Finding when $\mathbb{C}P^n / \mathbb{C}P^{n-2}$ is homotopy equivalent to $S^{2n} \vee S^{2n-2}$ using Steenrod squares Let $\mathbb{C}P^n$ denotes the complex projective space with real manifold dimension $2n.$  My question is for which values of $n$, the spaces $\mathbb{C}P^n / \mathbb{C}P^{n-2}$ is homotopy equivalent to $S^{2n} \vee S^{2n-2}?$
Attempt: The cohomology $H^{k}(\mathbb{C}P^n / \mathbb{C}P^{n-2}, \mathbb{Z}/2) \cong \mathbb{Z}/2$ for $n = 0, \; 2n-2, \; 2n$ and zero otherwise. Also the cohomology $H^{\ast}(S^{2n} \vee S^{2n-2})$ is same as the previous one. So we compute the Steenrod squares on the both the spaces. For the space $\mathbb{C}P^n / \mathbb{C}P^{n-2}$ we consider the map $q \colon \mathbb{C}P^n  \to \mathbb{C}P^n / \mathbb{C}P^{n-2}.$ Let the cohomology ring of $\mathbb{C}P^n $ is given by $\mathbb{Z}/2[x]/(x^{n+1})$ with $|x|=2.$ Also consider $H^{2n-2}(\mathbb{C}P^n / \mathbb{C}P^{n-2}) \cong \mathbb{Z}/2\{y\}$ and $H^{2n}(\mathbb{C}P^n / \mathbb{C}P^{n-2}) \cong \mathbb{Z}/2\{z\}$. Then $q^{\ast}(y) =x^{n-1}$ and $q^{\ast}(z) = x^n$ because $q^{\ast}$ is isomorphism at degrees $2n-2$ and $2n.$
$$q^*(Sq^2(y)) = Sq^2(x^{n-1})= nx^n = q^*({nz})$$ Therefore $Sq^2(y) =nz.$ But $Sq^2$ is zero on the space $S^{2n} \vee S^{2n-2}.$ So if $n$ is odd then these spaces are not homotopy equivalent.
Can someone help me to answer my question in details?
Thank you so much.
 A: Your calculation is ever-so-slightly incorrect. Actually, the formula is $\text{Sq}^2(x^{n-1}) = (n-1)x^{n}$. This is derived inductively; the problem with your formula is that its value on $x$ should be $x^2$, not $2x^2$, by the cup product of $\Bbb{CP}^n$.
So what one actually learns is that $\Bbb{CP}^n/\Bbb{CP}^{n-2} \not\simeq S^{2n} \vee S^{2n-2}$ when $n$ is even. (As a sanity check: $\Bbb{CP}^2 \not\simeq S^4 \vee S^2$, by looking at the cup product.)
How do we deal with the odd case? Well, $\Bbb{CP}^n/\Bbb{CP}^{n-2}$ is a CW complex with a cell of dimension $(2n-2)$ and $2n$, so it must be $X^{2n}_f = S^{2n-2} \cup_f D^{2n}$, where $f \in \pi_{2n-1}(S^{2n-2})$. When $n>2$ (which we may well assume, since we have easily dealt with the case $n=2$), the latter group is $\Bbb Z/2.$ So suppose $f$ is nonzero. If we can prove that $X^{2n}_f$ has nontrivial Steenrod squares, that will prove that when $n$ is odd, $\Bbb{CP}^n/\Bbb{CP}^{n-2} \simeq S^{2n} \vee S^{2n-2}$.
This is actually completely straightforward. In full generality, if $X$ is a CW complex, $Y \subset Z$ another, and $X \cup_f Z$ an adjunction space defined by a map $f: Y \to X$, then $\Sigma(X \cup_f Z) = \Sigma X \cup_{\Sigma f} \Sigma Z$. Applied to this case, we see that $X^{n+4}_f = \Sigma^n \Bbb{CP}^2$, because the nonzero element $f \in \pi_{n+3}(S^{n+2})$ is the $n$-fold suspension of the Hopf map. By suspension invariance of Steenrod squares, the Steenrod square map on $X^{n+4}_f$ is nonzero. Thus the desired result follows.
