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We are supposed to show that the spaces $X= \mathbb{R}P^2$ and $ Y= S^2 \times \mathbb{R}P^\infty$ have isomorphic homotopy groups but are not homotopy equivalent.
I already showed that all homotopy groups are isomorphic, but struggled to find a reason why $$X \sim_h Y $$ cannot be the case. Any help is appreciated. We have not introduced homology so far.

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  • $\begingroup$ Compute the homology of both spaces with coefficients in $F_2$. $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '17 at 16:10
  • $\begingroup$ yeah as I clearly stated, no homology so far @MarianoSuárez-Álvarez $\endgroup$ – Simonsays Nov 24 '17 at 16:18
  • $\begingroup$ Could you state the idea of your proof that they have the same homotopy groups. Do you use a fibration ? $\endgroup$ – Rene Schipperus Nov 24 '17 at 16:22
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    $\begingroup$ They have the same universal cover (up to homotopy) and the same fundamental group. $\endgroup$ – Giuseppe Bargagnati Nov 24 '17 at 16:25
  • $\begingroup$ What have you introduced so far? What can be used to prove the statement? $\endgroup$ – Giuseppe Bargagnati Nov 24 '17 at 16:26
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They can be distingushed by the action of $\pi_1 \cong \mathbb{Z}_2$ on $\pi_2 \cong \mathbb{Z}$. This action is trivial for $S^2 \times \mathbb{RP}^{\infty}$ but is given by inversion for $\mathbb{RP}^2$, because the antipode map on $S^2$ acts by $-1$ on $\pi_2$.

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  • $\begingroup$ thank you for the answer! And how do you conclude that they are not homotopic ? $\endgroup$ – Simonsays Nov 25 '17 at 9:04
  • $\begingroup$ I assume that by the action of $\pi_1$ on $\pi_2$ you mean the deck transformations? $\endgroup$ – Simonsays Nov 25 '17 at 9:25
  • $\begingroup$ @Simonsays: yes, you can think of it that way. The action of $\pi_1$ on the higher homotopy groups is a homotopy invariant. $\endgroup$ – Qiaochu Yuan Nov 25 '17 at 18:06

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