Prove that $\Sigma \mathbb CP^2\cong S^3\vee S^5$

Question

My question: Prove that $\Sigma \mathbb CP^2\cong S^3\vee S^5.$ The following is a proof of this result.

As well known, the complex projective plane $\mathbb{C}P^2$ can be obtained from the sphere $\mathbb S^2$ by attaching a 4-dimensional cell along the Hopf map $\rho: \mathbb S^3\to \mathbb S^2.$ In other words, we have the following CW complex structure for $\mathbb{C}P^2$: $$ \mathbb{C}P^2 = (\mathbb S^2\sqcup \mathbb D^4)/\sim\ = \mathbb S^2\cup_{\rho} \mathbb D^4$$ where the identification is given by $x\sim \rho (x)$ for $x\in \mathbb S^3$. Taking the suspension of $\mathbb{C}P^2$, we have $$\Sigma\mathbb{C}P^2 = (\mathbb S^3\sqcup S^1\sqcup \mathbb D^5)/\sim\ = \mathbb S^3\cup_{\rho} (\mathbb S^1\sqcup \mathbb D^5)$$ where the identification is given by identifying the basepoints of $\mathbb S^3$ and $\mathbb S^1$ and collapsing the $\mathbb S^1$ factor to a point. We can further simplify this space by noting that $\mathbb S^1\sqcup \mathbb D^5$ is homeomorphic to $\mathbb S^5$ via the map $(\theta,x)\mapsto(\cos\theta x,\sin\theta x)$ for $\theta\in[0,\pi]$ and $x\in \mathbb D^5$, which identifies the equator of $\mathbb S^5$ with $\mathbb S^1\subseteq \mathbb S^5$. Under this homeomorphism, the Hopf map $\rho$ becomes a map $g: \mathbb S^3\to \mathbb S^5$ that sends the equator of $\mathbb S^3$ to the equator of $\mathbb S^5$ and the north and south poles of $\mathbb S^3$ to the two points where the equator of $\mathbb S^5$ intersects the $\mathbb S^3$ factor of $\Sigma\mathbb{C}P^2$. Thus, we must have $\Sigma\mathbb{C}P^2\cong \mathbb S^3\cup_g \mathbb S^5\cong \mathbb S^3\vee \mathbb S^5.$

Give me your comments on this proof. Does it have any errors? I appreciate someone has another idea that allows proof.

Answer 1

Unfortunately, your proof can't be right since $\Sigma \mathbb{C}P^2$ is not homotopy equivalent to $S^3 \vee S^5$. This can be seen since the Steenrod square $\mathrm{Sq}^2$ is nontrivial on $H^3(\Sigma \mathbb{C}P^2; \mathbb{Z}/2)$. This is because there are in $H^2(\mathbb{C}P^2; \mathbb{Z}/2)$ with nontrivial squares.

The error in your proof is that you don't understand what attaching maps or CW structures on suspensions are. Particularly,

the Hopf map $\rho$ becomes a map $g:S^3→S^5$

does not make sense since (a suspension of) the Hopf map should decrease degree by 1. As well,

$$\Sigma\mathbb{C}P^2 = (\mathbb S^3\sqcup S^1\sqcup \mathbb D^5)/\sim\ = \mathbb S^3\cup_{\rho} (\mathbb S^1\sqcup \mathbb D^5)$$

is not a correct CW decomposition. If one is taking an unreduced suspension, the $S^1$ should be an interval. If one is taking a reduced suspension, it should just not appear.