---- > [!definition] Definition. ([[covering space]]) > Let $p: E \to B$ be a [[continuous]] [[surjection]] between [[topological space|topological spaces]]. If every $b \in B$ has a [[neighborhood]] $U$ that is [[evenly covered]] by $p$, then $p$ is called a **covering map**, and $E$ is called a **covering space** of $B$. > \ > If each $p ^{-1}(b)$ has [[cardinality]] $n$ for $n \in \mathbb{N} \cup \{ \infty \}$ ([[fibers of path-connected space under covering map are in bijection|e.g.]]), we call $p$ an **$n$-sheeted cover**. ($p=2$ a **double cover**, $p=3$ a **triple cover**, etc.) > [!basicproperties] > - If $p: E\to B$ is a covering map, then for each $b \in B$ the [[subspace topology|subspace]] $p ^{-1}(b)$ has the [[discrete topology]]. For it equals the disjoint union of open sets which are each [[homeomorphism]] to $\{ b \}$ and thus are singletons. > >- If $p: E\to B$ is a covering map, then $p$ is an [[open map]]. For suppose $A$ is an open subset of $E$. Given $x \in p(A)$, choose [[neighborhood]] $U \ni x$ that is [[evenly covered]] by $p$. Let $\{ V_{\alpha} \}$ be a partition of $p ^{-1}(U)$ into slices. There is a point $y \in A$ s.t. $p(y)=x$, let $V_{\beta}$ be the slice containing $y$. The set $V_{\beta} \cap A$ is open in $E$ and hence open in $V_{\beta}$; because $p$ maps $V_{\beta}$ [[homeomorphism|homeomorphically]] onto $U$, the set $p(V_{\beta} \cap A)$ is open in $U$ and hence open in $B$; it is thus a [[neighborhood]] of $x$ contained in $p(A)$, as desired. >- If $p: E \to B$ is a covering map, then $p$ is also a [[local homeomorphism]] of $E$ with $B$. Indeed, send $e \in E$ to $b \in B$ via $p$; using that $p$ is a covering map obtain a [[neighborhood]] $U \ni b$ s.t. $p ^{-1}(U)=\bigsqcup_{\alpha}V_{\alpha}$ for some partition $\{ V_{\alpha} \}$ of $p ^{-1}(U)$ into (open) slices. $e$ belongs to one of these (open) slices $V_{\alpha}$, which is [[homeomorphism]] to $U$ via $p$. > - The converse is false, see below. >- How do covering maps behave with [[subspace topology|subspaces]] and [[product topology|products]]? > - See [[condition for restriction of covering maps]] and [[the product of covering maps is a covering map]] ^0cf198 > [!basicexample] > - The [[identity map]] $i:X \to X$ is a covering map of the most trivial sort. More generally, let $E$ be the space $X \times \{ 1,\dots,n \}$ consisting of $n$ disjoint copies of $X$. Then map $p: E \to X$ given by $p(x,i)=x$ for all $i \in [n]$ is again a covering map, called the **trivial covering map**. In this case, we can picture *the entire space $E$* as a stack of pancakes over $X$. > - Any [[homeomorphism]] is a covering map, also of a trivial sort. The map $p: \mathbb{R} \to \mathbb{S}^{1}$ given by the equation $p(x):=(\cos 2\pi x, \sin 2 \pi x)$ is a covering map of the [[unit circle]]. Since $\mathbb{S}^{1}$ is [[homeomorphism]] to $(0,1]$ via the map $t \mapsto (\cos 2\pi t, \sin 2 \pi t)$, $t \in (0,1]$, a basic open set $U$ in $\mathbb{S}^{1}$ has the form $\{ (\cos 2\pi t, \sin 2\pi t): t \in (a,b) \}$ for some [[open interval]] $(a,b) \subset (0,1]$. In particular, note that $p |_{(a,b)}$ is a [[homeomorphism]] identifying $(a,b)$ with $U$, as is $p |_{(a,b) + k}$ for any integer $k$. ^706eb3 Let $x \in \mathbb{S}^{1}$. Obtain such a [[neighborhood]] $U \ni x$, $U=\{ (\cos 2\pi t, \sin 2\pi t): t \in (a_{0},b_{0}) \}$. Then $p ^{-1}(U)$ is $p ^{-1}(U)=\bigsqcup_{k \in \mathbb{Z}} \{(\cos 2 \pi t, \sin 2\pi t ): t \in (a_{0},b_{0})+k\}=:\bigsqcup_{k \in \mathbb{Z}} V_{k}.$ Let $\mathbb{S}^{1} \subset \mathbb{C}$ be the [[circle]] $\{ z \in \mathbb{C} : |z|=1 \}$ and consider the map $p:\mathbb{R} \to \mathbb{S}^{1}$ given by $p(t)=e^{2\pi i t}$. For the set $U_{y>0}:=\{ x+iy: y > 0 \}$, we have $p ^{-1}(U_{y>0})= \coprod_{k \in \mathbb{Z}}\left( k, k+ \frac{1}{2} \right),$ the disjoint union of $\mathbb{Z}$-many [[open interval|open intervals]]. Furthermore, $p |_{\left( k, k+\frac{1}{2} \right)} : \left( k, k+\frac{1}{2} \right) \to U_{y>0}$ is a [[continuous]] [[bijection]] with [[continuous]] [[inverse map|inverse]] $\begin{align} U_{y>0} \to \left( k,k+\frac{1}{2} \right) \\ x+iy \mapsto j + \frac{1}{2\pi} \cos ^{-1}(x) \end{align}$ ![[CleanShot 2024-06-11 at [email protected]]] Let $\mathbb{S}^{1} \subset \mathbb{C}$, and $p:\mathbb{S}^{1}\to \mathbb{S}^{1}$ be given by $p(z)=z^{n}$. Let $\eta=e^{2\pi i/n}$, and for $y \in \mathbb{S}^{1}$ let $\xi$ be some $n^{th}$ root of $y$. Given $z^{n}=y$, i.e. $e^{2\pi i n t_{1}}=e^{2\pi i t_{2}}$ for some $t_{1},t_{2} \in \mathbb{R}$. So $2\pi i n t_{1} = 2\pi i t_{2} + 2\pi i m \text{ for some } m \in \mathbb{Z}.$ Hence $nt_{1} = t_{2} + m$ from which it follows that $t_{2}=\frac{n}{m}t_{1}$. So $y=e^{2\pi i n t_{1} / m}=e^{2 \pi i }$ - [ ] [[TODO]] show equals xi times the thing - [ ] bring over covering space for $\mathbb{R}\mathbb{P}^{n}$ > [!warning] Basic Nonexamples. > - If $p: E \to B$ is a covering map, then $p$ is also a [[local homeomorphism]] of $E$ with $B$. But the converse is generally false. > - In general, covering maps do not restrict to covering maps. But do see the [[condition for restriction of covering maps]]. > > Below is a counterexample proving both these at once. > > If $p: E \to B$ is a covering map, then $p$ is also a [[local homeomorphism]] of $E$ with $B$. But the converse is false. The map $\begin{align} p: & \mathbb{R}_{+} \to \mathbb{S}^{1} \\ x \mapsto & (\cos 2\pi x, \sin 2\pi x) \end{align}$ is [[surjection|surjective]], and it is a [[local homeomorphism]]. But is not a covering map, for the point $b_{0}=(1,0)$ has no [[neighborhood]] $U$ that is [[evenly covered]] by $p$. The typical [[neighborhood]] $U$ of $b_{0}$ has preimage consisting of small [[neighborhood]]s $V_{n}$ for each $n \in \mathbb{N}$, along with a small interval $V_{0}$ of the form $(0,\varepsilon)$. Each of the intervals $V_{n}$ for $n >0$ is mapped [[homeomorphism|homeomorphically]] onto $U$ by $p$, but the interval $V_{0}$ is only [[topological embedding|embedded]] in $U$ by $p$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```