Worked Path Lifting and Endpoint Calculations

What Determines a Lifted Endpoint?

A path in a circle can return to its starting point while its lift in the real line ends at a different point from where it began. The covering map identifies real numbers that differ by an integer, so an endpoint on the circle does not by itself specify the endpoint of a lift. The chosen starting point of the lift matters. We will calculate lifts by writing a path with a continuous real-valued phase, then apply the same method to each coordinate of a torus path.

Give $\mathbb{R}$ its standard topology, give $\mathbb{C}$ its standard topology, and give $S^1=\{z\in\mathbb{C}:|z|=1\}$ the subspace topology. The standard exponential covering is $p:\mathbb{R}\to S^1$, defined by $p(u)=e^{2\pi i u}$. Thus $p(u)=p(v)$ exactly when $u-v\in\mathbb{Z}$. A path is a continuous map from $I=[0,1]$. A lift of a path $\gamma:I\to S^1$ is a path $\widetilde{\gamma}:I\to\mathbb{R}$ such that $p\circ\widetilde{\gamma}=\gamma$.

We use the Path Lifting Theorem and uniqueness of path lifts: for a covering map, a path $\gamma$, and a chosen $s$ in the fiber over $\gamma(0)$, there is a unique lift with $\widetilde{\gamma}(0)=s$. In every calculation below, the proposed formula is checked both at the starting point and after projection to the base.

The Phase Formula for the Circle

Theorem (Phase formula): Suppose $\gamma:I\to S^1$ has a continuous phase $\theta:I\to\mathbb{R}$, meaning $\gamma(t)=e^{2\pi i\theta(t)}$ for all $t\in I$. If $s\in\mathbb{R}$ satisfies $p(s)=\gamma(0)$, then the unique lift starting at $s$ is $$ \widetilde{\gamma}(t)=\theta(t)+s-\theta(0). $$ In particular, its endpoint is $s+\theta(1)-\theta(0)$.

Proof. Since $p(s)=p(\theta(0))$, the defining property of the exponential map gives $s-\theta(0)=k$ for some integer $k$. Define $\lambda(t)=\theta(t)+k$. This is continuous, $\lambda(0)=\theta(0)+k=s$, and $p(\lambda(t))=e^{2\pi i(\theta(t)+k)}=e^{2\pi i\theta(t)}=\gamma(t)$, because $e^{2\pi i k}=1$. Thus $\lambda$ is a lift with the required starting point. Uniqueness of path lifts makes it the unique such lift. Substituting $t=1$ gives $\lambda(1)=s+\theta(1)-\theta(0)$. $\square$

The formula separates two pieces of information: the change in phase, $\theta(1)-\theta(0)$, and the chosen starting point $s$. Changing the phase by an integer changes both $\theta(0)$ and $\theta(1)$ by that same integer, so the phase change and the resulting lift do not change.

Worked Example: A Circle Path That Makes More Than One Turn

Let $\gamma(t)=e^{2\pi i(1/6+5t/4)}$, and choose the lift to start at $s=7/6$. Find its formula and endpoint.

Here $\theta(t)=1/6+5t/4$, so $\theta(0)=1/6$. The initial point is valid because $s-\theta(0)=7/6-1/6=1$, an integer, and hence $p(7/6)=p(1/6)=\gamma(0)$. The phase formula gives $$ \widetilde{\gamma}(t)=\theta(t)+1=\frac{7}{6}+\frac{5t}{4}. $$ At $t=1$, the lift ends at $7/6+5/4=14/12+15/12=29/12$.

Check: the base path ends at phase $1/6+5/4=2/12+15/12=17/12$. The lift endpoint $29/12$ differs from $17/12$ by $1$, so $p(29/12)=p(17/12)=\gamma(1)$. The lift also starts at $7/6$, as required.

Crossing an Angle Cut

A continuous phase is often safer than choosing a preferred angle for each point of the circle. For example, representing angles by numbers in $[0,1)$ creates a jump when the path passes through $1$, even if the path itself is continuous. A lift must remain continuous and must also start at the specified point in the fiber.

Worked Example: Tracking the Starting Sheet Across the Circle

Let $\gamma(t)=e^{2\pi i(1/4+3t/2)}$, and choose the starting point $s=5/4$. Find the lift and compare its endpoint with the angle of $\gamma(1)$ in $[0,1)$.

Set $\theta(t)=1/4+3t/2$. Since $\theta(0)=1/4$, we have $s-\theta(0)=5/4-1/4=1$. Therefore $$ \widetilde{\gamma}(t)=\theta(t)+1=\frac{5}{4}+\frac{3t}{2}. $$ At $t=1/2$, the phase is $1/4+3/4=1$, so $\gamma(1/2)=1$, while the lift is $2$. This is consistent because $p(2)=1$. At $t=1$, the lift endpoint is $5/4+3/2=5/4+6/4=11/4$.

