Show $\mathbb{CP^2/CP^1}$ is not a retract of $\mathbb{CP^4/CP^1}$.

(I'm just answering this to get it off the unanswered list. It's CW because I'm not doing anything but expanding on information in the comments).

As Olivier notes, $H^\ast(\mathbb{C}P^4/\mathbb{C}P^1)\cong \mathbb{Z}[u,v]/u^3 = v^2 = uv = 0$ where $|u| = 4$ and $|v| = 6$. This follows, as Olivier says, via the long exact sequence of the pair $(\mathbb{C}P^1,\mathbb{C}P^4)$, together with the fact that the inclusion map $\mathbb{C}P^1\rightarrow \mathbb{C}P^4$ induces an isomorphism on $H^2$ and $H^0$, but is other wise the zero map.

Further, by the same technique, one easily sees that $H^\ast(\mathbb{C}^2/\mathbb{C}P^1)\cong \mathbb{Z}[t]/t^2$ where $|t| = 4$ (or one can also use the usual cell structure on $\mathbb{C}P^2$ to see that $\mathbb{C}P^2/\mathbb{C}P^1$ is homeomorphic to $S^4$).

Now, let $i:\mathbb{C}P^2/\mathbb{C}P^1\rightarrow \mathbb{C}P^4/\mathbb{C}P^1$ be the inclusion. Assume for a contradiction that $r:\mathbb{C}P^4/\mathbb{C}P^1\rightarrow \mathbb{C}P^2/\mathbb{C}P^1$ is a retraction.

Then we have the formula $r\circ i = Id_{\mathbb{C}P^2/\mathbb{C}P^1}$. In particular, $i^\ast r^\ast:H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)\rightarrow H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)$ is an isomorphism, which implies that $r^\ast:H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)\rightarrow H^\ast(\mathbb{C}P^4/\mathbb{C}P^1)$ is an injection. In particular, $r^\ast$ is not the zero map on $H^4$.

Now consider $r^\ast(t)$. It must be some multiple of $u$, $r^\ast(t) = ku$, because $u$ generates $H^4(\mathbb{C}P^4/\mathbb{C}P^1)\cong \mathbb{Z}$. Because $r^\ast$ is not the zero map, $k\neq 0$.

But now we have a contradiction: $t^2 = 0$ so $0 = r^\ast(t^2) =r^\ast(t)^2 = (ku)^2 = k^2 u^2$ which forces $k = 0$. Thus, there can not be a retraction $r$.