Math4202 Topology II (Lecture 13)

Algebraic Topology

Covering space

Definition of partition into slice

Let $p:E\to B$ be a continuous surjective map. The open set $U\subseteq B$ is said to be evenly covered by $p$ if it’s inverse image $p^{-1}(U)$ can be written as the union of disjoint open sets $V_\alpha$ in $E$ . Such that for each $\alpha$ , the restriction of $p$ to $V_\alpha$ is a homeomorphism of $V_\alpha$ onto $U$ .

The collection of $\{V_\alpha\}$ is called a partition $p^{-1}(U)$ into slice.

Stack of pancakes ( $\{V_\alpha\}$ ) on plate $U$ , each $V_\alpha$ is a pancake homeomorphic to $U$

Note that all the sets in the definition are open.

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 partitioned into slice, then $p$ is called a covering map and $E$ is called a covering space.

Examples of covering space

identity map is a covering map

Consider the $B\times \Gamma\to B$ with $\Gamma$ being the discrete topology with the projection map onto $B$ .

This is a covering map.

Let $S^1=\{z\mid |z|=1\}$ , then $p=z^n$ is a covering map to $S^1$ .

Solving the inverse image for the $e^{i\theta}$ with $\epsilon$ interval, we can get $n$ slices for each neighborhood of $e^{i\theta}$ , $-\epsilon< \theta< \epsilon$ .

You can continue the computation and find the exact $\epsilon$ so that the inverse image of $p^{-1}$ is small and each interval don’t intersect (so that we can make homeomorphism for each interval).

Usually, we don’t choose the $U$ to be the whole space.

Consider the projection for the boundary of mobius strip into middle circle.

This is a covering map since the boundary of mobius strip is winding the middle circle twice, and for each point on the middle circle with small enough neighborhood, there will be two disjoint interval on the boundary of mobius strip that are homeomorphic to the middle circle.

Proposition of covering map is open map

If $p:E\to B$ is a covering map, then $p$ is an open map.

Proof

Consider arbitrary open set $V\subseteq E$ , consider $U=p(V)$ , for every point $q\in U$ , with neighborhood $q\in W$ , the inverse image of $W$ is open, continue next lecture.