# 22. Covering maps

Consider a continuous map $p\colon X\to Y$. For each point $y\in Y$, we have a subset $p^{-1}\{y\}\subseteq X$, which we call the fibre of $p$ over $y$. We next define what it means for $p$ to be a covering map. The key points are that

• All the fibres $p^{-1}\{y\}$ must be discrete subsets of $X$;

• The fibre $p^{-1}\{y\}$ must depend continuously on $y$, in an appropriate sense.

It is not easy to formulate the second condition directly, so the formal definition looks rather different from this informal discussion.

###### Example 22.1.

Consider the exponential map $\exp\colon{\mathbb{C}}\to{\mathbb{C}}\setminus\{0\}$. The fibres are

 $\exp^{-1}\{r\,e^{i\theta}\}=\{\log(r)+i\theta+2n\pi i\;|\;n\in{\mathbb{Z}}\}$

Each fibre is a discrete set, which suggests that the map should be a covering. We will check that this is true once we have given the proper definition.

###### Definition 22.2.

Let $p\colon X\to Y$ be a continuous map of spaces. Consider an open subset $V\subseteq Y$. We say that $V$ is trivially covered by $p$ if there is a discrete space $F$ and a map $f\colon p^{-1}(V)\to F$ such that the combined map $\langle p,f\rangle\colon p^{-1}(V)\to V\times F$ is a homeomorphism. We say that $p$ is a covering map (or that $X$ is a covering space of $Y$) if for each point $y\in Y$ there is an open set $V$ that contains $y$ and is trivially covered.

###### Remark 22.3.

Suppose we have open subsets $V^{\prime}\subseteq V\subseteq Y$ and that $V$ is trivially covered, as witnessed by a map $f\colon p^{-1}(V)\to F$. We then note that $p^{-1}(V^{\prime})\subseteq p^{-1}(V)$, se we can restrict $f$ to get a map $f^{\prime}\colon p^{-1}(V^{\prime})\to F$. It is not hard to check that the combined map $\langle p,f^{\prime}\rangle\colon p^{-1}(V^{\prime})\to V^{\prime}\times F$ is a homeomorphism, so $V^{\prime}$ is also trivially covered.

###### Example 22.4.

Take $X={\mathbb{R}}\times{\mathbb{Z}}$ and $Y={\mathbb{R}}$, and let $p\colon X\to Y$ be the projection map, given by $p(x,n)=x$. We claim that the whole space $Y$ is trivially covered. Indeed, we can take $F={\mathbb{Z}}$ and define $f\colon X\to F$ by $f(x,n)=n$. Then the combined map $\langle p,f\rangle\colon X\to Y\times F$ is just the identity map ${\mathbb{R}}\times{\mathbb{Z}}\to{\mathbb{R}}\times{\mathbb{Z}}$, which is certainly a homeomorphism. From this it is clear that $p$ is a covering map.

###### Example 22.5.

Take $X$ and $Y$ to be the unit circle in ${\mathbb{C}}$, or in other words $X=Y=\{e^{i\theta}\;|\;\theta\in{\mathbb{R}}\}$. Define $p\colon X\to Y$ by $p(z)=z^{2}$, or equivalently $p(e^{i\theta})=e^{2i\theta}$. Each fibre $p^{-1}\{y\}$ is a discrete set consisting of the two square roots of $y$, which are negatives of each other. This suggests that $p$ should be a covering, which we can prove as follows. We first take $V_{0}=Y\setminus\{-1\}$, so each element $y\in V_{0}$ can be expressed in a unique way as $y=e^{i\theta}$ with $-\pi<\theta<\pi$. The two square roots of $y$ are then $x=e^{i\theta/2}$ (which has $\text{Re}(x)>0$) and $-x=-e^{i\theta/2}=e^{i(\theta/2+\pi)}$ (which has $\text{Re}(-x)<0$). We therefore have

 $p^{-1}(V_{0})=\{z\in X\;|\;z^{2}\neq-1\}=X\setminus\{i,-i\}=\{z\in X\;|\;\text% {Re}(z)\neq 0\},$

and we can define $f_{0}\colon p^{-1}(V_{0})\to\{1,-1\}$ by

 $f_{0}(z)=\begin{cases}1&\text{ if }\text{Re}(z)>0\\ -1&\text{ if }\text{Re}(z)<0.\end{cases}$

