Consider a continuous map $p:X\to Y$. 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.
Consider the exponential map $\mathrm{exp}:\u2102\to \u2102\setminus \{0\}$. The fibres are
$${\mathrm{exp}}^{-1}\{r{e}^{i\theta}\}=\{\mathrm{log}(r)+i\theta +2n\pi i|n\in \mathbb{Z}\}$$ |
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.
Let $p:X\to Y$ 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 $f:{p}^{-1}(V)\to F$ such that the combined map $\u27e8p,f\u27e9:{p}^{-1}(V)\to V\times F$ 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.
Suppose we have open subsets ${V}^{\prime}\subseteq V\subseteq Y$ and that $V$ is trivially covered, as witnessed by a map $f:{p}^{-1}(V)\to F$. We then note that ${p}^{-1}({V}^{\prime})\subseteq {p}^{-1}(V)$, se we can restrict $f$ to get a map ${f}^{\prime}:{p}^{-1}({V}^{\prime})\to F$. It is not hard to check that the combined map $\u27e8p,{f}^{\prime}\u27e9:{p}^{-1}({V}^{\prime})\to {V}^{\prime}\times F$ is a homeomorphism, so ${V}^{\prime}$ is also trivially covered.
Take $X=\mathbb{R}\times \mathbb{Z}$ and $Y=\mathbb{R}$, and let $p:X\to Y$ 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 $f:X\to F$ by $f(x,n)=n$. Then the combined map $\u27e8p,f\u27e9:X\to Y\times F$ 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.
Take $X$ and $Y$ to be the unit circle in $\u2102$, or in other words $X=Y=\{{e}^{i\theta}|\theta \in \mathbb{R}\}$. Define $p:X\to Y$ 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 $$. The two square roots of $y$ are then $x={e}^{i\theta /2}$ (which has $\text{Re}(x)>0$) and $-x=-{e}^{i\theta /2}={e}^{i(\theta /2+\pi )}$ (which has $$). We therefore have
$${p}^{-1}({V}_{0})=\{z\in X|{z}^{2}\ne -1\}=X\setminus \{i,-i\}=\{z\in X|\text{Re}(z)\ne 0\},$$ |
and we can define ${f}_{0}:{p}^{-1}({V}_{0})\to \{1,-1\}$ by
$$ |
We then find that the map $\u27e8p,{f}_{0}\u27e9:{p}^{-1}({V}_{0})\to {V}_{0}\times \{1,-1\}$ is a homeomorphism, showing that ${V}_{0}$ is trivially covered. A similar approach can be used to check that the set ${V}_{1}=Y\setminus \{1\}$ is also trivially covered. Any element $y\in {V}_{1}$ can be expressed uniquely as $y={e}^{i\theta}$ with $$, and the two square roots are then $x={e}^{i\theta /2}$ (which has $\text{Im}(x)>0$) and $-x$ (which has $$). We therefore have
$${p}^{-1}({V}_{1})=\{z\in X|{z}^{2}\ne 1\}=X\setminus \{1,-1\}=\{z\in X|\text{Im}(z)\ne 0\},$$ |
and we can define ${f}_{1}:{p}^{-1}({V}_{1})\to \{1,-1\}$ by
$$ |
We then find that the map $\u27e8p,{f}_{1}\u27e9:{p}^{-1}({V}_{1})\to {V}_{1}\times \{1,-1\}$ 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 $\pi :{S}^{n}\to \mathbb{R}{P}^{n}$, 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 ${S}^{n}$.
The map $\pi \mathrm{:}{S}^{n}\mathrm{\to}\mathbb{R}\mathit{}{P}^{n}$ is a covering map.
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
$${\pi}^{-1}({V}_{a})=\{x|\pi (x)\in \pi ({U}_{a})\}=\{x|x\in {U}_{a}\text{or}-x\in {U}_{a}\}=\{x|x.a\ne 0\}.$$ |
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 ${f}_{a}:{\pi}^{-1}({V}_{a})\to \{1,-1\}$ by
$$ |
It is easy to see that the combined map $\u27e8\pi ,{f}_{a}\u27e9:{\pi}^{-1}({V}_{a})\to {V}_{a}\times \{1,-1\}$ is a homeomorphism, so ${V}_{a}$ is trivially covered. The sets ${V}_{a}$ cover all of $\mathbb{R}{P}^{n}$ (because $\pi (a)\in {V}_{a}$) so $\pi $ is a covering map as claimed. ∎
The map $\mathrm{exp}\mathrm{:}\u2102\mathrm{\to}\u2102\mathrm{\setminus}\mathrm{\{}\mathrm{0}\mathrm{\}}$ is a covering map, as is the map $\mathrm{exp}\mathrm{:}i\mathit{}\mathbb{R}\mathrm{\to}{S}^{\mathrm{1}}$ (where we identify ${S}^{\mathrm{1}}$ with $\mathrm{\{}z\mathrm{\in}\u2102\mathrm{:}\mathrm{|}z\mathrm{|}\mathrm{=}\mathrm{1}\mathrm{\}}$).
