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.

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    $\begingroup$ treat $H^*(\mathbb{CP}^n/\mathbb{CP}^k)$ as a subring of $H^*(\mathbb{CP}^n)$ (included via $\pi^*$) $\endgroup$ – user8268 Jun 23 '14 at 20:47
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    $\begingroup$ $\Bbb CP^2/\Bbb CP^1$ is homeomorphic to the $4$ sphere, and thus its cohomology ring is $\Bbb Z[t]/(t^2)$ with $t$ of degree $4$. If you use the long exact cohomology sequence $$\cdots\to \tilde{H}^*(\Bbb CP^4/\Bbb CP^1)\to H^*(\Bbb CP^4)\to H^*(\Bbb CP^1)\to \cdots$$ you can see that the cohomology ring of $\Bbb CP^4/\Bbb CP^1$ is $\Bbb Z[u,v]/(u^3,v^2,uv=vu=0)$ with $u$ in degree $4$ and $v$ in degree $6$. I'm not certain you can solve the problem only using the ring structure of cohomology. $\endgroup$ – Olivier Bégassat Jun 23 '14 at 21:28
  • $\begingroup$ @OlivierBégassat As you wrote $\mathbb{C}P^{2}/\mathbb{C}P^{1}$ is homeomorphic to $S^{4}$ which does not satisfies the fixed point property. As a possible alternative proof for the main question can one prove that $\mathbb{C}P^{4}/\mathbb{C}P^{1}$ has the fixed point property?(Note that a retract of a fixed point space is a fixed point space) $\endgroup$ – Ali Taghavi Jun 29 '14 at 5:48
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    $\begingroup$ For what it's worth, I believe that, care of the first answer here: math.stackexchange.com/questions/308318/…, that $\mathbb{C}P^3/\mathbb{C}P^1$ does retract onto $\mathbb{C}P^2/\mathbb{C}P^1$. Sketch of proof: According to the link, $\mathbb{C}P^3/\mathbb{C}P^1$ has the homotopy type of $S^4\vee S^6$, which clearly does retract on $S^4$. (I don't know how to make all the details work). The point is, this question could be quite subtle! $\endgroup$ – Jason DeVito Jan 30 '15 at 19:51
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    $\begingroup$ I'm confused- why doesn't Olivier's answer just answer the question. There is no nonzero element in degree 4 that squares to zero, so there's not even a nonzero map $H^*(\mathbb{C}P^2/\mathbb{C}P^1) \rightarrow H^*(\mathbb{C}P^4/\mathbb{C}P^1)$ let alone a retract. $\endgroup$ – Dylan Wilson Apr 26 '15 at 12:59

(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 $r$.


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