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

Question

I'm currently studying for my qualifying exam in algebraic topology, and I'm looking over old exam questions. The first part of the question was to show that $\mathbb{R} P^3$ is not homotopy equivalent to $\mathbb{R} P^2 \vee S^3$, which is fairly straightforward to do using either covering spaces or the cohomology rings.

The second part of the question was to show that $\Sigma(\mathbb{R} P^3)$ is not homotopy equivalent to $\Sigma(\mathbb{R} P^2 \vee S^3)$. We can't use the cohomology rings to tell them apart any more (as the cup product structure is trivial), and I don't know of any tools to compute homotopy groups of suspensions without some amount of 'connectivity'.

Based on a hint given in the question, I attemped to to distinguish these spaces by computing the Bockstein homomorphism $\beta : H^*(X,\mathbb{Z}_2) \to H^{*+1}(X,\mathbb{Z}_2)$ for both spaces coming from the short exact sequence $0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0$, and using the fact that the Bockstein commutes with the suspension isomorphism on cohomology. However for both spaces I computed $\beta_1 : H^1(X,\mathbb{Z}_2) \to H^2(X,\mathbb{Z}_2)$ was an isomorphism, while $\beta_2 : H^2(X,\mathbb{Z}_2) \to H^3(X,\mathbb{Z}_2)$ was the zero map.

Is this the way I should be trying to distinguish these spaces, or is there something more obvious?

Answer 1

The suspensions $\Sigma({\mathbb R}P^3)$ and $\Sigma({\mathbb R}P^2\vee S^3)$ are not homotopy equivalent and can be distinguished by the Pontryagin square $H^2(X;{\mathbb Z}_2)\to H^4(X;{\mathbb Z}_4)$. One reference for this is the book "Combinatorial Homotopy and 4-Dimensional Complexes" by Hans Baues, where this fact is stated in an example on page 20 and proved later in the book using the theory developed there. Baues also cites Hilton's book "Homotopy Theory and Duality" for another proof. I would guess that this fact was known to J.H.C.Whitehead when he wrote his two papers "On simply connected 4-dimensional polyhedra" (1949) and "A certain exact sequence" (1950) which develop a general classification of simply-connected 4-complexes. It may also have been known earlier to Pontryagin. I wonder if there are other modern expositions of this material.

If one suspends again to $\Sigma^2({\mathbb R}P^3)$ and $\Sigma^2({\mathbb R}P^2\vee S^3)$ then these spaces become homotopy equivalent. The attaching map of the top cell of $\Sigma^n{\mathbb R}P^3$ gives an element of $\pi_{n+2}(\Sigma^n{\mathbb R}P^2)$ and for $n\geq 2$ this group can be computed to be ${\mathbb Z}_2$ generated by the suspended Hopf map $S^{n+2}\to S^{n+1}\subset \Sigma^n{\mathbb R}P^2$ by looking at the long exact sequence of (stable) homotopy groups for the cofiber sequence $S^{n+1}\to S^{n+1}\to \Sigma^n{\mathbb R}P^2$ where the first map has degree 2. Thus there are two possible attaching maps for the top cell of $\Sigma^n{\mathbb R}P^3$, up to homotopy, and they are distinguished by whether the Steenrod square $Sq^2$ is trivial on $H^{n+1}(\Sigma^n{\mathbb R}P^3;{\mathbb Z}_2)$ or not, using the fact that $Sq^2$ is nontrivial on the mapping cone of the suspended Hopf map. For ${\mathbb R}P^3$ the action of $Sq^2$ is trivial for dimension reasons, hence the action is also trivial for $\Sigma^n{\mathbb R}P^3$ so the top cell must be attached trivially.

Some further interesting information: The Hopf map $S^3\to S^2$ has infinite order, but when when we attach a 3-cell to $S^2$ by a map of degree 2 to form $\Sigma{\mathbb R}P^2$, the Hopf map still generates $\pi_3(\Sigma{\mathbb R}P^2)$ but now has order 4, rather than 2 as one might guess. More generally, if we attach the 3-cell by a map of degree $d$ then in the resulting 3-complex the Hopf map has order $2d$ if $d$ is even, but order $d$ if $d$ is odd. Going back to the case at hand, we have $\pi_3(\Sigma{\mathbb R}P^2)={\mathbb Z}_4$ and the attaching map of the 4-cell of $\Sigma{\mathbb R}P^3$ is the element of order 2 in this ${\mathbb Z}_4$. (It can't be a generator since $Sq^2$ acts trivially.)