10

We know that a necessary and sufficient condition for a path-connected, locally path-connected space to have a universal cover is that it is semi-locally simply connected.

Now since $\mathbb R^2\setminus\{0\}$ is such a space, it must have a universal cover. However I can't see what the universal cover of $\mathbb R^2\setminus\{0\}$ actually is. Can someone help me?

Thank you.

R_D
  • 7,312
  • 2
    You know the universal cover of $S^1$ is $\mathbb{R}$? Can you modify that to give a cover of the punctured plane? – Carl Mummert Feb 24 '16 at 12:03
  • Yes. And I also know that $\mathbb R^2\setminus {0}$ can be deformation retracted to $S^1$. But not sure what to do from there. – R_D Feb 24 '16 at 12:04
  • 2
    Maybe you should think of $\mathbb{R}^2\backslash {0}$ as a space parametrized with polar coordinates. – Clément Guérin Feb 24 '16 at 12:06
  • 3
    If you have a product space $X\times Y$, its universal cover is related to the universal covers of $X$ and $Y$. – Daniel Fischer Feb 24 '16 at 12:06
  • @DanielFischer, Can I identify $R^2\setminus{0}$ with $S^1\times R$ and hence get the the universal covering space as $R^2$? – R_D Feb 24 '16 at 12:13
  • Yes, precisely. – Daniel Fischer Feb 24 '16 at 12:15
  • 1
    @ClémentGuérin, I can define a map from $R_{>0}\times R\to R^2\setminus{0}$ by $(r,t)\mapsto re^{it}$ and if this is a covering map I am done? – R_D Feb 24 '16 at 12:17
  • 3
    Make sure that you can prove rigorously that this is a covering map (we can check your proof if you answer your own question) but this is the idea. – Clément Guérin Feb 24 '16 at 12:19
  • 1
    OBS: Your space is a Riemann surface. Therefore, the universal cover must also be one. By the Uniformization Theorem, there are only two options for it (topologically speaking). You don't have much to look for. – Aloizio Macedo Feb 24 '16 at 19:03

2 Answers2

7

Hint: Think of the mapping $z\mapsto e^z$ from $\Bbb{C}$ to $\Bbb{C}\setminus\{0\}$. Its derivative is also $e^z$, which is always non-zero. Therefore the mapping is conformal everywhere, i.e. a local homeomorphism.

Jyrki Lahtonen
  • 133,153
3

One (of many) possibilities to consider is as for all infinite cyclic covers of nice $X$ is to pull back the universal cover $\mathbb R \to S^1$ of some $f: X \to S^1$. In the smooth setting $f^{-1}(1)$ (let 1 be a regular value) is a 1-codimensional submanifold and the infinite cyclic cover is obtained by cutting along that submanifold and gluing $\mathbb Z$-many copies together.

In your case the map is the projection $X=\mathbb R^2 -0 \to S^1$. And the submanifold is a ray of a point $\mathbb R_+ x$, canonically $x=1$. So take $\sqcup_{\mathbb Z} ( \mathbb R^2 - \mathbb R_{\geq 0})$ and cross-glue them together along one boundary component $\mathbb R _{>0}$.

For more info cf. e.g. my answer here.

Edit: to explicitely write down what I wrote above you can write $$\mathbb R^2 -0 = \mathbb R_{>0} \times S^1 \stackrel{p_2} \to S^1,$$ where the pull back becomes obvious: $$\begin{array}{c}p^*(\mathbb R) = \mathbb R_{>0} \times \overbrace{id^*\mathbb R}^{\mathbb R} &\to &\mathbb R\\ \downarrow && \downarrow \\ \mathbb R_{>0}\times S^1 &\stackrel {p} \to & S^1\end{array} $$ and the covering map is giving by $id \times \pi$ where $\pi:\mathbb R \to S^1$. By using the isomorphism $\mathbb R_{>0}\cong \mathbb R$ you get the universal cover:

$$ \mathbb R^2 \cong \qquad\mathbb R_{>0} \times \mathbb R \stackrel {1\times\pi} \longrightarrow \mathbb R_{>0}\times S^1 \qquad \cong \mathbb R^2 -0.$$

Community
  • 1
Daniel Valenzuela
  • 6,305
  • 12
  • 20