How can I prove that a cylinder is diffeomorphic to a twice-punctured $n$-sphere?

Question

I'm having some trouble in solving this problem:

"Prove that the cylinder $S^{n-1}\times \mathbb{R}$ is diffeomorphic to $S^n\setminus\{N,S\}$, where $N$ and $S$ are the north pole and the south pole of $S^n$".

Can anyone help me?

Answer 1

Take the map$$\begin{array}{rccc}\Psi\colon&S^{n-1}\times(-1,1)&\longrightarrow&S^n\setminus\bigl\{(0,\ldots,0,\pm1)\bigr\}\\&(x_1,\ldots,x_n,y)&\mapsto&\left(x_1\sqrt{1-y^2},\ldots,x_n\sqrt{1-y^2},y\right)\end{array}$$It's a diffeomorphism. Can you use it to define the diffeomorphism that your're after?

Answer 2

I consider on $S^n\setminus\{N,S\}$ the atlas $\{(S^n\setminus\{N,S\},\varphi_N)\}$, where $\varphi_N$ is the stereographic projection from the north pole

$\varphi_N: S^{n}\setminus\{N,S\} \rightarrow \mathbb{R}^n,\ \ \ \ \ (y_1,\ldots,y_{n+1}) \mapsto \frac{1}{1-y_{n+1}}(y_1,\ldots,y_n)$, with $\varphi_N^{-1}:\mathbb{R}^n \rightarrow S^{n}\setminus\{N,S\} ,\ \ \ \ \ v=(v_1,\ldots,v_{n}) \mapsto \frac{1}{1+\|v\|^2}(2v_1,\ldots,2v_{n},\|v\|^2-1)$.

On $S^{n-1}\times \mathbb{R}$ I take the atlas $\{(U_1,\varphi_1),(U_2,\varphi_2)\}$, where $U_1=S^{n-1}\setminus\{(0,\ldots,0,1)\}\times\mathbb{R}$ and $U_2=S^{n-1}\setminus\{(0,\ldots,0,-1)\}\times\mathbb{R}$ and

$\varphi_1: U_1 \rightarrow \mathbb{R}^n,\ \ \ \ \ (x_1,\ldots,x_{n+1}) \mapsto \left(\frac{x_1}{1-x_{n}},\ldots,\frac{x_{n-1}}{1-x_{n}},x_{n+1}\right)$ $\varphi_2: U_2 \rightarrow \mathbb{R}^n,\ \ \ \ \ (x_1,\ldots,x_{n+1}) \mapsto \left(\frac{x_1}{1+x_{n}},\ldots,\frac{x_{n-1}}{1+x_{n}},x_{n+1}\right)$, with $\varphi_1^{-1}:\mathbb{R}^n \rightarrow U_1 ,\ \ \ \ \ (u,w)=(u_1,\ldots,u_{n-1},w) \mapsto \left(\frac{2u_1}{1+\|u\|^2},\ldots,\frac{2u_{n-1}}{1+\|u\|^2},\frac{\|u\|^2-1}{1+\|u\|^2},w\right)$ $\varphi_2^{-1}:\mathbb{R}^n \rightarrow U_2 ,\ \ \ \ \ (u,w)=(u_1,\ldots,u_{n-1},w) \mapsto \left(\frac{2u_1}{1+\|u\|^2},\ldots,\frac{2u_{n-1}}{1+\|u\|^2},\frac{1-\|u\|^2}{1+\|u\|^2},w\right)$.

Now we take the map $$\begin{array}{rccc}\psi\colon&S^{n-1}\times\mathbb{R}&\longrightarrow&S^n\setminus\bigl\{N,S\bigr\}\\&(x_1,\ldots,x_n,x_{n+1})&\mapsto&\left(\frac{x_1}{\|x\|},\ldots,\frac{x_{n+1}}{\|x\|}\right)\end{array}$$ with $$\begin{array}{rccc}\psi^{-1}\colon&S^n\setminus\bigl\{N,S\bigr\} &\longrightarrow& S^{n-1}\times\mathbb{R}\\&(y,z)=(y_1,\ldots,y_n,z)&\mapsto&\left(\frac{y_1}{\|y\|},\ldots,\frac{y_{n}}{\|y\|},\frac{z}{\|y\|}\right)\end{array}$$

Now we consider $(\varphi_N\circ\psi\circ\varphi_1^{-1})(u,w)=(\varphi_N\circ\psi\circ\varphi_1^{-1})(u_1,\ldots,u_{n-1},w)=(\varphi_N\circ\psi)\left(\frac{2u_1}{1+\|u\|^2},\ldots,\frac{2u_{n-1}}{1+\|u\|^2},\frac{\|u\|^2-1}{1+\|u\|^2},w\right)=\varphi_N\left(\frac{1}{1+w^2}\left(\frac{2u_1}{1+\|u\|^2},\ldots,\frac{2u_{n-1}}{1+\|u\|^2},\frac{\|u\|^2-1}{1+\|u\|^2},w\right)\right)=\frac{1}{1+w^2}\frac{1}{1-\frac{w}{1+w^2}}\left(\frac{2u_1}{1+\|u\|^2},\ldots,\frac{2u_{n-1}}{1+\|u\|^2},\frac{\|u\|^2-1}{1+\|u\|^2}\right)$ that is $C^{\infty}$ since $1-\frac{w}{1+w^2}\neq0$ because the equation $w^2-w+1=0$ has no solution.

Similarly we prove that $\varphi_N\circ\psi\circ\varphi_2^{-1}$ and $\varphi_i\circ\psi^{-1}\circ\varphi_N^{-1}$ for $i=1,2$ are $C^{\infty}$, so we can conclude that $\psi$ it's a diffeomorphism, and it means that $S^{n-1}\times\mathbb{R}$ is diffeomorphic to $S^{n}\setminus\{N,S\}$.

Is it all right?