What the Path Lifting Theorem Says
A covering map looks locally like a collection of separate copies of the same open set. A path in the base can therefore be followed in one of those copies, provided its starting point has been chosen. The path lifting theorem makes this precise: it guarantees a lift on the entire interval and says that the chosen starting point determines exactly one such lift.
We use \(I=[0,1]\) with its standard topology. A path in a topological space \(B\) is a continuous map \(\gamma:I\to B\). A covering map \(p:E\to B\) is a surjective continuous map such that for every \(b\in B\) there is an open neighborhood \(U\) of \(b\) for which \(p^{-1}(U)\) is a disjoint union of open sets \(V_\alpha\), and each restriction \(p|_{V_\alpha}:V_\alpha\to U\) is a homeomorphism. Such a neighborhood \(U\) is evenly covered. The topologies on \(E\) and \(B\) are part of this definition.
The starting-point condition is essential: it specifies which point of the initial fiber the lift must use. The endpoint conclusion says that the lifted path ends over the endpoint of the original path. It does not say that every point of the terminal fiber can be reached by lifting this particular path from this particular \(e_0\).
Uniqueness and Existence
We first establish uniqueness on any initial subinterval. We use the standard fact that intervals in \(\mathbb{R}\) are connected. We also use that a continuous image of a connected space is connected: if the image were separated into two disjoint nonempty relatively open sets, their preimages would separate the original space.
Uniqueness. Suppose \(\alpha,\beta:[0,b]\to E\) are lifts of the same path on \([0,b]\), where \(b\in[0,1]\), and \(\alpha(0)=\beta(0)\). Define \(A=\{t\in[0,b]:\alpha(t)=\beta(t)\}\). We show that both \(A\) and its complement are open relative to \([0,b]\). If \(t\in A\), take an evenly covered neighborhood \(U\) of the common image \(p(\alpha(t))\). By continuity, for all \(s\) sufficiently close to \(t\), both \(\alpha(s)\) and \(\beta(s)\) lie in the sheet over \(U\) that contains \(\alpha(t)=\beta(t)\). On that sheet \(p\) is one-to-one; since \(p(\alpha(s))=p(\beta(s))\), the two values are equal. Thus \(A\) is open. If \(t\notin A\), choose an evenly covered neighborhood \(U\) of \(p(\alpha(t))=p(\beta(t))\). The distinct points \(\alpha(t)\) and \(\beta(t)\) lie in distinct sheets over \(U\), since \(p\) is one-to-one on each sheet. Continuity keeps their nearby values in those respective, disjoint sheets, so they remain unequal. Thus the complement of \(A\) is open as well. Since \(A\) contains \(0\), it is nonempty; connectedness of \([0,b]\) implies \(A=[0,b]\). This argument includes \(b=0\), where the interval consists of its single point. \(\square\)
Existence. Let \(S\subseteq[0,1]\) consist of those \(s\) for which a lift on \([0,s]\) starting at \(e_0\) exists. The constant map on \([0,0]\) with value \(e_0\) is such a lift, so \(S\) is nonempty. Write \(a=\sup S\). By the uniqueness just proved, any two lifts on initial intervals agree on their common domain. We can therefore define \(\lambda(t)\) for \(0\le t<a\) by taking the value at \(t\) of any lift whose domain reaches beyond \(t\). Such a lift exists by the definition of supremum. The value is independent of the choice, and \(\lambda\) is continuous on \([0,a)\): near any \(t<a\), it agrees with one of the continuous lifts already defined on an initial interval containing \(t\).
If \(a=0\), choose an evenly covered neighborhood \(U\) of \(\gamma(0)\), and let \(V\) be the sheet over \(U\) containing \(e_0\). For a sufficiently short initial interval, \(\gamma\) stays in \(U\). Composing \(\gamma\) on that interval with the inverse of \(p|_V\) gives a lift starting at \(e_0\). Hence \(S\) contains a positive number, contradicting \(a=0\). Thus \(a>0\).
Choose an evenly covered neighborhood \(U\) of \(\gamma(a)\). Continuity of \(\gamma\) gives \(c\) with \(0\le c<a\) such that \(\gamma([c,a])\subseteq U\). The map \(\lambda\) is defined and continuous on \([c,a)\), and its image lies in \(p^{-1}(U)\). The interval \([c,a)\) is connected, so its image is connected and must lie in a single sheet \(V\) over \(U\). Indeed, the sheets are disjoint open subsets of \(p^{-1}(U)\); if the image met two of them, its intersections with one sheet and with the union of the others would separate it. On this sheet, \(p|_V\) has an inverse. Define \(\lambda(a)=(p|_V)^{-1}(\gamma(a))\). This extends \(\lambda\) continuously to \(a\), because on \([c,a]\) it is given by the continuous map \((p|_V)^{-1}\circ\gamma\). Therefore \(a\in S\).
If \(a<1\), continuity of \(\gamma\) lets us choose \(d>a\) such that \(\gamma([a,d])\subseteq U\). Using the inverse of \(p|_V\) on this interval extends the lift from \([0,a]\) to \([0,d]\). The extension agrees at \(a\) by its definition, so \(d\in S\), contradicting \(a=\sup S\). Consequently \(a=1\), and the lift exists on all of \(I\). Its initial value is \(e_0\); uniqueness follows from the uniqueness argument above with \(b=1\). Finally, applying \(p\circ\widetilde{\gamma}=\gamma\) at \(1\) gives the stated endpoint condition. \(\square\)
Checking the Theorem in Circle Coverings
In the examples below, \(S^1=\{z\in\mathbb{C}:|z|=1\}\) has the subspace topology from \(\mathbb{C}\). We use the standard covering maps \(p:\mathbb{R}\to S^1\), \(p(x)=e^{2\pi i x}\), and \(q:S^1\to S^1\), \(q(z)=z^2\). The real line has its standard topology. Their covering-map property is the standard local description of these circle coverings.
