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Is the stunted complex projective space $\mathbb{C}P^{n}/\mathbb{C}P^{m}$ coreducible for $m= 1,3$ and $n\geq m+2$ ? $\mathbb{C}P^{n}/\mathbb{C}P^{m}$ is said to be coreducible if $\exists$ a map $f:\mathbb{C}P^{n}/\mathbb{C}P^{m}\rightarrow S^{2m+2}$ such that the composition of $f$ with the inclusion $i:S^{2m+2}\hookrightarrow \mathbb{C}P^{n}/\mathbb{C}P^{m}$ is of degree one.

Also is it true that the stunted quaternionic projective space $\mathbb{H}P^{n}/\mathbb{H}P^{1}$, $n\geq 3$ coreducible?

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  • $\begingroup$ Excuse my ignorance, what definition of degree are you using here? $\endgroup$ May 26, 2017 at 16:39
  • $\begingroup$ If $h:S^{k}\rightarrow S^{k}$ then by deg$f$, I mean the usual definition of degree, i,e the integer '$d$' such that $f_{\ast}(\alpha)= d\alpha$ where $f_{\ast}:H_{k}(S^{k})\rightarrow H_{k}(S^{k})$ and $\alpha$ is a generator of $H_{k}(S^{k})$. $\endgroup$ May 27, 2017 at 8:47
  • $\begingroup$ I see, so you want $f\circ i$ to have degree one? The first time I read it, I thought you were asking about $i\circ f$, hence my question about degree. $\endgroup$ May 27, 2017 at 11:25
  • $\begingroup$ Definitely I am asking about 'f' composed with 'i'. Above I gave the defn of usual degree . $\endgroup$ May 27, 2017 at 12:44

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If $n\ge 2m+2$ then the generator of $H^{2m+2}({\mathbb C}P^n/{\mathbb C}P^m)$ has a non-trivial cup square, which easily implies that the space is not co-reducible. This leaves open the cases when $(m,n)$ is one of the following pairs $(1,3), (3,5), (3,6)$, or $(3,7)$. Let us consider them one by one.

Consider the action of mod $3$ Steenrod algebra on $H^*({\mathbb C}P^\infty; {\mathbb Z}/3)\cong {\mathbb F}_3[x]$. An easy calculation shows that ${\mathcal P}^1(x^4)=x^6$. It follows that in ${\mathbb C}P^{6}/{\mathbb C}P^3$ and ${\mathbb C}P^7/{\mathbb C}P^3$ the bottom cell (in dimension 8) is connected to the cell in dimension 12 by a Steenrod operation, and so these spaces are not co-reducible.

${\mathbb C}P^3/{\mathbb C}P^1$ and ${\mathbb C}P^5/{\mathbb C}P^3$ are two-cell complexes determined by attaching maps $S^5\to S^4$ and $S^9\to S^8$ respectively. Both sets of homotopy classes $[S^5, S^4]$ and $[S^9, S^8]$ have just one non-trivial element, which is the suspension of the Hopf map $S^3\to S^2$. The cofiber of the Hopf map has a non-trivial action of $Sq^2$. But $Sq^2$ acts trivially on $x^2$ and $x^4$ in the cohomology of ${\mathbb C}P^\infty$, which means that the attaching maps in our cases are nullhomotopic, and so these two spaces are in fact co-reducible.

Finally for the quaternionic space you also can use the action of ${\mathcal P}^1$ in the mod $3$ Steenrod algebra to show that the cell in dimension 8 is connected to the cell in dimension 12, and so ${\mathbb H}P^n/{\mathbb H}P^1$ is never co-reducible.

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