Homotopy equivalence of spaces involving spheres and projective space

Question

Let $Y = S^2 \times \mathbb{R} P^3$ and $X = S^3 \times \mathbb{R} P^2$. Let $y \in Y$ and $x \in X $ be basepoints.

$(a) $ Compute $\pi_{1}(Y,y) $ and $\pi_{1}(X,x)$.

$(b)$ Show that $\pi_{k}(Y,y) = \pi_{k}(X,x)$ for all integers $k \geq 1$

$(c)$ Are these spaces homotopy equivalent?

I know that the fundamental group of $S^2$ is $0$. But,

• how do I compute the fundamental group of it's product with $\mathbb{R} P^3$? Do I have to use the Van Kampen theorem? I'm also confused about how to compute $\pi_{1}(X,x)$.
• For part$(c)$, my intuition is that they are indeed homotopic but I'm not sure how to show that explicitly.
• I don't know where to start with part$(b)$ either.

Answer 1

For a), Mark Saving's comment gives that $\pi_1(S^3\times \mathbb{R}P^2)\cong\pi_1(\Bbb RP^2)\cong\pi_1(\Bbb RP^3)\cong\pi_1(S^2\times \Bbb RP^3)$, so they are isomorphic to $\Bbb Z/2$.

For b), we already showed that their fundamental groups are isomorphic in a), and we know that the universal covering $p:E\to B$ induces isomorphism on $p_\ast:\pi_n(E)\to\pi_n(B)$ for all $n\ge 2$, so let's construct their universal cover. Suppose $p_1:S^2\to S^2$ and $p_2:S^3\to\Bbb RP^3$ are two covering maps, then $p_1\times p_2$ gives the desired universal cover of $S^2\times \Bbb RP^3$. Similarly, $S^2\times S^3$ is also the universal cover of $S^3\times \Bbb RP^2$. They admit the same universal cover, and so their higher homotopy groups are isomorphic.

For c), yours is a good guess, but unfortunately it is not the case. These two spaces are not homotopy equivalent despite their isomorphic homotopy groups in all dimensions. You can deduce it by computing the top homology group using Künneth formula, which gives us \begin{align} H_5(S^2\times \Bbb RP^3;\Bbb Z)\cong\Bbb Z\\ H_5(S^3\times \Bbb RP^2;\Bbb Z)\cong0 \end{align}