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.

Proof

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.

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))

Examples of local homeomorphism that is not a covering map

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.

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.

Proof

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.

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.

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