We then find that the map $\langle p,f_{0}\rangle\colon p^{-1}(V_{0})\to V_{0}\times\{1,-1\}$ is a homeomorphism, showing that $V_{0}$ is trivially covered. A similar approach can be used to check that the set $V_{1}=Y\setminus\{1\}$ is also trivially covered. Any element $y\in V_{1}$ can be expressed uniquely as $y=e^{i\theta}$ with $0<\theta<2\pi$, and the two square roots are then $x=e^{i\theta/2}$ (which has $\text{Im}(x)>0$) and $-x$ (which has $\text{Im}(-x)<0$). We therefore have

 $p^{-1}(V_{1})=\{z\in X\;|\;z^{2}\neq 1\}=X\setminus\{1,-1\}=\{z\in X\;|\;\text% {Im}(z)\neq 0\},$

and we can define $f_{1}\colon p^{-1}(V_{1})\to\{1,-1\}$ by

 $f_{1}(z)=\begin{cases}1&\text{ if }\text{Im}(z)>0\\ -1&\text{ if }\text{Im}(z)<0.\end{cases}$

We then find that the map $\langle p,f_{1}\rangle\colon p^{-1}(V_{1})\to V_{1}\times\{1,-1\}$ is a homeomorphism, as required. As $Y=V_{0}\cup V_{1}$, this proves that $p$ is a covering.

Recall that the real projective space ${\mathbb{R}}P^{n}$ is defined to be the quotient space $S^{n}/\sim$, where $x\sim y$ iff $y=\pm x$. We therefore have a quotient map $\pi\colon S^{n}\to{\mathbb{R}}P^{n}$, for which $\pi(x)=\pi(y)$ iff $x=\pm y$. We give ${\mathbb{R}}P^{n}$ the quotient topology, which means that a subset $V\subseteq{\mathbb{R}}P^{n}$ is open iff $\pi^{-1}(V)$ is open in $S^{n}$.

###### Proposition 22.6.

The map $\pi\colon S^{n}\to{\mathbb{R}}P^{n}$ is a covering map.

###### Proof.

For each $a\in S^{n}$ we put $U_{a}=\{x\in S^{n}\;|\;x.a>0\}$ and $V_{a}=\pi(U_{a})\subseteq{\mathbb{R}}P^{n}$. We then find that

 $\pi^{-1}(V_{a})=\{x\;|\;\pi(x)\in\pi(U_{a})\}=\{x\;|\;x\in U_{a}\text{ or }-x% \in U_{a}\}=\{x\;|\;x.a\neq 0\}.$

This first shows that $\pi^{-1}(V_{a})$ is an open subset of $S^{n}$ and so (by the definition of the quotient topology) that $V_{a}$ is an open subset of ${\mathbb{R}}P^{n}$. It also allows us to define a continuous map $f_{a}\colon\pi^{-1}(V_{a})\to\{1,-1\}$ by

 $f_{a}(x)=\begin{cases}1&\text{ if }x.a>0\\ -1&\text{ if }x.a<0.\end{cases}$

It is easy to see that the combined map $\langle\pi,f_{a}\rangle\colon\pi^{-1}(V_{a})\to V_{a}\times\{1,-1\}$ is a homeomorphism, so $V_{a}$ is trivially covered. The sets $V_{a}$ cover all of ${\mathbb{R}}P^{n}$ (because $\pi(a)\in V_{a}$) so $\pi$ is a covering map as claimed. ∎

###### Proposition 22.7.

The map $\exp\colon{\mathbb{C}}\to{\mathbb{C}}\setminus\{0\}$ is a covering map, as is the map $\exp\colon i{\mathbb{R}}\to S^{1}$ (where we identify $S^{1}$ with $\{z\in{\mathbb{C}}:|z|=1\}$).

###### Proof.

This is closely related to the proof in Example 22.5. We put

 $\displaystyle V_{0}$ $\displaystyle={\mathbb{C}}\setminus(-\infty,0]$ $\displaystyle V_{1}$ $\displaystyle={\mathbb{C}}\setminus[0,\infty)$ $\displaystyle U_{0}$ $\displaystyle=\{x+iy\;|\;x\in{\mathbb{R}},-\pi $\displaystyle U_{1}$ $\displaystyle=\{x+iy\;|\;x\in{\mathbb{R}},0 $\displaystyle W_{0}$ $\displaystyle=\{x+iy\;|\;y\text{ is not an odd multiple of }\pi\}$ $\displaystyle W_{1}$ $\displaystyle=\{x+iy\;|\;y\text{ is not an even multiple of }\pi\}.$

