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.
 A: The intuition is simply that every complex line has a unit circle which is an orbit of $S^1$ in $S^{2n+1}$, and so the space $\mathbb{C}P^n$ of lines is "the same as" the space of circles in $S^{2n+1}$. 
To pin this down, we need mutually inverse maps and to show they're continuous. Naturally, we should map a complex line to its unit circle, and a circle in $S^{2n+1}$ to its span in $\mathbb{C}^{n+1}$, a complex line. These are certainly mutually inverse, and continuity should be intuitively apparent. To show it carefully, recall that $\mathbb{C}P^n$ has the quotient topology: its open sets are exactly the images of open sets of lines in $\mathbb{C}^{n+1}$. Similarly, an open set $\pi(U)$ in $S^{2n+1}/S^1$ is the image of an open union $U$ of circle orbits in $S^{2n+1}$. The inverse image of $\pi(U)$ under the map $\mathbb{C}P^n\to S^{2n+1}/S^1$ is just the set of lines corresponding to the span of $U$, which is open since the span of $U$ is open: thus $\mathbb{C}P^n\to S^{2n+1}/S^1$ is a continuous bijection. If you know that $\mathbb{C}P^n$ is compact Hausdorff then by a standard general topology lemma, we're done: otherwise you must show the inverse map is continuous, in a similar way.
A: $$
S^{2n+1} \subseteq \mathbb{R}^{2n+2} = \mathbb{C}^{n+1}
$$
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 $S^1$:
$$
w=\lambda v \quad\Rightarrow\quad 1=\|w\|=|\lambda|\|v\|=|\lambda|
$$
A: What is your definition of $\mathbb{C}P^n$? Is it

the quotient space of $\mathbb{C}^{n+1}− {0} $ under the equivalence
relation $ v \sim \lambda v$ for $ \lambda \neq 0 $

? I assume it is defined as above. I used $\pi$ to denote the quotient space. So  $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{R}P^n$.
Here we identify $\mathbb{C}^{n+1} = \mathbb{R}^{2n+2}$
You essentially just need to apply the universal property of quotients.
Consider the inclusion map $S^{
2n+1} \hookrightarrow \mathbb{R}^{2n+2}$. This is of course continuous, and thus the composition
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2} \backslash \{0\} &\\
 & \searrow{f} & \downarrow{\pi} \\
 & & \mathbb{C}P^n \end{array}$$
is continuous. Note that $f$ is defined on $S^{(2n+1)}$,  $f(x)=f(\lambda x)$, where $|\lambda| = 1$. So $f$ makes the same identifications as your map: $e^{i\theta}$.
Thus, based on passing to the quotient, $f$ induces a map $F$ in the quotient by
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2}\backslash \{0\} &\\
\downarrow{e^{i\theta}} & \searrow{f} & \downarrow{\pi} \\
S^{2n+1}/\sim & \xrightarrow{F} & \mathbb{C}P^n .\end{array}$$
We have that $F$ is continuous. Note that it is also a bijection. Since $S^{2n+1}/\sim$ is compact (it is the image by $e^{i\theta}$ of $S^{2n+1}$, which is compact) and $\mathbb{C}P^n$ is Hausdorff, it is a homeomorphism.
Actually, I just copied the proof from this answer, making some modifications.
