Short Arcs and the Exponential Map
Write \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology from \(\mathbb{C}\), and give \(\mathbb{R}\) its standard topology. Consider the map \(p:\mathbb{R}\to S^1\) defined by \(p(t)=e^{2\pi i t}\). The map repeats after every integer: \(p(s)=p(t)\) exactly when \(s-t\in\mathbb{Z}\). This periodicity means that an inverse cannot be chosen on the whole circle, but it can be chosen consistently on a sufficiently short open arc.
To see that \(U_{a,b}\) is open, note that \(p([b,a+1])\) is compact, since \([b,a+1]\) is compact and \(p\) is continuous. The circle is Hausdorff, so a compact subset of it is closed. Also \(p([b,a+1])\) is exactly the part of the circle not in \(p((a,b))\): every real parameter can be shifted by an integer into \([a,a+1]\), and the two endpoints \(a,b\) belong to the complementary arc. Thus \(U_{a,b}=S^1\setminus p([b,a+1])\), which is open.
Proof. A real number \(t\) belongs to \(p^{-1}(U_{a,b})\) exactly when \(p(t)=p(s)\) for some \(s\in(a,b)\). Since \(p(t)=p(s)\) exactly when \(t-s\) is an integer, this is equivalent to \(t\in(a+n,b+n)\) for some \(n\in\mathbb{Z}\). If \(n\ne m\), the intervals \((a+n,b+n)\) and \((a+m,b+m)\) are disjoint: their lengths are \(b-a<1\), so shifting one interval by a nonzero integer cannot make them overlap. On each interval, \(p\) is surjective onto \(U_{a,b}\), since \(p(s+n)=p(s)\). It is injective there because two parameters in that interval differ in absolute value by less than \(1\), and the only integer with absolute value less than \(1\) is \(0\). The restriction is continuous. Its inverse is continuous as well: it is the restriction of the inverse angular parametrization on the open arc, or, equivalently, \(p\) is a local homeomorphism on intervals of length less than \(1\). Hence each sheet maps homeomorphically onto the arc. \(\square\)
For each \(n\in\mathbb{Z}\), denote the inverse on the sheet \((a+n,b+n)\) by \(\theta_n:U_{a,b}\to(a+n,b+n)\). Its defining property is \(p(\theta_n(z))=z\). The subscript records which translated interval contains the chosen real angle; different choices differ by an integer.
Worked Example: Inverse Branches on an Arc
Take \(U=p((-1/6,1/6))\). Its sheets are \((-1/6+n,1/6+n)\), for integers \(n\), and \(\theta_n\) chooses the parameter in the \(n\)-th such interval. The point \(z=e^{\pi i/6}\) equals \(p(1/12)\), since \(2\pi(1/12)=\pi/6\). Because \(-1/6<1/12<1/6\), it lies in the \(n=0\) sheet, so \(\theta_0(z)=1/12\). In the sheet \(n=2\), the same point has inverse value \(\theta_2(z)=1/12+2=25/12\). Indeed, \(p(25/12)=p(1/12)\), because the parameters differ by the integer \(2\).
Check: \(25/12\) is in \((2-1/6,2+1/6)=(11/6,13/6)\), since \(22/12<25/12<26/12\). Thus the inverse value belongs to the specified sheet, and applying \(p\) returns \(z\).
Constructing a Lift by Local Continuation
A path in a space \(Y\) is a continuous map \(\gamma:I\to Y\), where \(I=[0,1]\). A lift of \(\gamma\) through \(p\) is a continuous map \(\widetilde{\gamma}:I\to\mathbb{R}\) such that \(p\circ\widetilde{\gamma}=\gamma\). A starting value \(r_0\in\mathbb{R}\) is compatible with the path when \(p(r_0)=\gamma(0)\). On any part of the path that stays inside an evenly covered arc, a lift is obtained by composing the path with one of the inverse branches. To continue along the whole path, choose successive branches so that the values agree at the subdivision points.
The subdivision can always be made finite. Here is the compactness fact that supplies it. If an open cover of \(I\) is given, then some \(\delta>0\) has the property that every subset of \(I\) of diameter less than \(\delta\) lies in a member of the cover. To verify this, suppose no such \(\delta\) exists. For each \(n\in\mathbb{N}\), choose a subset \(A_n\subseteq I\) of diameter less than \(1/n\) that lies in no cover member, and choose \(x_n\in A_n\). Compactness of \(I\) gives a subsequence \(x_{n_k}\) converging to some \(x\in I\). Choose a cover member containing \(x\); because it is relatively open in \(I\), it contains all points of \(I\) within some positive distance \(r\) of \(x\). For sufficiently large \(k\), both \(|x_{n_k}-x|<r/2\) and \(1/n_k<r/2\). Every point of \(A_{n_k}\) is then within distance \(r\) of \(x\), so \(A_{n_k}\) lies in that cover member, a contradiction.
Proof. For each \(t\in I\), choose an evenly covered open arc \(U_t\) containing \(\gamma(t)\). The sets \(\gamma^{-1}(U_t)\) form an open cover of \(I\). By the compactness fact above, there is \(\delta>0\) such that every subset of \(I\) of diameter less than \(\delta\) lies in one of these preimages. Choose a finite partition \(0=t_0<t_1<\cdots<t_m=1\) with \(t_j-t_{j-1}<\delta\) for each \(j\). Then \(\gamma([t_{j-1},t_j])\) lies in an evenly covered arc \(U_j\).
