The covering lifting criterion

Try

Can the map f:S¹→S¹, f(z)=z⁶, be lifted through the covering p:S¹→S¹, p(z)=z⁴? A lift would be a continuous map g with g(z)⁴=z⁶. Choosing fourth roots separately at each point is not enough: those choices must fit continuously around the circle.

With basepoints all equal to 1, a full turn under f winds six times, while a loop projected through p winds a multiple of four times. Predict whether a lift exists. Compare the degree-eight map, for which g(z)=z² is an explicit lift.

Predict

Suppose the domain is a disk rather than a circle. Every based loop in the disk contracts. Predict why any continuous disk-to-base map lifts to a covering once its starting point upstairs has been specified. The domain need not be compact for the general criterion, but it needs hypotheses that let paths determine a continuous map.

Experiment

For the degree-four circle covering, examine the image of the domain’s fundamental group.

Map f from S¹	Image subgroup in Z	Contained in 4Z?
z↦z⁶	6Z	No
z↦z⁸	8Z	Yes
Constant map to 1	Zero subgroup	Yes

For the middle row, construct all four lifts by multiplying z² by the fourth roots of unity. Specify which lift takes 1 to i. Distinguish four possible starting points from nonuniqueness after a starting point has been fixed.

Understand

Let p:E→B be a covering, f:Y→B continuous, and choose y₀ in Y and e₀ above f(y₀). Assume Y is path connected and locally path connected. Then a lift taking y₀ to e₀ exists exactly when the subgroup induced by f on π₁(Y,y₀) is contained in the subgroup induced by p on π₁(E,e₀). Use the component of e₀ if E is not connected. Induced homomorphisms preserve our convention that αβ travels α first.

Necessity follows by functoriality: if f=p∘g, every loop image under f is the projection of its image under g. Sufficiency requires both independence of choices and continuity. For y in Y, choose a path γ from y₀ to y and define g(y) as the endpoint of the lift of f∘γ starting at e₀.

If δ is another choice, the loop γδ⁻¹ projects under f to a class in the covering subgroup. Such a loop lifts closed: it is based-homotopic to the projection of a loop upstairs, and homotopy lifting preserves endpoints. Lifting the concatenation shows that the lifts of f∘γ and f∘δ end at the same point. This proves that g is well defined. It also proves uniqueness because any lift must carry each γ to its unique lifted path.

For continuity, fix y and choose an evenly covered neighborhood U of f(y). Local path connectedness gives a path-connected open neighborhood V of y inside f⁻¹(U). Extend one chosen path to y by paths inside V. Their lifts stay in the single sheet over U containing g(y), so on V the constructed g is the inverse sheet homeomorphism composed with f. Thus g is continuous locally and hence globally. This is the point where a bare set-theoretic endpoint construction needs a topological hypothesis.

For f(z)=zⁿ and p(z)=zᵈ with d positive, the condition is nZ contained in dZ, equivalently d divides n. It covers negative and zero n as well. A continuous logarithm through R→S¹ is the analogous case with covering subgroup zero: the obstruction is nonzero winding, not a failure to calculate a root at one point.

Apply

Take Y=S¹×[0,1] and f(z,s)=z³. Explain why it cannot lift to the degree-two circle covering, even though every small patch admits a lift. Then restrict the angular coordinate to a proper open arc: describe how the loss of the global loop obstruction permits an argument branch. State the domain change rather than claiming that local roots automatically glue.

Continue

Use the criterion in both directions: produce a lift when subgroup containment holds, and produce a loop obstruction when it fails. The next lesson packages loop endpoints as a group action on an entire fiber. The source theorem and its local hypotheses appear in Hatcher, Section 1.3.

Continue to Monodromy and deck transformations.