udpates

content/Math4202/Exam_reviews/Math4202_E1.md

@@ -149,4 +149,42 @@ $$
     $$
     $$
     [f]*[\bar{f}]=[e_{x_0}],\quad [\bar{f}]*[f]=[e_{x_1}]
-    $$
+    $$
+
+
+### 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.
+
+#### 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 $p(e)$)
+
+#### 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.
+
+#### 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.
+
+
+### Fundamental group of the circle
+
+Recall from previous lecture, we have unique lift for covering map.
+
+#### Lemma for unique lifting for covering map
+
+Let $p: E\to B$ be a covering map, and $e_0\in E$ and $p(e_0)=b_0$. Any path $f:I\to B$ beginning at $b_0$, has a unique lifting to a path starting at $e_0$.
+
+Back to the circle example, it means that there exists a unique correspondence between a loop starting at $(1,0)$ in $S^1$ and a path in $\mathbb{R}$ starting at $0$, ending in $\mathbb{Z}$.