Why One Shared Value Can Determine a Lift
A lift of a map is a map into the covering space whose composition with the covering map gives the original map. For a path, uniqueness says that the initial point determines its lift. The corresponding statement for maps on a connected domain is stronger in one respect: the domain need not be path-connected. Instead, local properties of a covering ensure that two lifts cannot agree in one region and differ in another without splitting the domain into disjoint open pieces.
Let \(p:E\to B\) be a covering map. Here this means that for each \(b\in B\), there is an open neighborhood \(U\) of \(b\) such that \(p^{-1}(U)\) is a union of pairwise disjoint open sets, called sheets over \(U\), and \(p\) maps each sheet homeomorphically onto \(U\). In particular, \(p\) is one-to-one on each sheet. The spaces \(E\) and \(B\) have their given topologies; no Hausdorff assumption is required below.
The Uniqueness Theorem
The local reason is that near any point of \(X\), both lifts must lie in sheets over the same evenly covered neighborhood. If they are in the same sheet, the covering map is one-to-one there, so the lifts agree nearby. If they are in different sheets, the lifts remain different nearby. Thus the domain splits into open regions of agreement and disagreement unless one of those regions is empty.
Proof. Define \(A=\{x\in X:f(x)=g(x)\}\). We show that both \(A\) and \(X\setminus A\) are open. Let \(x\in A\), and write \(e=f(x)=g(x)\) and \(b=h(x)=p(e)\). Choose an evenly covered open neighborhood \(U\) of \(b\), and let \(S\) be the sheet over \(U\) containing \(e\). Since \(f\) and \(g\) are continuous, \(f^{-1}(S)\cap g^{-1}(S)\) is an open neighborhood of \(x\). For any \(y\) in this neighborhood, \(f(y),g(y)\in S\) and \(p(f(y))=h(y)=p(g(y))\). Since \(p\) is one-to-one on \(S\), \(f(y)=g(y)\). Hence \(A\) is open.
Now let \(x\notin A\), and put \(b=h(x)=p(f(x))=p(g(x))\). Choose an evenly covered open neighborhood \(U\) of \(b\). The points \(f(x)\) and \(g(x)\) lie in different sheets over \(U\): if they lay in the same sheet, injectivity of \(p\) there would give \(f(x)=g(x)\). Call their sheets \(S_1\) and \(S_2\). The set \(f^{-1}(S_1)\cap g^{-1}(S_2)\) is an open neighborhood of \(x\). At every point \(y\) in it, \(f(y)\) and \(g(y)\) lie in disjoint sheets, so they differ. Thus \(X\setminus A\) is open as well.
Consequently, \(A\) is both open and closed. It is nonempty because \(x_0\in A\). If \(X\setminus A\) were nonempty, \(A\) and \(X\setminus A\) would be disjoint nonempty open sets whose union is \(X\), contradicting connectedness. Therefore \(A=X\), which means \(f=g\). \(\square\)
Notice what the proof does not require. It does not use a metric on \(X\), a path between \(x_0\) and every other point, or a Hausdorff condition on \(E\) or \(B\). Connectedness alone is enough.
Using the Theorem on an Interval
The interval \(I=[0,1]\), with its usual subspace topology from \(\mathbb{R}\), is connected; this is the standard connectedness theorem for intervals, which follows from the intermediate value theorem. Therefore, if two lifts of the same path agree at \(0\), the uniqueness theorem says they agree at every \(t\in I\). The same is true if they agree at any other point of the interval.
Worked Example: Lifts of a Loop into the Circle
Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology from \(\mathbb{C}\), and consider \(p:S^1\to S^1\) defined by \(p(z)=z^2\). This is the standard two-sheeted covering. To check its local covering structure, take any \(w_0=e^{i\theta_0}\) and the open arc \(U=\{e^{i\theta}:\theta_0-\pi/4<\theta<\theta_0+\pi/4\}\). Its preimage consists of two disjoint open arcs, described by angles \((\theta_0/2-\pi/8,\theta_0/2+\pi/8)\) and that interval shifted by \(\pi\). Squaring maps each arc homeomorphically onto \(U\).
Let \(h:I\to S^1\) be \(h(t)=\cos(2\pi t)+i\sin(2\pi t)\). Define \(f(t)=\cos(\pi t)+i\sin(\pi t)\). The double-angle identities give \(p(f(t))=f(t)^2=\cos(2\pi t)+i\sin(2\pi t)=h(t)\), so \(f\) is a lift. At the endpoints, \(f(0)=1\), \(f(1)=-1\), and \(p(f(0))=p(f(1))=1\), as required since \(h(0)=h(1)=1\).
