# Show that $\mathbb{R} P^3$ is not homotopy equivalent to $\mathbb{R} P^2 \vee S^3$.

I'm studying for an oral qualifying exam in algebraic topology, going through questions in various tests published on the interwebs. Here's a rather straightforward question from this exam that is giving me trouble.

Show that $\mathbb{R} P^3$ is not homotopy equivalent to $\mathbb{R} P^2 \vee S^3$.

Attempts:

• Use the fundamental groups of the spaces. This doesn't seem to work since $\pi_1(\mathbb{R} P^3) = \mathbb{Z}_2$, and by Van Kampen's, $\pi_1(\mathbb{R}P^2 \vee S^3) = \pi_1(\mathbb{R}^2) \ast \pi_1(S^3) = \mathbb{Z}_2$, and so the difference is not seen here.
• Use corresponding homology groups of the spaces. Intuitively, it seems like $H_3$ should do the trick, but, using the formula on Wikipedia for computing $H_3$ of each space and the wedge axiom for homology, I still don't see a difference at this level. Also, since $\pi_1$ is already abelian for both spaces, $H_1$ also doesn't see a difference.

What would be a more productive approach here? Does it generalize at all so that I can see when another approach might be more fruitful?

Thank you all for your suggestions!

• Cup product structure on cohomology would be the way to go. – Lee Mosher Jan 3 '15 at 16:14
• ...or compute $\pi_2$ – Grigory M Jan 3 '15 at 17:06
• Ah, okay. I haven't gone back over cup product yet, so I'll revisit this once I'm a bit farther along. Thank you! – Rachel Jan 3 '15 at 17:22

Hint: the cohomology rings with $\mathbf{Z}/2\mathbf{Z}$ coefficients are not isomorphic, in fact they are $H^\ast(\mathbf{RP}^3,\mathbf{Z}/2\mathbf{Z})=\mathbf{Z}/2\mathbf{Z}[x]/(x^4)$ with $|x|=1$ and $H^\ast(\mathbf{RP}^2\vee S^3,\mathbf{Z}/2\mathbf{Z})=\mathbf{Z}/2\mathbf{Z}[y]/(y^3)\times\mathbf{Z}/2\mathbf{Z}[z]/(z^2)$ with $|y|=1$ and $|z|=3$.