Homogeneous space and quotient space for complex projective spaces We know that 
$$
 O(n+1)/O(n) \simeq SO(n+1)/SO(n) \simeq  S^n,
$$
based on the result of homogeneous space.
Also
$$
 PO(n+1)/PO(n) \simeq  P^n,
$$
$P^n$ is the projective space.
These are in some sense spheres.
If we embed the complex projective space $\mathbb{C}P^n$ into $\mathbb{C}P^{n+1}$, we may be able to define the quotient space 
$$
{\mathbb{C}P^{n+1}}/{\mathbb{C}P^{n}} \simeq ?
$$


*

*Do we have simpler expressions for the above "manifolds" or "quotient space"?

*Homotopy group $\pi_i({\mathbb{C}P^{n+1}}/{\mathbb{C}P^{n}})=?$
Attempt: Note that ${\mathbb{C}P^{n}}=S^{2n+1}/U(1)$, so for 
 ${\mathbb{C}P^{1}}\simeq S^2$ and  ${\mathbb{C}P^{0}}\simeq 0$, 
so
$$\pi_i({\mathbb{C}P^{1}}/{\mathbb{C}P^{0}})=\pi_i(S^2/0)=\pi_i(S^2),$$
which homotopy group is known.
 A: The notation $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is typically understood to be the quotient space where we identify $\mathbb{C}P^n$ to a point. (But not other identifications are made).
For the usual embedding of $\mathbb{C}P^n$ into $\mathbb{C}P^{n+1}$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]\mapsto [z_0:z_1:...:z_n: z_{n+1}]$, $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is homeomorphic to $S^{2n+2}$.  One way to see this is as follows.
First, let $Z_i = \frac{z_i}{z_{n+1}}$ (assuming $z_{n+1}\neq 0$).  Also, let $Z = \sum_{i=0}^n |Z_i|^2$.
Then, we map $\mathbb{C}P^{n+1}\rightarrow S^{2n+2}$ via $$(z_0:...:z_{n+1})\mapsto \begin{cases} \left(\frac{2Z_0}{1+Z}, \frac{2Z_1}{1+Z}, ..., \frac{2Z_n}{1+Z}, \frac{-1+Z}{1+Z} \right) & z_{n+1}\neq 0 \\ (0,0,..,0,1) & z_{n+1} = 0  \end{cases}.$$
The case that $z_{n+1} = 0$ corresponds exactly to a point being in $\mathbb{C}P^n$.  As a result, $f$ is constant on $\mathbb{C}P^n$ so gives a map $\overline{f}:\mathbb{C}P^{n+1}/\mathbb{C}P^n\rightarrow S^{2n+2}$.  This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice.  For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2\wedge S^2$, which is not a manifold due to the wedge point.