The base path endpoint has phase $1/4+3/2=1/4+6/4=7/4$, which represents the same circle point as $3/4$, since $7/4-3/4=1$. The lift endpoint $11/4$ projects to that point as well: $11/4-7/4=1$. Its starting value $5/4$ projects to the initial point because $5/4-1/4=1$.

Common pitfall: taking the endpoint angle $3/4$ as the lifted endpoint ignores the starting point. It is wrong here: the lift starts at $5/4$, not at the phase $1/4$, and the phase formula gives endpoint $11/4$, not $3/4$. Indeed, a lift ending at $3/4$ would not even project to $\gamma(1)$, since $3/4$ differs from the path’s endpoint phase $7/4$ by $-1$, so it does project; but it would be the endpoint of the lift starting at $-3/4$, not $5/4$. The starting condition distinguishes these lifts.

For a loop, $\gamma(1)=\gamma(0)$, so any continuous phase satisfies $\theta(1)-\theta(0)\in\mathbb{Z}$. The phase formula then says that the lifted endpoint differs from the lifted starting point by that integer. The loop closes downstairs whether or not its lift closes upstairs.

Product Coverings and Torus Endpoints

Give $T^2=S^1\times S^1$ the product topology and $\mathbb{R}^2$ its standard topology. The standard product covering is $P:\mathbb{R}^2\to T^2$, defined by $P(u,v)=(p(u),p(v))$. A path in $T^2$ is a pair of circle paths. A lift is a pair of real-valued coordinate paths, and the chosen starting point must project to the chosen starting point of the torus path.

Theorem (Product phase formula): Suppose $\Gamma(t)=(e^{2\pi i\theta_1(t)},e^{2\pi i\theta_2(t)})$, where $\theta_1,\theta_2:I\to\mathbb{R}$ are continuous. If $P(s_1,s_2)=\Gamma(0)$, then the unique lift starting at $(s_1,s_2)$ is $$ \widetilde{\Gamma}(t)= \bigl(\theta_1(t)+s_1-\theta_1(0),\, \theta_2(t)+s_2-\theta_2(0)\bigr). $$

Proof. Apply the circle phase formula to the first coordinate path with starting point $s_1$, and to the second coordinate path with starting point $s_2$. The resulting coordinate functions are continuous, so together they give a continuous path into $\mathbb{R}^2$. Applying $P$ to this pair gives $\Gamma(t)$ in each coordinate, and at $t=0$ the pair is $(s_1,s_2)$. Thus it is a lift with the required starting point. If another lift had the same starting point, each of its coordinate paths would be a lift of the corresponding circle path with the same initial value. Uniqueness for circle path lifts makes both coordinate paths equal to the displayed ones, proving uniqueness. $\square$

Worked Example: A Lift in the Torus

Let $\Gamma(t)=(e^{2\pi i(1/3+2t)},e^{2\pi i(3/4-5t/2)})$, and choose $(s_1,s_2)=(4/3,-1/4)$. Compute the lift and its endpoint.

For the first coordinate, $s_1-\theta_1(0)=4/3-1/3=1$. For the second, $s_2-\theta_2(0)=-1/4-3/4=-1$. Both differences are integers, so the chosen point projects to $\Gamma(0)$. The product phase formula gives $$ \widetilde{\Gamma}(t)= \left(\frac{4}{3}+2t,\,-\frac{1}{4}-\frac{5t}{2}\right). $$ At $t=1$, the endpoint is $\left(4/3+2,-1/4-5/2\right)=\left(10/3,-11/4\right)$.

Check: the first base phase at $t=1$ is $1/3+2=7/3$, and $10/3-7/3=1$, so the first coordinates project to the same circle point. The second base phase is $3/4-5/2=3/4-10/4=-7/4$, and $-11/4-(-7/4)=-1$, so the second coordinates also project to the same circle point. At $t=0$, the coordinate differences from the base phases are $4/3-1/3=1$ and $-1/4-3/4=-1$, confirming the chosen start.

A Reliable Endpoint Check

For a circle path given by a continuous phase, compute the lift by adding the constant $s-\theta(0)$ to the whole phase, not just to its endpoint. For a torus path, do this independently in each coordinate. In either case, the endpoint must project to the specified base endpoint, and the lift at $t=0$ must equal the chosen starting point.

These checks catch two different errors. Projecting the endpoint tests whether the calculated point lies in the correct fiber; checking the start tests whether it lies on the lift selected by the problem. The first check alone cannot select the correct lift, because many points in the covering space project to the same base point.

Check Your Understanding

In each problem, keep the chosen starting point explicit and verify the endpoint by applying the covering map.

For $\gamma(t)=e^{2\pi i(2/5-t)}$, find the lift starting at $7/5$ and its endpoint.
For $\gamma(t)=e^{2\pi i(3t)}$, find the endpoint of the lift starting at $0$. Does the lifted path close?
For $\Gamma(t)=(e^{2\pi i(1/5+t)},e^{2\pi i(1/2+2t)})$, find the lift starting at $(6/5,-1/2)$.
Explain why a circle path’s endpoint alone cannot determine the endpoint of its lift.
State the two checks that should be made after computing a proposed lift.