Taking $π:C^{n+1} \setminus \{ 0 \}→CP^n$, does the continuous application $q:CP^n→C^{n+1} \setminus \{ 0 \}$ exist such that $π∘q = Id$?

Question

I'm trying to show that the continuous application $q:CP^n→C^{n+1} \setminus \{ 0 \}$ exists such that $π∘q = Id$ in $CP^n$.

To show they are homotopy equivalent, I have been trying to use the lifting Homotopy property, but I have been completely lost for a couple hours.

Can anybody give help?

Answer 1

I give three possible arguments:

I) I claim such a map $q\colon\mathbb{CP}^n\to\mathbb C^{n+1}\setminus 0$ is nullhomotopic. Thus, the composition $\pi\circ q$ must also be null-homotopic, which is impossible, since $\mathbb{CP}^n$ is not contractible.

Indeed, $\mathbb C^{n+1}\setminus 0\simeq S^{2n+1}$, so $q$ can be thought of as a map between CW complexes. Thus, by the cellular approximation theorem it is homotopic to a cellular map. However, $\mathbb{CP}^n$ only has cells in dimension $\le 2n$, so in fact, any cellular map must be constant. (Alternatively, approximate $q\colon\mathbb{CP}^n\to S^{2n+1}$ by a smooth map, which by Sard's theorem is not surjective. But $S^{2n+1}\setminus\{pt\}$ is contractible, so $q$ must be null-homotopic.)

II) Suppose there is such a map $q\colon\mathbb{CP}^n\to \mathbb C^{n+1}\setminus \{0\}$, for $n\ge1$. Then, by the functoriality of $\pi_2$, the composition $$\pi_2(\mathbb C\mathbb P^n)\xrightarrow{q_*}\pi_2(\mathbb C^{n+1}\setminus 0)\xrightarrow{\pi_*}\pi_2(\mathbb C\mathbb P^n)$$ is the identity.

However, I claim $\pi_2(\mathbb{CP}^n)=\mathbb Z$ and $\pi_2(\mathbb C^{n+1}\setminus 0)=0$, so the composition cannot be the identity. Indeed, $\mathbb C^{n+1}\setminus 0$ is homotopic to $S^{2n+1}$, which has $\pi_2(S^{2n+1})=0$ since $2<2n+1$.

Moreover, there is a fiber bundle $S^1\to S^{2n+1}\to \mathbb{CP}^n$, and the homotopy long exact sequence is $$\pi_2(S^{2n+1})=0\to\pi_2(\mathbb{CP}^n)\to\pi_1(S^1)\cong\mathbb Z\to \pi_1(S^{2n+1})=0,$$ so $\pi_2(\mathbb{CP}^n)\cong\mathbb Z$.

III) One can also compute the homology groups instead (these two approaches are basically equivalent due to Hurewicz). Indeed, we have $H_2(S^{2n+1})=0$ and $H_2(\mathbb{CP}^n;\mathbb Z)=\mathbb Z$. Thus, again, the composition$$H_2(\mathbb C\mathbb P^n)\xrightarrow{q_*}H_2(\mathbb C^{n+1}\setminus 0)\xrightarrow{\pi_*}H_2(\mathbb C\mathbb P^n)$$ cannot be the identity.