TITLE: No retract $X \wedge \mathbb{R}P^2 \to X \wedge \mathbb{R}P^1$
QUESTION [9 upvotes]: Let $X$ be a finite CW complex, and suppose $\Sigma X \cong X \wedge \mathbb{R}P^1$ is not contractible.  By considering the fundamental group or otherwise, it is easy to see that there can be no retraction $\mathbb{R}P^2 \to \mathbb{R}P^1$.  But how about a retraction $X \wedge \mathbb{R}P^2 \to X \wedge \mathbb{R}P^1$?  I'm told that this too is impossible, but I don't know how to show it.  
Using the Kunneth formula for homology, we see that $\tilde{H}_*(X)$ is 2-torsion, and that the top homology group is zero.  But this does not rule out spaces like suspensions of even-dimensional real projective spaces.  I think we need to look at some other algebraic invariants/structures.  
(For those in the know, this question is related to the assertion that the suspension order of a finite complex is not two.)

REPLY [2 votes]: You can prove the $X = S^0$ case using cohomology, since $H^1(RP^2;Z) = 0$ and $ H^1(RP^1;Z) = Z$. We can apply this to the general case. Note that since $RP^1 = S^1,$ we have $ X \wedge RP^1 = \Sigma X$, and so let $k$ be minimal such that $\tilde{H}^k(X \wedge RP^1;Z) \cong \tilde{H}^{k-1}(X; Z) \neq 0$. If the splitting map existed, this nonzero group would split off of $\tilde{H}^k(X \wedge RP^2;Z)$. Since $RP^2$ is the cofiber of $\times 2:S^2 \rightarrow S^2$, we can smash with $X$ and consider the LES in cohomology associated to the cofiber sequence $X \wedge S^2 \rightarrow X \wedge S^2 \rightarrow X \wedge RP^2$. We find in it that $\tilde{H}^k(X \wedge RP^2;Z)$ sits in between $\tilde{H}^k(X \wedge S^2; Z) \cong \tilde{H}^{k-2}(X;Z) = 0$ and $\tilde{H}^{k-1}(X \wedge S^2; Z) \cong \tilde{H}^{k-3}(X;Z) = 0$. 
This implies that the reduced cohomology groups of $X$ are all zero. Hence $X$ is connected, and so $\Sigma X$ is a simply connected acyclic space, hence contractible.