From 3a493e5e20c0855f777fbd83259428f2c55123b3 Mon Sep 17 00:00:00 2001 From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com> Date: Fri, 13 Feb 2026 12:29:10 -0600 Subject: [PATCH] updates --- content/Math4202/Math4202_L14.md | 74 ++++++++++++++++++++++++++++++++ content/Math4202/_meta.js | 1 + 2 files changed, 75 insertions(+) create mode 100644 content/Math4202/Math4202_L14.md diff --git a/content/Math4202/Math4202_L14.md b/content/Math4202/Math4202_L14.md new file mode 100644 index 0000000..4f45e9c --- /dev/null +++ b/content/Math4202/Math4202_L14.md @@ -0,0 +1,74 @@ +# 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. diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index c9726de..cd99f91 100644 --- a/content/Math4202/_meta.js +++ b/content/Math4202/_meta.js @@ -16,4 +16,5 @@ export default { Math4202_L11: "Topology II (Lecture 11)", Math4202_L12: "Topology II (Lecture 12)", Math4202_L13: "Topology II (Lecture 13)", + Math4202_L14: "Topology II (Lecture 14)", }