This is closely related to the proof in Example 22.5. We put
${V}_{0}$ | $=\u2102\setminus (-\mathrm{\infty},0]$ | ${V}_{1}$ | $=\u2102\setminus [0,\mathrm{\infty})$ | ||
${U}_{0}$ | $$ | ${U}_{1}$ | $$ | ||
${W}_{0}$ | $=\{x+iy|y\text{is not an odd multiple of}\pi \}$ | ${W}_{1}$ | $=\{x+iy|y\text{is not an even multiple of}\pi \}.$ |
If $y\in {V}_{0}$ then there is a unique choice of $r$ and $\theta $ with $$ and $r>0$ and $y=r{e}^{i\theta}$. It follows that the number $x=\mathrm{log}(r)+i\theta $ lies in ${U}_{0}$ and has $\mathrm{exp}(x)=y$. This means that the restricted map $\mathrm{exp}:{U}_{0}\to {V}_{0}$ is bijective. Standard complex analysis shows that the inverse is also continuous, so we have a homeomorphism $\mathrm{exp}:{U}_{0}\to {V}_{0}$. Similarly, the restricted map $\mathrm{exp}:{U}_{1}\to {V}_{1}$ is also a homeomorphism. From this we see that
${\mathrm{exp}}^{-1}({V}_{0})$ | $=\{{x}_{0}+2n\pi i|{x}_{0}\in {U}_{0},n\in \mathbb{Z}\}={W}_{0}$ | ||
${\mathrm{exp}}^{-1}({V}_{1})$ | $=\{{x}_{1}+2n\pi i|{x}_{1}\in {U}_{1},n\in \mathbb{Z}\}={W}_{1}.$ |
We can thus define continuous maps ${f}_{i}:{W}_{i}\to \mathbb{Z}$ (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 $\u27e8\mathrm{exp},{p}_{0}\u27e9:{W}_{0}\to {V}_{0}\times \mathbb{Z}$ and $\u27e8\mathrm{exp},{p}_{1}\u27e9:{W}_{1}\to {V}_{1}\times \mathbb{Z}$ are homeomorphisms, so ${V}_{0}$ and ${V}_{1}$ are trivially covered. We also have $\u2102\setminus \{0\}={V}_{0}\cup {V}_{1}$, so $\mathrm{exp}:\u2102\to \u2102\setminus \{0\}$ is a covering map. The proof for the restricted map $\mathrm{exp}:i\mathbb{R}\to {S}^{1}$ is essentially the same.
∎
Let $p:X\to Y$ be a covering map.
Consider a point $y\in Y$. A lift of $y$ means a point $\stackrel{~}{y}\in X$ with $p(\stackrel{~}{y})=y$.
Consider a continuous path $u:[0,1]\to Y$. A lift of $u$ means a continuous path $\stackrel{~}{u}:[0,1]\to X$ such that $p\circ \stackrel{~}{u}=u$. This means in particular that $\stackrel{~}{u}(t)$ is a lift of $u(t)$ for all $t$.
More generally, let $T$ be any space and let $u:T\to Y$ be a continuous map. A lift of $u$ is a continuous map $\stackrel{~}{u}:T\to X$ with $p\circ \stackrel{~}{u}=u$.
For a map $u:T\to Y$ as in (c), we say that $u$ is small if there is a trivially covered open set $V\subseteq Y$ such that $u(T)\subseteq V$.
Suppose we have a path-connected space $T$ and a small continuous map $u\mathrm{:}T\mathrm{\to}Y$. Suppose we also have points ${t}_{\mathrm{0}}\mathrm{\in}T$ and ${x}_{\mathrm{0}}\mathrm{\in}X$ with $u\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}\mathrm{=}p\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{)}$. Then there is a unique lift $\stackrel{\mathrm{~}}{u}\mathrm{:}K\mathrm{\to}X$ such that $\stackrel{\mathrm{~}}{u}\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}\mathrm{=}{x}_{\mathrm{0}}$.
