Prove that $\Bbb R^2 - \{0\}$ is homeomorphic to $S^1 \times \Bbb R$.

Hint: The plane minus the origin can be written as

$$\Bbb{R}^2 \setminus \{0\} = \{(r \cos{t}, r \sin{t}) : 0 < r < \infty, 0 \le t < 2\pi\}$$

Do you see how this looks like a circle cross a line?

HINTS: Use polar coordinates for $\Bbb R^2\setminus\{\langle 0,0\rangle\}$ and $S^1$.

• For each $\theta\in[0,2\pi)$ the ray $\{\langle r,\theta\rangle:r>0\}$ corresponds to the line $\{\langle 1,\theta\rangle\}\times\Bbb R$.

• The map $$\Bbb R^+\to\Bbb R:x\mapsto\begin{cases}x-1,&\text{if }x\ge 1\\\\1-\dfrac1x,&\text{if }0<x<1\end{cases}$$ is a homeomorphism.