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, e0 lies in only one of them. so when we pick the "slice" Vα to restrict p to (for the interval [0,s1]), we pick the one containing e0. 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 = S1, suppose we choose e0 = 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:
U1 = p((-π/2, π/2))
U2 = p((0,π))
U3 = p((π/2,3π/2))
U4 = p((π,2π))
now we need to pick our subintervals of [0,1].
we can pick s1 = 3/16, so that h([0,3/16]) ⊂ U1.
we can pick s2 = 7/16, so that h([3/16,7/16]) ⊂ U2.
we can pick s3 = 11/16, so that h([7/16,11/16]) ⊂ U3.
we can pick s4 = 15/16, so that h([11/16,15/16]) ⊂ U4.
finally, we see that h([15/16,1]) ⊂ U1.
now let's look at p-1(U1), 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
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 U2 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 U1, now we want the interval in p-1(U1) 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.