Show $\mathbb{CP^2/CP^1}$ is not a retract of $\mathbb{CP^4/CP^1}$. So, I have shown that the natural projection $\pi\colon \mathbb{CP^n}\rightarrow \mathbb{CP^n/CP^k}$ induces a monomorphism $\pi^*\colon H^*(\mathbb{CP^n/CP^k},\mathbb Z)\rightarrow H^*(\mathbb{CP^n},\mathbb Z) $. I would like to use this and the cohomology ring structure to show that we can't have a retract, but I am not exactly sure what the ring structure of $\mathbb{CP^2/CP^1}$ and  $\mathbb{CP^4/CP^1}$ are. 
I know that $H^*(\mathbb{CP^n},\mathbb Z) \cong \mathbb Z[\gamma]/(\gamma^{n+1})$, where $|\gamma|=2$, so is $H^*(\mathbb{CP^n/CP^k},\mathbb Z) \cong \mathbb Z[\gamma^{k+1},\ldots ,\gamma^{n}]/(\gamma^{n+1})$? 
This would give me $H^*(\mathbb{CP^2/CP^1},\mathbb Z) \cong \mathbb Z[\gamma^{2}]/(\gamma^{4})$ and $\pi^*(\gamma^2)\neq 0 \in H^*(\mathbb{CP^4/CP^1},\mathbb Z)$.
If this is true, then I think I can use that:
$0=\pi^*(0)=\pi^*(\gamma^2 \cup \gamma^2)=\pi^*(\gamma^2) \cup \pi^*(\gamma^2)\neq0$, which gives a contradiction.
 A: (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 $r$.