By assumption, we can choose an open subset $V\subseteq Y$ containing $u(K)$, and a homeomorphism $\u27e8p,f\u27e9:{p}^{-1}(V)\to V\times F$ as in Definition 22.2. If $\stackrel{~}{u}:T\to X$ is a lift of $u$, then we have $p(\stackrel{~}{u}(t))=u(t)\in V$, so $\stackrel{~}{u}(t)\in {p}^{-1}(V)$ for all $t\in T$, so we have a well-defined and continuous composite $f\circ \stackrel{~}{u}:T\to F$. As $T$ is path connected and $F$ is discrete, this must be constant. Thus, if $\stackrel{~}{u}({t}_{0})={x}_{0}$, then $f(\stackrel{~}{u}(t))=f({x}_{0})$ for all $t$. It follows that the only possibility is
$$\stackrel{~}{u}(t)={\u27e8p,f\u27e9}^{-1}(u(t),f({x}_{0})).$$ |
∎
Let $p\mathrm{:}X\mathrm{\to}Y$ be a covering map. Let $u$ be a path from $a$ to $b$ in $Y$, and let $\stackrel{\mathrm{~}}{a}\mathrm{\in}X$ be a lift of $a$. Then there is a unique lift $\stackrel{\mathrm{~}}{u}$ of $u$ such that $\stackrel{\mathrm{~}}{u}\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\stackrel{\mathrm{~}}{a}$.
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 $\u03f5>0$ (by Proposition 8.31). Choose $n>1/\u03f5$ and divide $[0,1]$ into subintervals ${T}_{k}=[(k-1)/n,k/n]$ for $k=0,\mathrm{\dots},n$. 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 $\stackrel{~}{a}\in X$ with $p(\stackrel{~}{a})=a=u(0)$. We put ${x}_{0}=\stackrel{~}{a}$, and use Lemma 22.9 to show that there is a unique map ${\stackrel{~}{u}}_{1}:{T}_{1}\to X$ with $p({\stackrel{~}{u}}_{1}(t))=u(t)$ and ${\stackrel{~}{u}}_{1}(0)={x}_{0}$. We now define ${x}_{1}={\stackrel{~}{u}}_{1}(1/n)$, so $p({x}_{1})=u(1/n)$. Applying Lemma 22.9 again, we see that there is a unique map ${\stackrel{~}{u}}_{2}:{T}_{2}\to X$ with $p({\stackrel{~}{u}}_{2}(t))=u(t)$ and ${\stackrel{~}{u}}_{2}(1/n)={x}_{1}$. We put ${x}_{2}={\stackrel{~}{u}}_{2}(2/n)\in X$, so $p({x}_{2})=u(2/n)$. We then repeat the process in the obvious way, to get a family of maps ${\stackrel{~}{u}}_{k}:[(k-1)/n,k/n]\to X$ and points ${x}_{k}\in X$ with ${\stackrel{~}{u}}_{k}(k/n)={x}_{k}={\stackrel{~}{u}}_{k+1}(k/n)$. It follows that the maps ${\stackrel{~}{u}}_{k}$ can be patched together to give a continuous map $\stackrel{~}{u}:[0,1]\to X$ with $p\circ \stackrel{~}{u}=u$ and $\stackrel{~}{u}(0)={x}_{0}=\stackrel{~}{a}$. The same kind of induction shows that this is unique. ∎
Note that Proposition 22.10 does not say anything about $\stackrel{~}{u}(1)$. We know that $p(\stackrel{~}{u}(t))=u(t)$ for all $t$, so in particular $p(\stackrel{~}{u}(1))=u(1)=b$, so $\stackrel{~}{u}(1)\in {p}^{-1}\{b\}$. However, if we have two different paths $u,v:a\rightsquigarrow b$ in $Y$ and we use the same starting point $\stackrel{~}{a}\in {p}^{-1}\{a\}$ in both cases, then it can easily happen that the endpoints $\stackrel{~}{u}(1),\stackrel{~}{v}(1)$ 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.
Let $p\mathrm{:}X\mathrm{\to}Y$ be a covering map. Suppose we have a path-connected space $T$ and a continuous map $u\mathrm{:}T\mathrm{\to}Y$. Suppose that $m\mathrm{,}n\mathrm{:}T\mathrm{\to}X$ are continuous lifts of $u$, and that there is at least one point ${t}_{\mathrm{0}}\mathrm{\in}T$ with $m\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}\mathrm{=}n\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}$; then $m\mathrm{=}n$.
