Definition (covering space):
Let be a topological space. Another topological space is called a covering space of if and only along with comes a continuous function such that the following condition is satisfied:
- For each , there exists a neighbourhood of so that is the disjoint union of open subsets such that for all the function is a homeomorphism between and .
Definition (covering map):
Let be a covering space of the topological space and let be the continuous map that goes along with it. Then is called the covering map of the covering space .
Note that more formally, one could have defined a covering space to be equal to a pair , where is the covering map that belongs to the covering space; indeed, many covering maps may be possible if is not specified. But as with group operations, we will mostly suppress from the notation for brevity.
Definition (evenly covered neighbourhood):
Let be a topological space and let be a covering space of , where is the covering map. Let . The set guaranteed by the definition of a covering space (so that , restricted to adequate subsets of the preimage, is a homeomorphism) is called an evenly covered neighbourhood of .
Let be topological spaces, and let be a continuous function. If is a covering space of , then a lift of to is a continuous function so that, when is the covering map that belongs to , .
When the domain of the function that is to be lifted is connected, then lifts are, in a certain sense, unique:
Proposition (maps of connected domain lift uniquely):
Let be a continuous function between topological spaces where is connected, and suppose that is a covering space of . Then if two lifts of coincide in one point , they are equal.
Proof: Let be the set on which and agree. Since adn agree on , is nonempty. Further, if , set and let be an evenly covered neighbourhood of . Let be the disjointed component in which lies, so that is a homeomorphism between and . Then define , both of which are open since , are continuous. We claim that in fact on ; indeed, we may note that implies that , from which identity of follows upon composing on the left with . We conclude that is open. However, suppose that , suppose that is an evenly covered neighbourhood of and let respectively be the (disjoint) components of , so that and are homeomorphisms. Then define . Whenever , we will have and , so that on the functions and disagree on every point. Thus, we get that is open, so that is open and closed, and since is connected, .
Proposition (lifting paths):
Let be a topological space, let be a path, and let (together with a suitable covering map ) be a covering space of . Set . Then for each , there exists a unique curve such that .
Proof: Note that is compact. For each , pick an evenly covered neighbourhood of . Since is compact, upon defining and then writing each as
where the are intervals (note that the intervals form a basis of the Euclidean topology) and then considering the open cover , we may pick a finite number of intervals which cover and whose images via are evenly covered. We assume that the intervals are ordered increasingly according to their starting points. Now we define successively on these intervals. For , we note that is evenly covered and contains . Suppose that is an evenly covered neighbourhood of , and let be the components of so that restricts to a homeomorphism on them. Let so that . Then define for and observe that , since and and are both in . (Note that a homeomorphism is in particular bijective and that the definition of is independent of the choice of interval, therefore is well defined even on the intersection .) Proceed similarly for the ensuing intervals . Then will be continuous on all intervals of the form , where is the beginning point of and , since the interval is contained in . Hence, since a function which is continuous on two closed sets is continuous on their union, we obtain that is continuous on , then on , and so on, and finally on . is then the desired lift, and it is unique since maps of connected domain lift uniquely.
Proposition (lifting homotopies):
Let be topological spaces and let be a continuous function. Suppose that is a covering space of , and that we are given a lift of the function defined by . Then there exists a unique continuous function so that for one thing for all and further .
Proof: For each , lifts uniquely to a path so that . Define . Obviously, , and further, we claim that is continuous. Let be arbitrary. Let be an evenly covered neighbourhood of ; then is a homeomorphism that has a continuous inverse . Now by definition of the product topology, take open and so that . By uniqueness of path lifting, we have on , which is continuous. We conclude that is continuous, since it is continuous in an open set about an arbitrary point.