# Covering Spaces, Different Lifts

## Different Lifts

Let S~ be a covering space for S, and let e(D) be a map from a connected domain D into S. Let f(D) and g(D) be maps into S~ that are both lifts of e. We will prove that f and g are everywhere equal or nowhere equal.

We usually think of D as the unit interval, whence e becomes a path. Now f and g are lifts of the same path, and if f = g at any point, the unique path lifting theorem makes the paths equal everywhere. But sometimes we can lift more than just a path. We already talked about lifting the image of a hypercube. Whatever the domain, two lifts are everywhere equal or nowhere equal.

Select v in D such that f(v) = g(v) = x in S~. Let x project to y in S, so that e(v) = y. Find an open set U in S such that U contains y, and U is evenly covered. Thus U~ in S~ is homeomorphic to U, through p.

Across U~, f has to be e composed with p inverse. Yet the same is true of g. So f and g agree on the open set U~. Since these are continuous functions, take the preimages of U~ under f and g, and find two open sets in D. Intersect these open sets to find a smaller open set containing v. Note that f and g agree on this intersection. A similar open set can be created for every v where f(v) = g(v). Therefore the set of points in D where f and g agree is open.

Next we will prove that this same set is closed. In other words, the set of points in D where f and g disagree is open.

Remember that f and g are both lifts of e, so even if f(v) ≠ g(v), the two images are in the fiber of y. In other words, p(f(v)) = p(g(v)) = y.

Again, find an evenly covered open set U in S that contains y. Let Uf and Ug be the preimages of U in S~, for the lifts f and g respectively. If these intersect, then their union is a path connected open set that is in the preimage of U. In other words, their union is one of the path connected components in the preimage. It is the union that must be homeomorphic to U. However, two different points map to y, so we have a contradiction, and Uf and Ug are disjoint.

As before, f is e composed with p inverse (back to Uf), and g is e composed with p inverse (back to Ug). The functions have disjoint images when they pass through U. Take the intersection of the preimages of Uf and Ug, under f and g respectively, and find an open set in D, containing v, where f and g disagree. Therefore the set of points in D where f and g disagree is open.

The subspace of D where f and g agree is both open and closed. Since D is connected, this is all of D, or nothing. The two lifts agree everywhere or nowhere.

## Fiber Correspondence

Assume there is a path from x to y. By the path lifting theorem, each x~ in S~ above x implies a y~ over y. This induces a map from the fiber of x to the fiber of y. If we lift the same path starting at y, the lifts agree at y~, so they agree everywhere, pulling us back to x~. The map is a bijection. The cardinality of the fiber does not change as we move about within a path connected region.