Start with \(\widetilde{\gamma}(0)=r_0\). On the first subinterval, select the sheet over \(U_1\) containing \(r_0\), and define \(\widetilde{\gamma}(t)=\theta(\gamma(t))\) using that sheet's inverse branch. This formula is defined throughout the subinterval and agrees with the starting value, because the inverse of \(\gamma(0)\) in that sheet is \(r_0\). At \(t_1\), select the sheet over \(U_2\) containing the value just obtained and use its inverse branch on \([t_1,t_2]\). Continue through the finite partition. At each joining time the formulas agree by the choice of sheet. Each formula is continuous, and finitely many continuous formulas that agree at their shared endpoints define a continuous map on \(I\). Each formula also satisfies \(p(\widetilde{\gamma}(t))=\gamma(t)\), so this constructs a lift with the required starting value.
For uniqueness, suppose \(\widetilde{\gamma}_1\) and \(\widetilde{\gamma}_2\) are lifts with the same starting value, and let \(E=\{t\in I:\widetilde{\gamma}_1(t)=\widetilde{\gamma}_2(t)\}\). The set \(E\) is nonempty, since \(0\in E\), and closed, since both lifts are continuous. It is also open relative to \(I\): if the lifts agree at \(t\), choose an evenly covered arc around \(\gamma(t)\), and let \(J\) be the sheet containing their common value. By continuity, on a sufficiently small relative neighborhood of \(t\), both lifts stay in \(J\) and \(\gamma\) stays in the arc. On that neighborhood each lift must be the same inverse branch applied to \(\gamma\), so they agree there. The interval \(I\) is connected, and a nonempty subset that is both open and closed in \(I\) is all of \(I\). Therefore the lifts agree everywhere. \(\square\)
Worked Example: Continuing Across the Angle \(1\)
Let \(\gamma(s)=p(3s/2)\) for \(s\in I\), and prescribe \(\widetilde{\gamma}(0)=0\). The path has a lift that can be built using two arcs. For \(0\le s\le4/9\), the parameter \(3s/2\) runs from \(0\) to \(2/3\), which lies in \((-1/5,3/4)\). Use the inverse branch on that interval to obtain \(\widetilde{\gamma}(s)=3s/2\).
For \(4/9\le s\le1\), the parameter \(3s/2\) runs from \(2/3\) to \(3/2\). This lies in the translated sheet \((3/5,31/20)\), which is the \(n=1\) sheet over the arc \(p((-2/5,11/20))\), since \(3/5=-2/5+1\) and \(31/20=11/20+1\). The inverse branch on this sheet again gives \(\widetilde{\gamma}(s)=3s/2\). At the joining point \(s=4/9\), both formulas give \(3(4/9)/2=2/3\), so they agree. Applying \(p\) on either piece gives \(p(3s/2)=\gamma(s)\); the lift starts at \(0\), as prescribed.
Check: At \(s=1\), the lift value is \(3/2\), and \(p(3/2)=e^{3\pi i}=-1\). The path endpoint is \(\gamma(1)=p(3/2)=-1\), as required.
Worked Example: A Constant Path and Its Starting Values
Let \(\gamma(s)=1\) for every \(s\in I\). Since \(p(n)=1\) for every \(n\in\mathbb{Z}\), each integer \(n\) is a compatible starting value. For the starting value \(n\), the constant function \(\widetilde{\gamma}(s)=n\) is a lift: \(p(\widetilde{\gamma}(s))=p(n)=1=\gamma(s)\) at every \(s\). Uniqueness says that this is the only lift with that particular starting value. Thus the starting value matters; uniqueness does not say that a path has only one lift without specifying its initial point.
Check: If \(n\ne m\), the two constant lifts have different values at \(s=0\), so they do not contradict uniqueness for a fixed starting value.
Why the Branch Choice Matters
A common mistake is to use a preferred angle at each point independently, without checking that the choices agree as the path moves. For example, for the path \(\gamma(s)=p(3s/2)\), the first local construction reaches the value \(2/3\) at \(s=4/9\). At that point the circle value also has the representation \(p(-1/3)\), because \(2/3-(-1/3)=1\). If the second piece is assigned the inverse value \(-1/3\) instead, the proposed real-valued function jumps from \(2/3\) to \(-1/3\) at the joining time. It is not continuous, so it is not a lift. The correction is to choose the inverse branch whose sheet contains the current lift value: here the \(n=1\) branch gives \(2/3\), not \(-1/3\).
The whole circle is not an evenly covered arc for \(p\). Its preimage is all of \(\mathbb{R}\), and \(p\) is not injective on \(\mathbb{R}\), since \(p(0)=p(1)=1\). Short arcs avoid this obstruction by splitting the preimage into disjoint intervals, each with its own continuous inverse. Local inverse branches are therefore the practical tool for following a path continuously; the essential rule is to continue from the sheet containing the value already reached.
Check Your Understanding
Use the arc sheets and the starting-value condition in your answers.
- Describe \(p^{-1}(p((1/4,1/2)))\), and give the inverse branch on the sheet \((5/4,3/2)\).
- For \(\gamma(s)=p(s/3)\) with \(\widetilde{\gamma}(0)=0\), find a lift and verify both defining conditions.
- Why does \(b-a<1\) ensure that the intervals \((a+n,b+n)\) are disjoint?
- For the constant path at \(1\), list three different compatible starting values and their corresponding lifts.
- Explain why choosing the inverse value \(-1/3\) at the joining point in the second worked example fails to continue the lift starting at \(0\).