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 $\mathbb{R} P^2 \vee S^3$.


  • 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!

  • $\begingroup$ Cup product structure on cohomology would be the way to go. $\endgroup$ – Lee Mosher Jan 3 '15 at 16:14
  • $\begingroup$ ...or compute $\pi_2$ $\endgroup$ – Grigory M Jan 3 '15 at 17:06
  • $\begingroup$ 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! $\endgroup$ – 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 $|z|=3$.

| cite | improve this answer | |
  • $\begingroup$ This doesn't seem to be correct: a vee product splits over the reduced cohomology ring, not over the non-reduced one. $\endgroup$ – Markus Klyver Aug 17 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.