homeomorphism projective space cell

Question

Define $\Phi: D^n \rightarrow \mathbb{R}P^n$ by $\Phi(x_1,........,x_n) = \left[x_1,.......,x_n,\sqrt{1-||x||^2}\right]$ .

Where $\mathbb{R}P^n$ is quotient space obtained by $x\sim y$ in $\mathbb{R}^{n+1}\backslash \{0\}$ if and only if $y= \lambda x$ for non zero real $\lambda$

Problem: Show that $\Phi|_{D^n\backslash \mathbb{S}^{n-1}}: D^n\backslash \mathbb{S}^{n-1}\rightarrow \mathbb{R}P^n\backslash \mathbb{R}P^{n-1}$ is a homeomorphism.

Clearly $\Phi$ is continuous and so its restriction to $D^n\backslash \mathbb{S}^{n-1}$ is continuous. I tried defining $f: \mathbb{R}P^n\backslash \mathbb{R}P^{n-1}\rightarrow D^n\backslash \mathbb{S}^{n-1}$ by $f([x_1,.......,x_n,x_{n+1}])= [x_1,.......,x_n]$ or $f\left([x_1,.......,x_n,x_{n+1}]\right)= \left[\frac{x_1}{x_{n+1}},.......,\frac{x_n}{x_{n+1}}\right]$ but none seemed to work. Note, $x_{n+1}\neq 0$

May someone help? How do I show the restriction is a homeomorphism?

Answer 1

Ted Shifrin has already answered your question, but let me give some supplementary explanations.

Let us write $x = (x_1,\ldots, x_n)$. Then $\Phi(x) = [x,\sqrt{1- \lVert x \rVert^2}]$. Your attemps to define $f$ are absolutely reasonable. Let us write $f([x,x_{n+1}]) = x'$. In order to achieve that $\phi(f([x,x_{n+1}]) = [x,x_{n+1}]$ we must have $\Phi(x') = ([x',\sqrt{1- \lVert x' \rVert^2}]) = [x,x_{n+1}]$, i.e. $x' = \lambda x$ and $\sqrt{1- \lVert x' \rVert^2} = \lambda x_{n+1}$ for some $\lambda \in \mathbb R$. This requires $$\lambda x_{n+1} = \sqrt{1- \lVert \lambda x \rVert^2} = \sqrt{1- \lambda^2\lVert x \rVert^2}.$$ Noting that $\lambda x_{n+1}$ must be positive, we get $$\lambda = \frac{\operatorname{sgn} x_{n+1}}{\sqrt{\lVert x \rVert^2 + x_{n+1}^2}} = \frac{x_{n+1}}{\lvert x_{n+1} \rvert \sqrt{\lVert x \rVert^2 + x_{n+1}^2}}.$$ This means $$f([x,x_{n+1}]) = \frac{x_{n+1}}{\lvert x_{n+1} \rvert \sqrt{\lVert x \rVert^2 + x_{n+1}^2}}x .$$ It is now easily verified that $f(\Phi(x)) = x$. Therefore $\Phi$ is a bijection with inverse $f$.

It remains to show that $f$ is continuous. Let $p : \mathbb R^{n+1} \setminus \{0\} \to \mathbb RP^n$ denote the quotient map. The set $W = \mathbb{R}P^n\setminus \mathbb{R}P^{n-1}$ is open in $\mathbb RP^n$, thus $p_W : p^{-1}(W) \stackrel{p}{\to} W$ is a quotient map. We have $p^{-1}(W) = \{ (x_1,\ldots,x_{n+1}) \in \mathbb R^{n+1} \mid x_{n+1} \ne 0\}$. The map $f \circ p_W : p^{-1}(W) \to D^n\backslash \mathbb{S}^{n-1}$ is clearly continuous, thus also $f$ is continuous.

Answer 2

You want to find the formula for $(\Phi|_{D^n-S^{n-1}})^{-1}$. Start with $n=1$ to see the algebra: You want $$\big[y,1\big] = \big[x,\sqrt{1-x^2}\big],$$ so you solve $\dfrac x{\sqrt{1-x^2}} = y$ and get $x=\dfrac y{\sqrt{1+y^2}}$.

In general, then, set $$f([y_1,\dots,y_n,1]) = (x_1,\dots,x_n) \quad\text{with}\quad x_j = \frac{y_j}{\sqrt{1+\|y\|^2}}.$$ Since $[y_1,\dots,y_n,1] = [x_1,\dots,x_n,\sqrt{1-\|x\|^2}]$, there's no problem checking that $f$ is the desired inverse and is continuous.