Why a Single Agreement Determines the Whole Lift
A path lift records a path upstairs in a space that covers the space where the original path travels. The covering map controls the choices locally: over a suitable neighborhood, each sheet maps homeomorphically onto that neighborhood. The uniqueness theorem shows that these local choices cannot change partway along a path. If two lifts agree at even one parameter value, they agree throughout the interval.
We use the standard topology on \(I=[0,1]\). A path in a space \(X\) is a continuous map \(\alpha:I\to X\). A covering map \(p:E\to X\) is a continuous surjection such that every \(x\in X\) has an open neighborhood \(U\) for which \(p^{-1}(U)\) is a union of pairwise disjoint open sets \(V_\lambda\), and each restriction \(p|_{V_\lambda}:V_\lambda\to U\) is a homeomorphism. Such a neighborhood \(U\) is evenly covered, and the sets \(V_\lambda\) are its sheets.
The equality \(p\circ\widetilde{\alpha}=\alpha\) means that the two lifts always lie over the same point of \(X\). They may initially lie on different sheets, but if they meet at one parameter, the local one-to-one property of a sheet forces them to agree nearby. Connectedness will then extend that agreement over all of \(I\).
The Connectedness Fact We Need
A topological space is connected if it cannot be written as the union of two disjoint, nonempty open subsets. The uniqueness proof uses the connectedness of \(I\); here is a direct verification for this interval.
Proof. Suppose, to the contrary, that \(I=A\cup B\), where \(A\) and \(B\) are disjoint, nonempty, relatively open subsets of \(I\). Choose one point from each. By interchanging the names of \(A\) and \(B\) if needed, there are \(a<b\) with \(a\in A\) and \(b\in B\). The sets \(A\cap[a,b]\) and \(B\cap[a,b]\) are relatively open in \([a,b]\). Since \(A\) is relatively open and contains \(a\), there is a point \(x>a\) such that \([a,x]\subseteq A\). Define \(S=\{x\in[a,b]:[a,x]\subseteq A\}\) and let \(c=\sup S\). The preceding observation gives \(c>a\). Also \(c<b\): relative openness of \(B\) at \(b\) gives an interval immediately to the left of \(b\) contained in \(B\), so no point sufficiently close to \(b\) can belong to \(S\).
For each \(y\) with \(a\le y<c\), the definition of supremum gives an \(x\in S\) with \(y<x\), and then \(y\in[a,x]\subseteq A\). If \(c\in B\), relative openness of \(B\) gives a point \(y<c\), sufficiently close to \(c\), with \(y\in B\). This contradicts \(y\in A\). Therefore \(c\in A\). Relative openness of \(A\) at \(c\), together with \([a,c)\subseteq A\), now gives some \(x>c\) with \([a,x]\subseteq A\). That makes \(x\in S\), contradicting \(c=\sup S\). No such separation exists, so \(I\) is connected. \(\square\)
The Uniqueness Theorem
Proof. Define the agreement set \(C=\{t\in I:\widetilde{\alpha}(t)=\widehat{\alpha}(t)\}\). We show that both \(C\) and \(I\setminus C\) are open in \(I\). Since \(t_0\in C\), this will make \(C\) a nonempty subset that is both open and closed. Connectedness of \(I\) will then imply \(C=I\).
First let \(t\in C\), and write \(e=\widetilde{\alpha}(t)=\widehat{\alpha}(t)\) and \(x=\alpha(t)\). Choose an evenly covered open neighborhood \(U\) of \(x\). There is a sheet \(V\subseteq p^{-1}(U)\) containing \(e\). By continuity, \(\widetilde{\alpha}^{-1}(V)\cap\widehat{\alpha}^{-1}(V)\) is an open neighborhood of \(t\) in \(I\). For every \(s\) in this neighborhood, both lifted points belong to \(V\), and both map under \(p\) to \(\alpha(s)\). The map \(p|_V\) is injective, so \(\widetilde{\alpha}(s)=\widehat{\alpha}(s)\). Thus \(C\) is open.
Now let \(t\in I\setminus C\). The distinct points \(\widetilde{\alpha}(t)\) and \(\widehat{\alpha}(t)\) lie over the same point \(x=\alpha(t)\). Choose an evenly covered open neighborhood \(U\) of \(x\). The two points belong to different sheets \(V\) and \(W\), because the sheets are disjoint and each sheet maps injectively onto \(U\). Continuity gives an open neighborhood \(\widetilde{\alpha}^{-1}(V)\cap\widehat{\alpha}^{-1}(W)\) of \(t\) in \(I\). At every parameter in this neighborhood the lifted points lie in the disjoint sets \(V\) and \(W\), so they are unequal. Hence \(I\setminus C\) is open. Therefore \(C\) is both open and closed. As \(I\) is connected and \(C\ne\varnothing\), we have \(C=I\), as required. \(\square\)
The proof uses two different features of a covering. The sheets give local injectivity, which makes agreement persist near an agreement parameter and inequality persist near a disagreement parameter. Connectedness rules out a switch between those two behaviors. Continuity of the lifts is essential for keeping them in the chosen sheets near a given parameter.
Worked Applications
Worked Example: Two Sheets Over the Interval
Give \(I\) its standard topology and \(\{0,1\}\) the discrete topology. Give \(E=I\times\{0,1\}\) the product topology, let \(X=I\), and define \(p:E\to X\) by \(p(s,i)=s\). This is a covering map: the two open sheets \(I\times\{0\}\) and \(I\times\{1\}\) are disjoint, and the restriction of \(p\) to either sheet is a homeomorphism onto \(I\).
