How to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$ under a given identification We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.

$$
S^{2n+1} \subseteq \mathbb{R}^{2n+2} = \mathbb{C}^{n+1}
$$
The usual map restricts to a surjection (check this)
$$
S^{2n+1} \rightarrow \mathbb{C}P^n
$$
Now check that two points $w,v$ on the sphere produce the same 1-dimensional subspace in 
$\mathbb{C}^{n+1}$ iff they differ by an element of $S^1$:
$$
w=\lambda v \quad\Rightarrow\quad 1=\|w\|=|\lambda|\|v\|=|\lambda|
$$
