# Math4202 Topology II (Lecture 14)

## Algebraic Topology

### Covering space

#### Definition of covering space

Let $p:E\to B$ be a continuous surjective map.

If every point $b$ of $B$ has a neighborhood **evenly covered** by $p$, which means $p^{-1}(U)$ is a union of disjoint open sets, then $p$ is called a covering map and $E$ is called a covering space.

#### Theorem exponential map gives covering map

The map $p:\mathbb{R}\to S^1$ defined by $x\mapsto e^{2\pi ix}$ or $(\cos(2\pi x),\sin(2\pi x))$ is a covering map.

<details>
<summary>Proof</summary>

Consider $(1,0)\in S^1$, we choose a neighborhood of $(1,0)\in S^1$ of the form $U=\{e^{2\pi ix}|x\in (-\frac{1}{2}, \frac{1}{2})\}$. (punctured circle)

$$
p^{-1}(U)=\{x\in \mathbb{R}|e^{2\pi ix}\neq -1\}=\{x\neq k+\frac{1}{2}, k\in \mathbb{Z}\}=\dots\cup (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2},\frac{3}{2})\cup (\frac{3}{2},\frac{5}{2})\cup \dots
$$

Are disjoint union of open sets.

When we restrict our map on each interval, the exponential map gives a homeomorphism.

Check using $\ln$ function (continuous) and show bijective with inverse.

$p^{-1}(U)\to U$ evenly covered, and for $(-1,0)$ choose the neighborhood of $(-1,0)$ is $V=\{e^{2\pi ix}|x\in (0,1)\}$ Shows $p|_{p^{-1}(V)}$ is also evenly covered.

</details>

#### Definition of local homeomorphism

A continuous map $p:E\to B$ is called a local homeomorphism if for **every $e\in E$** (note that for covering map, we choose $b\in B$), there exists a neighborhood $U$ of $b$ such that $p|_U:U\to p(U)$ is a homeomorphism on to an open subset $p(U)$ of $B$.

Obviously, every open map induce a local homeomorphism. (choose the open disk around $p(e)$)

<details>
<summary>Examples of local homeomorphism that is not a covering map</summary>

Consider the projection of open disk of different size, the point on the boundary of small disk. There is no $u\in U$ with neighborhood homeomorphic to small disks.

</details>

#### Theorem for subset covering map

Let $p: E\to B$ be a covering map. If $B_0$ is a subset of $B$, the map $p|_{p^{-1}(B_0)}: p^{-1}(B_0)\to B_0$ is a covering map.

<details>
<summary>Proof</summary>

For every point $b\in B_0$, $\exists U$ neighborhood of $b$ such that $p^{-1}(U)$ is a partition into slices, $p^{-1}(U)=\bigcup_{\alpha} V_\alpha$, where $V_\alpha$ is a open set in $E$ and homeomorphic to $U$.

Take $V=U\cup B_0$, then

$$
\begin{aligned}
p^{-1}(V)&=p^{-1}(U)\cup p^{-1}(B_0)\\
&=\left(\bigcup_{\alpha} V_\alpha\right)\cup p^{-1}(B_0)\\
&=\bigcup_{\alpha} V_\alpha\cup p^{-1}(B_0)
\end{aligned}
$$

Therefore $p|_{p^{-1}(V)}:V_\alpha\cap p^{-1}(B_0)\to U\cup B_0$ is a homeomorphism.

</details>

#### Theorem for product of covering map

If $p:E\to B$ and $p':E'\to B'$ are covering maps, then $p\times p':E\times E'\to B\times B'$ is a covering map.