A few questions on $S^3 \times \mathbb{R}P^2$ and $\mathbb{R}P^3 \times S^2$.

Question

Consider $S^3 \times \mathbb{R}P^2$ and $\mathbb{R}P^3 \times S^2$. Both are smooth $5$-manifolds.

1. Why do they both have fundamental group $\mathbb{Z}/2$?

Progress. I can show that $\pi_1(X \times Y)$ is isomorphic to $\pi_1(X) \times \pi_1(Y)$ if $X$ and $Y$ are path-connected.

2. Why do they both have universal cover $S^3 \times S^2$?

Progress. I can show that the universal cover of $\mathbb{R}P^n$ is $S^n$.

3. Why do they have different homology groups?

Progress. Only $\mathbb{R}P^3 \times S^2$ is orientable since $\mathbb{R}P^3$ is orientable but $\mathbb{R}P^2$ is not, so perhaps they have different values on $H_k$ for some $k$?

Answer 1

1. All four spaces are path-connected, and one has $\pi_1(\mathbb{RP}^2) = \pi_1(\mathbb{RP}^3) = \mathbb{Z}/2\mathbb{Z}$ while $\pi_1(S^2) = \pi_1(S^3) = 0$.

2. If $\tilde X \to X$ and $\tilde Y \to Y$ are universal covers, then the product $\tilde{X} \times \tilde{Y} \to X \times Y$ is the universal cover of $X \times Y$. This is more or less immediate: the product of covering maps is a covering map, and $\pi_1(\tilde X \times \tilde Y) = \pi_1(\tilde X) \times \pi_1(\tilde Y) = 0 \times 0 = 0$.

3. If you remember, a closed $n$-manifold $M$ is orientable iff $H_n(M) = \mathbb{Z}$, and otherwise $H_n(M) = 0$. Thus $H_2(\mathbb{RP}^2) = 0$ and $H_3(\mathbb{RP}^3) = \mathbb{Z}$. By Künneth's formula and the well-known description of the homology of $S^n$, it follows that $H_5(\mathbb{RP}^3 \times S^2) = \mathbb{Z}$ while $H_5(S^3 \times \mathbb{RP}^2) = 0$.