Lift paths one sheet at a time

Try

Follow a point moving around the circle while recording its angle continuously on the real line. After one revolution the point returns to its starting circle position, but the recorded angle has increased by one turn. This is a path lift. Its starting sheet matters; once that sheet is chosen, continuity determines the rest.

Predict

Predict whether a base loop must lift to a closed loop. A single turn under R→S¹ ends one integer away, so it does not. Predict whether a lift may jump to a different sheet midway while keeping the same projected path. Such a jump would violate continuity. Local uniqueness turns this intuition into a proof.

Experiment

Consider the separate worked path α(t)=exp(2πi(2t+t²)) and choose lift start −1.

t	Real lift −1+2t+t²	Projected argument modulo one
0	−1	0
1/2	1/4	1/4
1	2	0

The circle path closes, but the lift travels three turns. Change the start to another integer and verify that the entire lift shifts by that integer while its displacement stays three. No finite sampling rule is being used to infer winding here; the parameterization is exact.

Understand

Let p:E→B be a covering and α:[0,1]→B a path. Given e₀ with p(e₀)=α(0), there is a unique path α̃ beginning at e₀ and satisfying p∘α̃=α.

For existence, pull back evenly covered neighborhoods along α. They form an open cover of the compact interval. Choose a finite subcover and a Lebesgue number, then subdivide the interval finely enough that each closed subinterval maps into one evenly covered neighborhood. On the first subinterval use the inverse of the sheet containing e₀. Its final point lies above the next starting value, selecting a sheet for the next subinterval. Continue finitely many times. The formulas agree at subdivision endpoints, so the pasting lemma gives a continuous lifted path.

For uniqueness, compare two lifts with the same start. Near any parameter time, the base path stays in an evenly covered neighborhood, and each lift stays in its respective sheet after shrinking the time neighborhood. If the sheet is the same, injectivity makes the lifts equal there. If the sheets differ, they are disjoint and the lifts remain unequal. Thus both the agreement set and its complement are open in [0,1]. Connectedness and agreement at zero force agreement everywhere. This proof does not require a global metric or Hausdorff assumption on the covering spaces.

The lift of a concatenation is obtained by lifting the first path, then lifting the second from the first endpoint. A reversed lift is the lift of the reversed path from the former endpoint. Both facts follow by constructing a candidate and invoking uniqueness. These operations will later turn loop classes into permutations of a fiber.

Apply

For p(z)=z^m, a base path exp(2πiθ(t)) with θ(0)=0 lifts from exp(2πij/m) to exp(2πi(j+θ(t))/m). A loop with integer θ(1)=w closes upstairs exactly when m divides w. At this stage this formula is checked for a supplied continuous angle; the next lessons prove that arbitrary circle paths admit the required continuous lifted angle.

Continue

A lift is a continuous global choice assembled from local inverses. The finite subdivision proves existence, while connectedness proves uniqueness. Next, lift an entire homotopy square so that the endpoint of a lifted loop cannot change during an endpoint-preserving deformation.

Continue to Lift homotopies and detect injectivity.