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 $u\circ v:[0,1]\to Y$, 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. ∎
Let $T$ be a compact convex subset of ${\mathbb{R}}^{N}$, and suppose that ${t}_{\mathrm{0}}\mathrm{\in}T$. Let $p\mathrm{:}X\mathrm{\to}Y$ be a covering map, and let $u\mathrm{:}T\mathrm{\to}Y$ be continuous. Suppose that ${x}_{\mathrm{0}}\mathrm{\in}X$ with $p\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{)}\mathrm{=}u\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}$. Then there is a unique continuous lift $\stackrel{\mathrm{~}}{u}\mathrm{:}T\mathrm{\to}X$ with $p\mathrm{\circ}\stackrel{\mathrm{~}}{u}\mathrm{=}u$ and $\stackrel{\mathrm{~}}{u}\mathit{}\mathrm{(}{t}_{\mathrm{0}}\mathrm{)}\mathrm{=}{x}_{\mathrm{0}}$.
Roughly speaking, the idea is as follows: to define $\stackrel{~}{u}(t)$, we move a short distance from $t$ towards ${t}_{0}$ to reach a point ${t}^{\prime}$, then $\stackrel{~}{u}({t}^{\prime})$ will already be defined and we define $\stackrel{~}{u}(t)$ to be the unique lift of $u(t)$ that is close to $\stackrel{~}{u}({t}^{\prime})$. The rest of this proof should be seen as a more complete and rigorous version of this idea.
We first claim that there exists $\u03f5>0$ such that for all $t\in T$, the restricted map $u:OB(t,\u03f5)\to Y$ is small. (Here and elsewhere in this proof, notation for balls should be interpreted relative to $T$, so $$.) 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 $\u03f5>0$. This has the required property.
Next, for $j>0$ we put ${T}_{j}=\{t\in T|\parallel t-{t}_{0}\parallel \le j\u03f5/2\}$. We will prove by induction on $j$ that there is a unique continuous map ${\stackrel{~}{u}}_{j}:{T}_{j}\to X$ with ${\stackrel{~}{u}}_{j}({t}_{0})={x}_{0}$ and $p({\stackrel{~}{u}}_{j}(t))=u(t)$ for all $t\in {T}_{j}$. To start with, the map $u:{T}_{1}\to Y$ is small by our choice of $\u03f5$, so Lemma 22.9 gives ${\stackrel{~}{u}}_{1}$. Suppose we have already constructed ${\stackrel{~}{u}}_{j}$. For each $a\in {T}_{j}$, we note that the map $u:OB(a,\u03f5)\to Y$ is small, so there is a unique continuous ${v}_{a}:OB(a,\u03f5)\to X$ lifting $u$ with ${v}_{a}(a)={\stackrel{~}{u}}_{j}(a)$. Both ${\stackrel{~}{u}}_{j}$ and ${v}_{a}$ restrict to give lifts of $u$ over the convex set $OB(a,\u03f5)\cap {T}_{j}$, and they agree at the point $a$, and the restricted map $u:OB(a,\u03f5)\cap {T}_{j}\to Y$ is small; it follows that ${v}_{a}$ agrees with ${\stackrel{~}{u}}_{j}$ on $OB(a,\u03f5)\cap {T}_{j}$.
Now suppose that $a,b\in {T}_{j}$ and that the set $U=OB(a,\u03f5)\cap OB(b,\u03f5)$ 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 ${\stackrel{~}{u}}_{j}$ at $c$, so they agree with each other. The restricted map $u:U\to Y$ 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 $v:{\bigcup}_{a\in {T}_{j}}OB(a,\u03f5)\to X$. As the sets $OB(a,\u03f5)$ 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 ${\stackrel{~}{u}}_{j}$ on ${T}_{j}$, we see that $v$ agrees with ${\stackrel{~}{u}}_{j}$ on ${T}_{j}$.
It is also easy to see that ${T}_{j+1}$ is contained in the domain of $v$, so we can define ${\stackrel{~}{u}}_{j+1}$ to be the restriction of $v$ to ${T}_{j+1}$. This is a continuous lift of $u$ extending ${\stackrel{~}{u}}_{j}$ and therefore satisfying ${\stackrel{~}{u}}_{j+1}({t}_{0})={x}_{0}$ 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. ∎