cong M$ or more generally $M/IM \cong M \otimes_R R/I$ for any ideal $I \subset R$. You can also see $0 \otimes_R M \cong 0$ directly as every elementary tensor is $0$. That is because $$0 \otimes m = (0+0) \otimes m = 0 \otimes m + 0 \otimes m,$$ such that we get $0 \otimes m = 0$ by subtracting $0 \otimes m$ from both sides.
Edit: I did not check your computation btw.