If $y\in V_{0}$ then there is a unique choice of $r$ and $\theta$ with $-\pi<\theta<\pi$ and $r>0$ and $y=r\,e^{i\theta}$. It follows that the number $x=\log(r)+i\theta$ lies in $U_{0}$ and has $\exp(x)=y$. This means that the restricted map $\exp\colon U_{0}\to V_{0}$ is bijective. Standard complex analysis shows that the inverse is also continuous, so we have a homeomorphism $\exp\colon U_{0}\to V_{0}$. Similarly, the restricted map $\exp\colon U_{1}\to V_{1}$ is also a homeomorphism. From this we see that

 $\displaystyle\exp^{-1}(V_{0})$ $\displaystyle=\{x_{0}+2n\pi i\;|\;x_{0}\in U_{0},\;n\in{\mathbb{Z}}\}=W_{0}$ $\displaystyle\exp^{-1}(V_{1})$ $\displaystyle=\{x_{1}+2n\pi i\;|\;x_{1}\in U_{1},\;n\in{\mathbb{Z}}\}=W_{1}.$

We can thus define continuous maps $f_{i}\colon W_{i}\to{\mathbb{Z}}$ (for $i=0,1$) by $f_{i}(x_{i}+2n\pi)=n$. Equivalently, $f_{0}(x+iy)$ is the closest integer to $y/(2\pi)$; this is well-defined and continuous on $W_{0}$ because points where $y$ is an odd multiple of $\pi$ have been removed from $W_{0}$. Similarly, $f_{1}(x+iy)$ is the closest integer to $(y-\pi)/(2\pi)$. We find that the maps $\langle\exp,p_{0}\rangle\colon W_{0}\to V_{0}\times{\mathbb{Z}}$ and $\langle\exp,p_{1}\rangle\colon W_{1}\to V_{1}\times{\mathbb{Z}}$ are homeomorphisms, so $V_{0}$ and $V_{1}$ are trivially covered. We also have ${\mathbb{C}}\setminus\{0\}=V_{0}\cup V_{1}$, so $\exp\colon{\mathbb{C}}\to{\mathbb{C}}\setminus\{0\}$ is a covering map. The proof for the restricted map $\exp\colon i{\mathbb{R}}\to S^{1}$ is essentially the same.

###### Definition 22.8.

Let $p\colon X\to Y$ be a covering map.

• (a)

Consider a point $y\in Y$. A lift of $y$ means a point $\widetilde{y}\in X$ with $p(\widetilde{y})=y$.

• (b)

Consider a continuous path $u\colon[0,1]\to Y$. A lift of $u$ means a continuous path $\widetilde{u}\colon[0,1]\to X$ such that $p\circ\widetilde{u}=u$. This means in particular that $\widetilde{u}(t)$ is a lift of $u(t)$ for all $t$.

• (c)

More generally, let $T$ be any space and let $u\colon T\to Y$ be a continuous map. A lift of $u$ is a continuous map $\widetilde{u}\colon T\to X$ with $p\circ\widetilde{u}=u$.

• (d)

For a map $u\colon T\to Y$ as in (c), we say that $u$ is small if there is a trivially covered open set $V\subseteq Y$ such that $u(T)\subseteq V$.

###### Lemma 22.9.

Suppose we have a path-connected space $T$ and a small continuous map $u\colon T\to Y$. Suppose we also have points $t_{0}\in T$ and $x_{0}\in X$ with $u(t_{0})=p(x_{0})$. Then there is a unique lift $\widetilde{u}\colon K\to X$ such that $\widetilde{u}(t_{0})=x_{0}$.

###### Proof.

By assumption, we can choose an open subset $V\subseteq Y$ containing $u(K)$, and a homeomorphism $\langle p,f\rangle\colon p^{-1}(V)\to V\times F$ as in Definition 22.2. If $\widetilde{u}\colon T\to X$ is a lift of $u$, then we have $p(\widetilde{u}(t))=u(t)\in V$, so $\widetilde{u}(t)\in p^{-1}(V)$ for all $t\in T$, so we have a well-defined and continuous composite $f\circ\widetilde{u}\colon T\to F$. As $T$ is path connected and $F$ is discrete, this must be constant. Thus, if $\widetilde{u}(t_{0})=x_{0}$, then $f(\widetilde{u}(t))=f(x_{0})$ for all $t$. It follows that the only possibility is

 $\widetilde{u}(t)=\langle p,f\rangle^{-1}(u(t),f(x_{0})).$

###### Proposition 22.10.

Let $p\colon X\to Y$ be a covering map. Let $u$ be a path from $a$ to $b$ in $Y$, and let $\widetilde{a}\in X$ be a lift of $a$. Then there is a unique lift $\widetilde{u}$ of $u$ such that $\widetilde{u}(0)=\widetilde{a}$.

