The universal cover of $S^1 \times S^2$

Question

I'm trying to prove that the universal cover of $S^1 \times S^2$ is $\mathbb{R}^3 \setminus \{0\}$. I know that the universal cover of $S^1$ is $\mathbb{R}$ and the universal cover of $S^2$ is $S^2 $. But I don't know how I can use this for prove it. Thanks for the help!

Answer 1

Hint: Try to find an explicit homeomorphism between $\Bbb R^3\setminus \{0\}$ and $\Bbb S^2\times \Bbb S^1$.

Any line passing through the origin splits in two half-lines in $\Bbb R^3\setminus\{0\}$. Any half-lines is a copy of $\Bbb R$ and contains a unique point of $\Bbb S^2$. Can you get from here?

Answer 2

Consider the map $f:\mathbb{R}^3-\{0\}\rightarrow S^2\times \mathbb{R}^+-\{0\}$ defined by $f(x)=({x\over{\|x\|}},\|x\|)$ it is a diffeomorphism

since $\mathbb{R}^+-\{0\}$ is diffeomorphic to $\mathbb{R}$ (use the logarithm) and $S^2$ is $1$-connected, you deduce that the universal cover of $S^2\times S^1$ is the product of the universal cover of $S^2$ and $S^1$ is diffeomorphic to $\mathbb{R}^3-\{0\}$.