Worked Example: A Path That Winds Three Times
Take \(p(x)=e^{2\pi i x}\), the path \(\gamma(t)=e^{2\pi i(3t)}\), and the initial point \(e_0=2\). Since \(p(2)=e^{4\pi i}=1=\gamma(0)\), the initial-point condition holds. Define \(\widetilde{\gamma}(t)=2+3t\). Then \(p(\widetilde{\gamma}(t))=e^{2\pi i(2+3t)}=e^{4\pi i}e^{6\pi i t}=e^{6\pi i t}=\gamma(t)\), and \(\widetilde{\gamma}(0)=2\). The theorem says this is the unique lift from \(2\). At the endpoint, \(\widetilde{\gamma}(1)=5\), while \(\gamma(1)=e^{6\pi i}=1\); checking gives \(p(5)=e^{10\pi i}=1\), as required.
Worked Example: Squaring on the Circle
Let \(q(z)=z^2\), let \(\gamma(t)=e^{2\pi i t}\), and choose \(z_0=1\). Since \(q(1)=1=\gamma(0)\), this is an allowed initial point. Set \(\widetilde{\gamma}(t)=e^{\pi i t}\). Then \(q(\widetilde{\gamma}(t))=(e^{\pi i t})^2=e^{2\pi i t}=\gamma(t)\), and \(\widetilde{\gamma}(0)=e^0=1\). Thus it is the unique lift from \(1\). Its endpoint is \(e^{\pi i}=\cos(\pi)+i\sin(\pi)=-1\), and \(q(-1)=(-1)^2=1=\gamma(1)\).
The other point \(-1\) in the initial fiber gives a different lift, namely \(-e^{\pi i t}\). It starts at \(-1\), and its square is again \(e^{2\pi i t}\). Its endpoint is \(1\). This does not contradict uniqueness: the initial points differ.
Worked Example: A Constant Path and an Unspecified Start
Use \(p(x)=e^{2\pi i x}\) and the constant path \(\gamma(t)=1\). Choose \(e_0=4\), for which \(p(4)=e^{8\pi i}=1\). The constant map \(\widetilde{\gamma}(t)=4\) is a lift: \(p(\widetilde{\gamma}(t))=e^{8\pi i}=1=\gamma(t)\) for every \(t\), and it starts at \(4\). Its endpoint is \(4\), in the fiber \(p^{-1}(\{1\})=\mathbb{Z}\).
If no initial point is specified, there are many lifts of this same path: for every integer \(n\), the constant map with value \(n\) is a lift, since \(e^{2\pi i n}=1\). The theorem gives uniqueness only after the starting point has been fixed.
Endpoint Conditions and a Common Mistake
For fixed \(p\), \(\gamma\), and \(e_0\), the theorem determines one lifted endpoint. It must lie in the terminal fiber, but membership in that fiber alone does not establish that a point is the endpoint of the lift from \(e_0\). If both endpoints are prescribed, a lift with those endpoints exists precisely when the unique lift from the prescribed initial point has the prescribed terminal point.
Common mistake. For the map \(q(z)=z^2\), one might try to lift \(\gamma(t)=e^{2\pi i t}\) from \(1\) by setting \(\widetilde{\gamma}(t)=\gamma(t)\). This has the correct initial value, but its image under \(q\) is \(q(\widetilde{\gamma}(t))=(e^{2\pi i t})^2=e^{4\pi i t}\), which is not \(\gamma(t)=e^{2\pi i t}\) for every \(t\). For example, at \(t=\tfrac14\), these values are \(e^{\pi i}=-1\) and \(e^{\pi i/2}=i\), respectively. The correct lift from \(1\) is \(e^{\pi i t}\), whose square is \(e^{2\pi i t}\). A lift must satisfy the composition equation at every time, not merely have the right starting point and an endpoint in the right fiber.
The theorem is useful because it lets us track a path in a covering space without making arbitrary choices at intermediate times. Once the initial point in the fiber is fixed, both the entire lift and its endpoint are determined. The proof uses the local homeomorphisms in the definition of a covering map and the connectedness of an interval; it does not require \(E\) to be Hausdorff.
Check Your Understanding
Use the composition condition and the initial-point requirement to check your answers.
- For \(p(x)=e^{2\pi i x}\), \(\gamma(t)=e^{2\pi i(2t)}\), and initial point \(3\), write the lift and find its endpoint.
- For \(q(z)=z^2\) and \(\gamma(t)=e^{2\pi i t}\), what is the endpoint of the lift starting at \(-1\)?
- Why does the path lifting theorem not assert that every point of \(p^{-1}(\{\gamma(1)\})\) is the endpoint of a lift starting at a fixed \(e_0\)?
- State the two conditions a map \(\widetilde{\gamma}:I\to E\) must satisfy to be the lift specified by the theorem.
- In the uniqueness proof, why is it not necessary to assume that \(E\) is Hausdorff?