# Covering Space Examples R2 \ {(0, 0)}

Recall from the Covering Spaces page that if $X$ is a topological space then a covering space of $X$ is a pair $(\tilde{X}, p)$ where $\tilde{X}$ is a path connected and locally path connected topological space and $p : \tilde{X} \to X$ is a continuous map such that for every $x \in X$ there exists a path connected open neighbourhood $U$ of $x$ (called an elementary neighbourhood of $x$) such that $p$ restricted to every path component of $p^{-1}(U)$ is a homeomorphism onto $U$.

We will now look at an example of a covering space of $\mathbb{R}^2 \setminus \{ (0, 0) \}$.

Let $\tilde{X} = \mathbb{R}^2$ and let $p : \mathbb{R}^2 \to \mathbb{R}^2 \setminus \{ (0, 0) \}$ be defined for all $(x, y) \in \mathbb{R}^2$ by:

(1)Clearly $p$ is a continuous map since each component of $p$ is continuous.

Observe that $(0, 0)$ is mapped to the point $(1, 0)$. In general, $(0, 2k \pi)$ is mapped to the point $(1, 0)$ for every $k \in \mathbb{Z}$.

Also observe that for a fixed $x_0$, $(x_0, 0)$ is mapped to the point $(e^{x_0}, 0)$.

As we can see, $p$ maps $\mathbb{R}^2$ onto $\mathbb{R}^2 \setminus \{ (0, 0) \}$ an infinite number of times.

If $(x_0, y_0) \in \mathbb{R}^2 \setminus \{ (0, 0) \}$ then there exists an open neighbourhood of $(x_0, y_0)$ which does not contain the puncture from the removal of the origin. In particular, there exists an open disk $U$ centered at $(x_0, y_0)$ that is contained in $\mathbb{R}^2 \setminus \{ (0, 0) \}$. If we take this open disk small enough, then $p^{-1}(U)$ will contain an infinite number of squashed open disks which are disjoint from each other. Clearly the restriction of $p$ onto any of these open disks will be a homeomorphism onto $U$.