###### Proof.

Because $p$ is a covering map, we can find a family of trivially covered open sets $V_{i}\subseteq Y$ such that $Y=\bigcup_{i}V_{i}$. The preimages $u^{-1}(V_{i})$ then form an open covering of $[0,1]$. Because $[0,1]$ is a compact metric space, this covering has a Lebesgue number $\epsilon>0$ (by Proposition 8.31). Choose $n>1/\epsilon$ and divide $[0,1]$ into subintervals $T_{k}=[(k-1)/n,k/n]$ for $k=0,\dotsc,n$. By the Lebesgue number property, we can choose an index $i_{k}$ such that $T_{k}\subseteq u^{-1}(V_{i_{k}})$, so $u(T_{k})\subseteq V_{i_{k}}$. This means that the restriction of $u$ to $T_{k}$ is small, so Lemma 22.9 is applicable. We are given $\widetilde{a}\in X$ with $p(\widetilde{a})=a=u(0)$. We put $x_{0}=\widetilde{a}$, and use Lemma 22.9 to show that there is a unique map $\widetilde{u}_{1}\colon T_{1}\to X$ with $p(\widetilde{u}_{1}(t))=u(t)$ and $\widetilde{u}_{1}(0)=x_{0}$. We now define $x_{1}=\widetilde{u}_{1}(1/n)$, so $p(x_{1})=u(1/n)$. Applying Lemma 22.9 again, we see that there is a unique map $\widetilde{u}_{2}\colon T_{2}\to X$ with $p(\widetilde{u}_{2}(t))=u(t)$ and $\widetilde{u}_{2}(1/n)=x_{1}$. We put $x_{2}=\widetilde{u}_{2}(2/n)\in X$, so $p(x_{2})=u(2/n)$. We then repeat the process in the obvious way, to get a family of maps $\widetilde{u}_{k}\colon[(k-1)/n,k/n]\to X$ and points $x_{k}\in X$ with $\widetilde{u}_{k}(k/n)=x_{k}=\widetilde{u}_{k+1}(k/n)$. It follows that the maps $\widetilde{u}_{k}$ can be patched together to give a continuous map $\widetilde{u}\colon[0,1]\to X$ with $p\circ\widetilde{u}=u$ and $\widetilde{u}(0)=x_{0}=\widetilde{a}$. The same kind of induction shows that this is unique. ∎

###### Remark 22.11.

Note that Proposition 22.10 does not say anything about $\widetilde{u}(1)$. We know that $p(\widetilde{u}(t))=u(t)$ for all $t$, so in particular $p(\widetilde{u}(1))=u(1)=b$, so $\widetilde{u}(1)\in p^{-1}\{b\}$. However, if we have two different paths $u,v\colon a\rightsquigarrow b$ in $Y$ and we use the same starting point $\widetilde{a}\in p^{-1}\{a\}$ in both cases, then it can easily happen that the endpoints $\widetilde{u}(1),\widetilde{v}(1)$ are different elements of $p^{-1}\{b\}$. However, this cannot happen if there is a pinned homotopy between $u$ and $v$, as we will show later.

###### Corollary 22.12.

Let $p\colon X\to Y$ be a covering map. Suppose we have a path-connected space $T$ and a continuous map $u\colon T\to Y$. Suppose that $m,n\colon T\to X$ are continuous lifts of $u$, and that there is at least one point $t_{0}\in T$ with $m(t_{0})=n(t_{0})$; then $m=n$.

###### Proof.

Consider a point $t\in T$; we must show that $m(t)=n(t)$. As $T$ is path connected, we can choose a path $v$ from $t_{0}$ to $t$ in $T$. Now $m\circ v$ and $n\circ v$ are both lifts of the path $u\circ v\colon[0,1]\to Y$, and they satisfy $(m\circ v)(0)=m(t_{0})=n(t_{0})=(n\circ v)(0)$. Thus, the uniqueness clause in Proposition 22.10 tells us that $m\circ v=n\circ v$. In particular, we have $(m\circ v)(1)=(n\circ v)(1)$, or in other words $m(t)=n(t)$ as required. ∎

###### Proposition 22.13.

Let $T$ be a compact convex subset of ${\mathbb{R}}^{N}$, and suppose that $t_{0}\in T$. Let $p\colon X\to Y$ be a covering map, and let $u\colon T\to Y$ be continuous. Suppose that $x_{0}\in X$ with $p(x_{0})=u(t_{0})$. Then there is a unique continuous lift $\widetilde{u}\colon T\to X$ with $p\circ\widetilde{u}=u$ and $\widetilde{u}(t_{0})=x_{0}$.