Take the path \(\alpha(t)=t/2\). Two lifts are \(\widetilde{\alpha}(t)=(t/2,0)\) and \(\widehat{\alpha}(t)=(t/2,1)\), since \(p(t/2,0)=t/2=\alpha(t)\) and \(p(t/2,1)=t/2=\alpha(t)\). At \(t=0\) their values are \((0,0)\) and \((0,1)\), so they do not agree there; indeed, they remain in different sheets for every \(t\).
If a third lift \(\gamma\) satisfies \(\gamma(0)=(0,0)\), then \(\gamma\) and \(\widetilde{\alpha}\) agree at \(0\). The theorem gives \(\gamma(t)=\widetilde{\alpha}(t)=(t/2,0)\) for all \(t\). At the endpoint, \(p(\gamma(1))=p(1/2,0)=1/2=\alpha(1)\). This example shows why the theorem requires a specified agreement value: a path can have different lifts starting at different points of the fiber.
Worked Example: A Lift Through the Circle Covering
Give \(\mathbb{R}\) its standard topology and \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology from \(\mathbb{C}\). Use the standard covering \(p:\mathbb{R}\to S^1\), \(p(s)=e^{2\pi i s}\). Consider \(\alpha(t)=e^{2\pi i(2t+1/3)}\). The map \(\widetilde{\alpha}(t)=2t+1/3\) is a lift, because \(p(\widetilde{\alpha}(t))=e^{2\pi i(2t+1/3)}=\alpha(t)\).
At \(t=0\), this lift has value \(1/3\), and at \(t=1\), it has value \(2+1/3=7/3\). Its projected endpoint is \(p(7/3)=e^{2\pi i(7/3)}=e^{2\pi i/3}=-1/2+i\sqrt{3}/2\), which is also \(\alpha(1)\). If \(\widehat{\alpha}\) is any other lift with \(\widehat{\alpha}(0)=1/3\), it agrees with \(\widetilde{\alpha}\) at \(0\), so uniqueness gives \(\widehat{\alpha}(t)=2t+1/3\) for every \(t\). In particular its endpoint must be \(7/3\).
The initial value condition cannot be omitted here. The formula \(\eta(t)=2t+4/3\) also gives a lift, since \(p(\eta(t))=e^{2\pi i(2t+4/3)}=e^{2\pi i(2t+1/3)}=\alpha(t)\); the equality holds because \(4/3-1/3=1\) and \(e^{2\pi i}=1\). But \(\eta(0)=4/3\ne1/3\), so this does not contradict the theorem. At \(t=1\), \(\eta(1)=10/3\), and \(p(10/3)=e^{2\pi i/3}=-1/2+i\sqrt{3}/2\), the same projected endpoint as before.
Worked Example: Why the Covering Hypothesis Matters
Give both copies of \(\mathbb{R}\) the standard topology and define \(p:\mathbb{R}\to\mathbb{R}\) by \(p(s)=s^2\). Let \(\alpha:I\to\mathbb{R}\) be \(\alpha(t)=t\). The two continuous maps \(\beta(t)=\sqrt{t}\) and \(\gamma(t)=-\sqrt{t}\) are lifts: for every \(t\in I\), \(p(\beta(t))=(\sqrt{t})^2=t=\alpha(t)\) and \(p(\gamma(t))=(-\sqrt{t})^2=t=\alpha(t)\).
At \(t=0\), both lifts equal \(0\). At \(t=1/4\), however, \(\beta(1/4)=1/2\) and \(\gamma(1/4)=-1/2\), so they differ. The map \(p\) is not a covering map at \(0\). In fact, if an open neighborhood \(U\) of \(0\) were evenly covered, the sheet containing \(0\) would be an open neighborhood \(V\) of \(0\) on which \(p\) is injective. But every open neighborhood \(V\) of \(0\) in \(\mathbb{R}\) contains some \(s>0\) and \(-s\), and \(p(s)=s^2=(-s)^2=p(-s)\). Thus \(p|_V\) is not injective. The calculation exhibits precisely the failure of the local sheet condition that the uniqueness proof needs.
A Common Mistake and the Practical Test
Common mistake. One might argue, “The two lifts agree at one point, and both are continuous, so they must agree everywhere.” Continuity alone does not imply that conclusion: in the square-map example, \(\beta(0)=\gamma(0)=0\), while \(\beta(1/4)=1/2\) and \(\gamma(1/4)=-1/2\). The corrected argument uses the covering hypothesis to show that the agreement set and its complement are both open. Only then does connectedness force the agreement set to be the whole interval.
For a particular pair of lifts, the theorem therefore reduces a global comparison to a single value. First check that both maps are continuous lifts of the same path under a covering map. Then check that they agree at one parameter. The theorem determines the entire lift, not merely its endpoint. It does not say that every starting point gives the same lift: distinct points in the fiber may produce distinct lifts, as the two-sheet and circle examples show.
Check Your Understanding
Use the agreement-set proof and the examples to answer these questions.
- In the uniqueness proof, why must the two unequal lifted points over \(\alpha(t)\) lie in different sheets of an evenly covered neighborhood?
- Which step in the proof uses continuity of the lifts, and what open sets does it produce?
- For the path \(\alpha(t)=e^{2\pi i(2t+1/3)}\), what is the initial value of the lift \(2t+1/3\), and why does that value determine the lift uniquely?
- For \(p(s)=s^2\) and \(\alpha(t)=t\), compute both lifts at \(t=1/4\). Which hypothesis of the uniqueness theorem fails?
- Why does connectedness apply to the agreement set once both it and its complement have been shown open?