On the homotopy type of $\mathbb R P^3$ and $\mathbb R P^2 \vee S^3$ I want to prove $\mathbb R P^3$ and $\mathbb R P^2 \vee S^3$ are not homotopy equivalent. After some work I've shown that they have the same fundamental group and homology, and $X = S^3 \vee S^2 \vee S^3$ is a universal covering for $\mathbb R P^2 \vee S^3$ (just attach one copy of $S^3$ to each one of the poles in $S^2$ and you get a $2$-fold covering).
I know this question has already been answered here before, but everyone I ask uses some variation of an argument based on the cohomology ring of these spaces, and I don't know nothing about cohomology or cup products yet (neither has the text book said anything about it when this exercise was proposed). Apparently the second homotopy group of $\mathbb R P^2 \vee S^3$ is $\mathbb Z$ while we know $\pi_2(\mathbb R P^3)$ is trivial, so that's  the way to go. Unfortunately, I do know how to compute the homotopy groups of the wedge sum $X = S^3 \vee S^2 \vee S^3$ as I can't use van Kampen, so I'm honestly quite lost here.
Could someone show me how does one prove they are not homotopy equivalent using just the homology and maybe homotopy and the covering spaces?
 A: As you observed, $$\pi_2(\Bbb{RP}^3) = \pi_2(\widetilde{\Bbb{RP}^3}) = \pi_2(S^3) = 0.$$ On the other hand, the wedge of two spaces $X \vee Y$ retracts onto either of the summands, by sending every point in $Y$ (or $X$, depending) to the common basepoint. In particular, $\pi_2(X) \to \pi_2(X \vee Y)$ is injective, because composing it with the map $\pi_2(X \vee Y) \to \pi_2(X)$ induced by the retraction gives the identity. So $\Bbb Z = \pi_2(\Bbb{RP}^2) \hookrightarrow\pi_2(\Bbb{RP}^2 \vee S^3)$, and that's enough to get a contradiction.
A: Here's another solution.
If $ \mathbb R \mathbb P^3 \simeq \mathbb R \mathbb P ^2 \lor S^3 $ then their universal covers are also homotopy equivalent. The first one has universal cover $ S^3$ while the second has universal cover $S^3 \lor S^2 \lor  S^3 $ and hence these two should be homotopy equivalent.
But $ H_2 ( S^3 )= 0 \neq H_2 ( S^3 \lor S^2 \lor  S^3 ) = \mathbb Z $
This is a contradiction. 
A: Hint: look the multiplicative structure of the cohomology groups. $H^*(\mathbb{R}P^3,\mathbb{Z}/2)$ is generated by an element of degree $1$, it is not the case of $H^*(\mathbb{R}P^2\vee S^3,\mathbb{Z}/2)$.