At \(t=0\), the fiber over \(h(0)=1\) consists of \(1\) and \(-1\), because \(z^2=1\) has exactly those two solutions on \(S^1\). Thus a lift beginning at \(1\) must equal \(f\), by the theorem. A lift beginning at \(-1\) must equal the lift \(-f\), since \(p((-f(t)))=p(f(t))=h(t)\) and \((-f)(0)=-1\). These are distinct lifts: for every \(t\), \(f(t)\ne -f(t)\) because \(f(t)\in S^1\) is never zero.
Check: at \(t=1\), \(f(1)=-1\), so \(p(f(1))=(-1)^2=1=h(1)\). For the other lift, \((-f)(1)=1\), so \(p((-f)(1))=1^2=1=h(1)\). Both endpoint values therefore satisfy the lift equation.
Applying Uniqueness to a Disk
A domain can be connected without requiring that the proof follow paths through it. For a useful way to apply the theorem to a disk, recall that \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}\) has the subspace topology from \(\mathbb{R}^2\). It is path-connected: for \(u,v\in D^2\), the segment \(\gamma(t)=(1-t)u+tv\) remains in \(D^2\), since \(\|\gamma(t)\|\leq (1-t)\|u\|+t\|v\|\leq 1\) for \(0\leq t\leq 1\). Path-connected spaces are connected: a separation into disjoint nonempty open sets would, along a path between them, give a separation of \(I\) by their preimages, contradicting connectedness of \(I\).
Worked Example: A Lift Defined on the Disk
Use the same covering \(p:S^1\to S^1\), \(p(z)=z^2\). For \((x,y)\in D^2\), define \(h(x,y)=\cos(2\pi x)+i\sin(2\pi x)\) and \(f(x,y)=\cos(\pi x)+i\sin(\pi x)\). Both maps are continuous, since their coordinate functions are continuous. Squaring \(f(x,y)\) and applying the double-angle identities gives \(p(f(x,y))=\cos(2\pi x)+i\sin(2\pi x)=h(x,y)\); hence \(f\) is a lift.
At the center \((0,0)\), \(f(0,0)=1\). If \(g:D^2\to S^1\) is any other lift of this same \(h\) and \(g(0,0)=1\), the uniqueness theorem applies because \(D^2\) is connected. It follows that \(g=f\) everywhere on the disk. In contrast, \(-f\) is also a lift, since \(p(-f)=(-f)^2=f^2=h\), but \((-f)(0,0)=-1\). The theorem does not say that lifts with different values at the chosen point must agree.
Check: at \((1,0)\in D^2\), \(h(1,0)=\cos(2\pi)+i\sin(2\pi)=1\) and \(f(1,0)=\cos(\pi)+i\sin(\pi)=-1\). Therefore \(p(f(1,0))=(-1)^2=1=h(1,0)\).
Why Connectedness Matters
Worked Example: Failure on a Disconnected Domain
Let \(X=\{a,b\}\) have the discrete topology, so every subset is open. Use again \(p:S^1\to S^1\), \(p(z)=z^2\), and let \(h:X\to S^1\) be constant with value \(1\). Define \(f(a)=f(b)=1\), while \(g(a)=1\) and \(g(b)=-1\). Both maps are continuous because every map from a discrete space is continuous. Also, \(p(f(a))=p(f(b))=1\), \(p(g(a))=1^2=1\), and \(p(g(b))=(-1)^2=1\). Thus \(p\circ f=h=p\circ g\).
The lifts agree at \(a\), but they differ at \(b\). The domain is not connected: \(\{a\}\) and \(\{b\}\) are disjoint nonempty open sets whose union is \(X\). The example shows why the connectedness hypothesis cannot simply be removed.
Common pitfall. From \(p(f(x))=p(g(x))\), it is wrong to conclude that \(f(x)=g(x)\) by treating \(p\) as globally one-to-one. In the last example, \(p(1)=1\) and \(p(-1)=(-1)^2=1\), although \(1\ne -1\). The correction is to use injectivity only after showing that both values lie in the same sheet over an evenly covered neighborhood. The theorem’s connectedness argument then propagates local agreement across the entire domain.
In practice, the theorem is useful whenever a lift has already been constructed by one method and a second lift is presented by another. Rather than comparing their formulas at every point, verify that both are continuous lifts of the same map, find one point where their values agree, and check that the domain is connected. For an interval or a disk, these conditions give equality everywhere.
Check Your Understanding
For each question, identify which hypotheses of the uniqueness theorem are being used.
- Two lifts of the same map from \(I\) agree at \(t=1\). What can be concluded?
- Why does the proof not require \(E\) or \(B\) to be Hausdorff?
- For the disk example, verify directly that \(p(-f(x,y))=h(x,y)\) for every \((x,y)\in D^2\).
- In the disconnected-domain example, which pair of values lies in the same fiber but is unequal?
- Can two lifts of a map from a connected domain be different if they have different values at every point? Explain what the theorem does and does not assert.