the uniqueness derives from the fact that we are considering "evenly-covered" open neighborhoods U in B. this means the pre-images p^{-1}(U) are disjoint.

in particular, e_{0}lies in only one of them. so when we pick the "slice" V_{α}to restrict p to (for the interval [0,s_{1}]), we pick the one containing e_{0}. so not just any of the slices will do.

when we extend the lifted map to "the next subinterval", we pick the ONLY pre-image of that U that contains .

for example, given the path h(s) = (cos(4πs),sin(4πs)) in B = S^{1}, suppose we choose e_{0}= 0 in E = R, under the covering map:

p(x) = (cos(x),sin(x)).

for our open cover of the circle, all we need to do is choose "arc-lengths" less than 2pi radians. so let's use:

U_{1}= p((-π/2, π/2))

U_{2}= p((0,π))

U_{3}= p((π/2,3π/2))

U_{4}= p((π,2π))

now we need to pick our subintervals of [0,1].

we can pick s_{1}= 3/16, so that h([0,3/16]) ⊂ U_{1}.

we can pick s_{2}= 7/16, so that h([3/16,7/16]) ⊂ U_{2}.

we can pick s_{3}= 11/16, so that h([7/16,11/16]) ⊂ U_{3}.

we can pick s_{4}= 15/16, so that h([11/16,15/16]) ⊂ U_{4}.

finally, we see that h([15/16,1]) ⊂ U_{1}.

now let's look at p^{-1}(U_{1}), which is:

only one of these intervals contains the real number 0, namely the one with k = 0. so that's where we start, and since

we have:

on (-π/2,π/2), p is injective, so we see that on [0,3/16], that

(it's the injectivity of p on (-π/2,π/2) that allows us to uniquely "undo" the trig functions).

so that gives us the lift of h on [0,3/16]. to extend this to the next sub-interval [3/16,7/16],

we want to find the interval in the pre-image of U_{2}that contains ,

which is (0,π). again we see that , on this interval as well.

we continue in this way, obtaining a path in R that goes from 0 to 4π. the only interesting part is when we pick the slice for the last pre-image.

note that although we want a pre-image of the same open set U_{1}, now we want the interval in p^{-1}(U_{1}) that contains ,

which is no longer (-π/2,π/2) but rather (7π/2,9π/2) instead (k = 2, instead of k = 0).

yes, there are other paths in R that p would map to h, but this is the only continuous one.