Ytukyg

So, I have shown that the natural projection $picolon mathbb{CP^n}rightarrow mathbb{CP^n/CP^k}$ 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.

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

Then we have the formula $rcirc i = Id_{mathbb{C}P^2/mathbb{C}P^1}$. 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, $kneq 0$.

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 $r$.