Showing that $(\Bbb C^{n+1} - 0) {/} (\Bbb C - 0) \cong \Bbb C P^n.$

Question

Show that the complex projective space $\Bbb C P^n \cong \Bbb C^{n+1} - 0 / \Bbb C - 0.$

$\textbf {My thought} :$ We know that $\Bbb C P^n : = \Bbb S^{2n+1}/\Bbb S^1,$ where $\Bbb S^1$ is the unit circle. Now we have a surjective continuous map $f$ from $\Bbb C^{n+1} - 0$ onto $\Bbb S^{2n+1}$ defined by $$x \mapsto \frac {x} {\|x\|},\ x \in \Bbb C^{n+1} - 0.$$ Let $p : \Bbb S^{2n+1} \longrightarrow \Bbb S^{2n+1}/ \Bbb S^1 \cong \Bbb CP^n$ be the quotient map. Then the map $g : = p \circ f : \Bbb C^{n+1} - 0 \longrightarrow \Bbb C P^n$ gives us a surjective continuous map. Let $X^* = \{g^{-1} \{z\}\ |\ z \in \Bbb C P^n \}$ and endow $X^*$ with the quotient topology. Then $g$ induces a bijective continuous map $\overline {g} : X^* \longrightarrow \Bbb C P^n.$ Furthermore, $\overline {g}$ is a homeomorphism if and only if $g$ is a quotient map.

So if we can able to show that $X^* \cong \Bbb C^{n+1} - 0 / \Bbb C - 0$ and $g$ is a quotient map then we are through by universal property of quotient topology. But I am unable to show these things. Would anybody give me some suggestion in this regard?

Thanks in advance.

EDIT $:$ For any $z \in \Bbb S^{2n+1}$ there exists $z' \in C^{n+1} - 0$ such that $z = \frac {z'} {\|z'\|}.$ Now let $w \in g^{-1} \{[z]\} = g^{-1} \left \{\left [\frac {z'} {\|z'\|} \right ] \right \} ,$ where $[z]$ is the equivalence class $z \in \Bbb S^{2n+1}$ in $\Bbb C P^n : = \Bbb S^{2n+1} / \Bbb S^1.$ Then there exists $\lambda_0 \in \Bbb S^1$ such that $$\frac {w} {\|w\|} = \lambda_0 \frac {z'} {\|z'\|} \implies w = \frac {\lambda_0 \|w\|} {\|z\|} z = \lambda z$$ where $\lambda = \frac {\lambda_0 \|w\|} {\|z\|} \in \Bbb C - 0.$ Conversely, if $\lambda \in \Bbb C - 0$ then $$g(\lambda z') = \left [ \frac {\lambda} {\|\lambda\|} \frac {z'} {\|z'\|} \right ] = [\lambda' z] = [z]$$ since $\lambda' = \frac {\lambda} {\|\lambda\|} \in \Bbb S^1.$ Therefore $g^{-1} \{[z]\} = \{\lambda z'\ |\ \lambda \in \Bbb C - 0 \},$ which is nothing but the orbit of $z' \in \Bbb C^{n+1} - 0$ under the action of $\Bbb C - 0$ on $\Bbb C^{n+1} - 0$ which is given as follows $:$

For any given $z \in \Bbb C^{n+1} - 0$ and for any $z \in \Bbb C - 0$ we define $$z \cdot (z_0,z_1, \cdots, z_n) : = (zz_0, zz_1, \cdots, zz_n).$$

So we have $$X^* = (\Bbb C^{n+1} - 0) / (\Bbb C - 0).$$

Also since $f$ is a retraction it is quotient map and thus $g = p \circ f,$ is a quotient map being the composition of two quotient maps. Thus we are the done with the proof!

Answer 1

In the literature on can find various definitions of $\mathbb CP^n$.

1. $\mathbb CP^n = (\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$, where the group $(\mathbb C \setminus \{0\},\cdot)$ operates via scalar multiplication on $\mathbb C^{n+1} \setminus \{0\}$. This seems to be the standard (or at least most popular) definition.

2. $\mathbb CP^n = S^{2n+1}/S^1$, where the group $(S^1,\cdot)$ operates via scalar multiplication on $S^{2n+1} \subset \mathbb C^{n+1}$.

3. $\mathbb CP^n = \mathbf{Gr}_1(\mathbb C^{n+1})$. Here $\mathbf{Gr}_k(\mathbb C^m)$ denotes the set of all $k$-dimensional (complex) linear subspaces of $\mathbb C^m$. This is known as a Grassmann manifold. It can be endowed with a natural topology and a natural structure of a differentiable manifold. We shall not go into details here.

For the equivalence of 1. and 2. see When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map? Morever, on the level of sets it should be clear that $(\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$ and $\mathbf{Gr}_1(\mathbb C^{n+1})$ can be identified naturally: Take each equivalence class $\zeta \in (\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$ to $\zeta \cup \{0\} \in \mathbf{Gr}_1(\mathbb C^{n+1})$.

Answer 2

Each complex vector $v$ in $C^{n+1}$ has a unique norm $\|v\|: = \sqrt{\sum_i |v_i|^2}$ and whenever $v\neq 0$, then $v = \|v\|\frac{v}{\|v\|}$ is a unique decomposition to a positive real number $\|v\|$ and an point on the $2n+1$ sphere.

The map $v (C\setminus\{0\}) \mapsto \frac{v}{\|v\|} S^1$ is a well defined map from $C^{n+1}\setminus\{0\} \,\, / \,\,C\setminus\{0\}$ to $S^{2n+1} / S^1$, because a different choice $v' = z v$ representing the same element is mapped to the same element $\frac{v'}{\|v'\|} S^1=\frac{v}{\|v\|} \frac{z}{|z|}S^1 = \frac{v}{\|v\|} S^1$.

Vice versa, the map $v S^1 \mapsto v (C\setminus\{0\})$ is also well defined.

You easily verify that those are inverse to each other and continuous (if you need it).