$RP^n\times S^m$ and $RP^m\times S^n$ not homotopy equivalent ($n\not=m)$ I'm trying to see why $RP^n\times S^m$ and $RP^m\times S^n$ are not homotopy equivalent. The tricky part is that they have the same homotopy groups(if n,m>1). I have spent countless hours on this question without any luck, I know that if $n<m$ and $n$ is even, then we can consider the action of $\pi_1$ on $\pi_n$:
$\pi_1(RP^n\times S^m)$ acting on $\pi_n(RP^n\times S^m)$ is nontrivial but
$\pi_1(RP^m\times S^n)$ acting on $\pi_n(RP^m\times S^n)$ is trivial. 
But I couldn't think of a way to deal with the case $n$ is odd.
We haven't introduced homology/cohomology yet, otherwise I would have just used the Künneth formula.
Any hint is appreciated, if possible I would like to complete the argument myself. Thanks a lot!
 A: If $m=n$ then the spaces are homotopy equivalent. If $m,n$ are different, say $n>m$, then $S^m$ is not a retract (up to homotopy) of $\mathbb{R}P^m\times S^n$, whilst it is obviously a retract of $\mathbb{R}P^n\times S^m$. To see the first statment this observe that
$$[S^m,\mathbb{R}P^m\times S^n]\cong [S^m,\mathbb{R}P^m]\times[S^m,S^n]=[S^m,\mathbb{R}P^m]$$
since $[S^m,S^n]$ is trivial by cellular approximation. Hence any map $S^m\rightarrow \mathbb{R}P^m\times S^n$ factors up to homotopy through the inclusion $\mathbb{R}P^m\hookrightarrow \mathbb{R}P^m\times S^n$. Therefore if $S^m$ is a retract of $\mathbb{R}P^m\times S^n$, then it is also a retract of $\mathbb{R}P^m$, and we know that this is not true. Hence the statement.
On the other hand, if we could find a homotopy equivalence between $\mathbb{R}P^m\times S^n$ and $\mathbb{R}P^n\times S^m$ then we could simply insert it and its inverse between the inclusion $S^m\hookrightarrow \mathbb{R}P^n\times S^m$ and the projection $\mathbb{R}P^n\times S^m\rightarrow S^m$ to get a homotopy retract of $S^m$ off of $\mathbb{R}P^m\times S^n$. Hence such a homotopy equivalence cannot exist.
