Homeomorphism between a subspace of the complex projective space and $\mathbb C^n.$

Question

Is it true that $\mathbb CP^n - \mathbb CP^{n-1} \simeq \mathbb C^n\ $? Our professor used this result in showing that $$\mathbb CP^n/\mathbb CP^{n-1} \simeq S^{2n}.$$ My idea in this regard is the following $:$

There is a natural inclusion $i : \mathbb C P^{n-1} \longrightarrow \mathbb CP^{n}$ given by $$[z_0,\cdots,z_n] \mapsto [z_0,\cdots, z_n,0].$$ In view of this inclusion one can think of $$\mathbb CP^n - \mathbb CP^{n-1} = \left \{[z_0,\cdots, z_n] \in \mathbb CP^n\ \big |\ z_n \neq 0 \right \}.$$

Now since $\mathbb CP^n : = \frac {\mathbb C^{n+1} - \{0\}} {\mathbb C - \{0\}}$ it turns out that the elements of $\mathbb C P^n - \mathbb C P^{n-1}$ are of the form $[z_0,\cdots,z_{n-1},1].$ So in order a get a map $f : \mathbb CP^n - \mathbb CP^{n-1} \longrightarrow \mathbb C^n$ the obvious thing to do is to take the following map $:$ $$[z_0,\cdots, z_{n-1}, 1] \longmapsto (z_0, \cdots, z_{n-1}).$$ This map is clearly well-defined, one-one and onto. So if we can prove that $f$ is continuous and $\mathbb CP^n - \mathbb CP^{n-1}$ is compact then we are done with the proof because $\mathbb C^n$ (the range of $f$) is known to be Hausdorff.

But I am unable to prove these two facts. That's why I am coming here in this site to get some help. Any help will be greatly appreciated at this stage.

Thank you all!

Answer 1

We think $\mathbb{C}^n$ as a subspace embedded in $\mathbb{C}^{n+1}$ by the inclusion. Now, Since the usual map $\mathbb{C}^{n+1}\setminus \mathbb{C}^n\to \mathbb CP^{n+1}\setminus\mathbb CP^n$ is a quotient map, it follows from the universal property of quotient map, $\frac{\mathbb{C}^{n+1}\setminus \mathbb{C}^n}{\mathbb{C}\setminus 0}$ is homeomorphic to $\mathbb CP^{n+1}\setminus\mathbb CP^n$

Now, we will show that, $\frac{\mathbb{C}^{n+1}\setminus \mathbb{C}^n}{\mathbb{C}\setminus 0}$ is homeomorphic to $\mathbb{C}^n$

Consider, $f:\mathbb{C}^{n+1}\setminus \mathbb{C}^n \to\mathbb{C}^n$ via, $f(z_0,z_1,..,z_{n+1})=(\frac{z_0}{z_{n+1}},..)$

So, by universal property of quotient map, there exists a continuous map, $\phi:\frac{\mathbb{C}^{n+1}\setminus \mathbb{C}^n}{\mathbb{C}\setminus 0}\to\mathbb{C}^n$

On the other hand, $q\circ g$ gives a continuous map $\psi:\mathbb{C}^n\to\frac{\mathbb{C}^{n+1}\setminus \mathbb{C}^n}{\mathbb{C}\setminus 0},$ where $g:\mathbb{C}^n\to\mathbb{C}^{n+1}\setminus \mathbb{C}^n$ given by, $g(z_0,..,z_{n-1})=(z_0,..,z_{n-1},1)$ and $q$ is the usual quotient map from $\mathbb{C}^{n+1}\setminus \mathbb{C}^n\to\frac{\mathbb{C}^{n+1}\setminus \mathbb{C}^n}{\mathbb{C}\setminus 0}$

Now, it's not hard to see that $\phi,\psi$ are inverse of each other and hence result follows.

Answer 2

$\mathbb CP^n \setminus \mathbb CP^{n-1}$ is not compact; how could a compact space be homeomorphic to $\mathbb C^n$?

We define $$j : \mathbb C^n \to \mathbb CP^n, j(z_0,\ldots,z_{n-1}) = [z_0,\ldots,z_{n-1},1]$$ which is clearly a continuous injection whose image is $\mathbb CP^n \setminus \mathbb CP^{n-1}$ (which is an open subset of $\mathbb CP^n$).

We shall prove that $j$ is an open map; this shows that $j$ maps $\mathbb C^n$ homeomorphically onto $\mathbb CP^n \setminus \mathbb CP^{n-1}$.

So let $U \subset \mathbb C^n$ be open. To show that $j(U)$ is open it suffices to show that $U^* = p^{-1}(j(U))$ is open in $\mathbb C^{n+1} \setminus \{0\}$, where $p : \mathbb C^{n+1} \setminus \{0\} \to \mathbb CP^n$ is the quotient map. The set $U^*$ is the set of all points in $\mathbb C^{n+1} \setminus \{0\}$ which are equivalent to some $(\zeta,1)$ with $\zeta \in U$. Therefore $$U^* = \{\lambda(\zeta,1) = (\lambda \zeta, \lambda) \in \mathbb C^n \times (\mathbb C \setminus \{0\}) \subset \mathbb C^{n+1} \setminus \{0\} \mid \zeta \in U, \lambda \in \mathbb C \setminus \{0\}\} .$$ Define $\phi, \psi: \mathbb C^n \times (\mathbb C \setminus \{0\}) \to \mathbb C^n \times (\mathbb C \setminus \{0\})$ by $\phi(\zeta,\lambda) = (\lambda \zeta, \lambda)$ and $\psi(\zeta,\lambda) = (\frac{1}{\lambda}\zeta,\lambda)$. These maps are continuous and we have $\psi \circ \phi = id$ and $\phi \circ \psi = id$. Thus $\phi, \psi$ are homeomorphisms which are inverse to each other.

We conclude that $U^* = \phi(U \times (\mathbb C \setminus \{0\})$ is open in $\mathbb C^n \times (\mathbb C \setminus \{0\})$. Since the latter is open in $\mathbb C^{n+1} \setminus \{0\}$, we see that $U^*$ is open in $\mathbb C^{n+1} \setminus \{0\}$.