###### Proof.

Roughly speaking, the idea is as follows: to define $\widetilde{u}(t)$, we move a short distance from $t$ towards $t_{0}$ to reach a point $t^{\prime}$, then $\widetilde{u}(t^{\prime})$ will already be defined and we define $\widetilde{u}(t)$ to be the unique lift of $u(t)$ that is close to $\widetilde{u}(t^{\prime})$. The rest of this proof should be seen as a more complete and rigorous version of this idea.

We first claim that there exists $\epsilon>0$ such that for all $t\in T$, the restricted map $u\colon OB(t,\epsilon)\to Y$ is small. (Here and elsewhere in this proof, notation for balls should be interpreted relative to $T$, so $OB(t,\epsilon)=\{t^{\prime}\in T\;|\;\|t-t^{\prime}\|<\epsilon\}$.) Indeed, for each $t\in T$ we can choose a trivially covered open set $V_{t}\subseteq Y$ containing $u(t)$. The set $u^{-1}(V_{t})$ is then open in $T$ and contains $t$. This means that the sets $u^{-1}(V_{t})$ form an open cover of the compact metric space $T$, so there is a Lebesgue number $\epsilon>0$. This has the required property.

Next, for $j>0$ we put $T_{j}=\{t\in T\;|\;\|t-t_{0}\|\leq j\epsilon/2\}$. We will prove by induction on $j$ that there is a unique continuous map $\widetilde{u}_{j}\colon T_{j}\to X$ with $\widetilde{u}_{j}(t_{0})=x_{0}$ and $p(\widetilde{u}_{j}(t))=u(t)$ for all $t\in T_{j}$. To start with, the map $u\colon T_{1}\to Y$ is small by our choice of $\epsilon$, so Lemma 22.9 gives $\widetilde{u}_{1}$. Suppose we have already constructed $\widetilde{u}_{j}$. For each $a\in T_{j}$, we note that the map $u\colon OB(a,\epsilon)\to Y$ is small, so there is a unique continuous $v_{a}\colon OB(a,\epsilon)\to X$ lifting $u$ with $v_{a}(a)=\widetilde{u}_{j}(a)$. Both $\widetilde{u}_{j}$ and $v_{a}$ restrict to give lifts of $u$ over the convex set $OB(a,\epsilon)\cap T_{j}$, and they agree at the point $a$, and the restricted map $u\colon OB(a,\epsilon)\cap T_{j}\to Y$ is small; it follows that $v_{a}$ agrees with $\widetilde{u}_{j}$ on $OB(a,\epsilon)\cap T_{j}$.

Now suppose that $a,b\in T_{j}$ and that the set $U=OB(a,\epsilon)\cap OB(b,\epsilon)$ is nonempty. We then find that the point $c=(a+b)/2$ must lie in $U$ and it also lies in $T_{j}$ because $T_{j}$ is convex. The maps $v_{a}$ and $v_{b}$ both agree with $\widetilde{u}_{j}$ at $c$, so they agree with each other. The restricted map $u\colon U\to Y$ is small, so we conclude that $v_{a}|_{U}=v_{b}|_{U}$. Because of this consistency property, we see that the maps $v_{a}$ can be combined to give a map $v\colon\bigcup_{a\in T_{j}}OB(a,\epsilon)\to X$. As the sets $OB(a,\epsilon)$ are all open, an open patching argument shows that $v$ is continuous. As each map $v_{a}$ is a lift of $u$, we see that $v$ is a lift of $u$. As $v_{a}$ agrees with $\widetilde{u}_{j}$ on $T_{j}$, we see that $v$ agrees with $\widetilde{u}_{j}$ on $T_{j}$.

It is also easy to see that $T_{j+1}$ is contained in the domain of $v$, so we can define $\widetilde{u}_{j+1}$ to be the restriction of $v$ to $T_{j+1}$. This is a continuous lift of $u$ extending $\widetilde{u}_{j}$ and therefore satisfying $\widetilde{u}_{j+1}(t_{0})=x_{0}$ as required.

As $T$ is assumed to be compact, it must be bounded. We therefore have $T_{j}=T$ for sufficiently large $j$, and this completes the proof. ∎