Consider a continuous map . For each point , we have a subset , which we call the fibre of over . We next define what it means for to be a covering map. The key points are that
All the fibres must be discrete subsets of ;
The fibre must depend continuously on , in an appropriate sense.
It is not easy to formulate the second condition directly, so the formal definition looks rather different from this informal discussion.
Consider the exponential map . The fibres are
Each fibre is a discrete set, which suggests that the map should be a covering. We will check that this is true once we have given the proper definition.
Let be a continuous map of spaces. Consider an open subset . We say that is trivially covered by if there is a discrete space and a map such that the combined map is a homeomorphism. We say that is a covering map (or that is a covering space of ) if for each point there is an open set that contains and is trivially covered.
Suppose we have open subsets and that is trivially covered, as witnessed by a map . We then note that , se we can restrict to get a map . It is not hard to check that the combined map is a homeomorphism, so is also trivially covered.
Take and , and let be the projection map, given by . We claim that the whole space is trivially covered. Indeed, we can take and define by . Then the combined map is just the identity map , which is certainly a homeomorphism. From this it is clear that is a covering map.
Take and to be the unit circle in , or in other words . Define by , or equivalently . Each fibre is a discrete set consisting of the two square roots of , which are negatives of each other. This suggests that should be a covering, which we can prove as follows. We first take , so each element can be expressed in a unique way as with . The two square roots of are then (which has ) and (which has ). We therefore have
and we can define by
We then find that the map is a homeomorphism, showing that is trivially covered. A similar approach can be used to check that the set is also trivially covered. Any element can be expressed uniquely as with , and the two square roots are then (which has ) and (which has ). We therefore have
and we can define by
We then find that the map is a homeomorphism, as required. As , this proves that is a covering.
Recall that the real projective space is defined to be the quotient space , where iff . We therefore have a quotient map , for which iff . We give the quotient topology, which means that a subset is open iff is open in .
The map is a covering map.
For each we put and . We then find that
This first shows that is an open subset of and so (by the definition of the quotient topology) that is an open subset of . It also allows us to define a continuous map by
It is easy to see that the combined map is a homeomorphism, so is trivially covered. The sets cover all of (because ) so is a covering map as claimed. ∎
The map is a covering map, as is the map (where we identify with ).
This is closely related to the proof in Example 22.5. We put
If then there is a unique choice of and with and and . It follows that the number lies in and has . This means that the restricted map is bijective. Standard complex analysis shows that the inverse is also continuous, so we have a homeomorphism . Similarly, the restricted map is also a homeomorphism. From this we see that
We can thus define continuous maps (for ) by . Equivalently, is the closest integer to ; this is well-defined and continuous on because points where is an odd multiple of have been removed from . Similarly, is the closest integer to . We find that the maps and are homeomorphisms, so and are trivially covered. We also have , so is a covering map. The proof for the restricted map is essentially the same.
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Let be a covering map.
Consider a point . A lift of means a point with .
Consider a continuous path . A lift of means a continuous path such that . This means in particular that is a lift of for all .
More generally, let be any space and let be a continuous map. A lift of is a continuous map with .
For a map as in (c), we say that is small if there is a trivially covered open set such that .
Suppose we have a path-connected space and a small continuous map . Suppose we also have points and with . Then there is a unique lift such that .
By assumption, we can choose an open subset containing , and a homeomorphism as in Definition 22.2. If is a lift of , then we have , so for all , so we have a well-defined and continuous composite . As is path connected and is discrete, this must be constant. Thus, if , then for all . It follows that the only possibility is
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Let be a covering map. Let be a path from to in , and let be a lift of . Then there is a unique lift of such that .
Because is a covering map, we can find a family of trivially covered open sets such that . The preimages then form an open covering of . Because is a compact metric space, this covering has a Lebesgue number (by Proposition 8.31). Choose and divide into subintervals for . By the Lebesgue number property, we can choose an index such that , so . This means that the restriction of to is small, so Lemma 22.9 is applicable. We are given with . We put , and use Lemma 22.9 to show that there is a unique map with and . We now define , so . Applying Lemma 22.9 again, we see that there is a unique map with and . We put , so . We then repeat the process in the obvious way, to get a family of maps and points with . It follows that the maps can be patched together to give a continuous map with and . The same kind of induction shows that this is unique. ∎
Note that Proposition 22.10 does not say anything about . We know that for all , so in particular , so . However, if we have two different paths in and we use the same starting point in both cases, then it can easily happen that the endpoints are different elements of . However, this cannot happen if there is a pinned homotopy between and , as we will show later.
Let be a covering map. Suppose we have a path-connected space and a continuous map . Suppose that are continuous lifts of , and that there is at least one point with ; then .
Consider a point ; we must show that . As is path connected, we can choose a path from to in . Now and are both lifts of the path , and they satisfy . Thus, the uniqueness clause in Proposition 22.10 tells us that . In particular, we have , or in other words as required. ∎
Let be a compact convex subset of , and suppose that . Let be a covering map, and let be continuous. Suppose that with . Then there is a unique continuous lift with and .
Roughly speaking, the idea is as follows: to define , we move a short distance from towards to reach a point , then will already be defined and we define to be the unique lift of that is close to . The rest of this proof should be seen as a more complete and rigorous version of this idea.
We first claim that there exists such that for all , the restricted map is small. (Here and elsewhere in this proof, notation for balls should be interpreted relative to , so .) Indeed, for each we can choose a trivially covered open set containing . The set is then open in and contains . This means that the sets form an open cover of the compact metric space , so there is a Lebesgue number . This has the required property.
Next, for we put . We will prove by induction on that there is a unique continuous map with and for all . To start with, the map is small by our choice of , so Lemma 22.9 gives . Suppose we have already constructed . For each , we note that the map is small, so there is a unique continuous lifting with . Both and restrict to give lifts of over the convex set , and they agree at the point , and the restricted map is small; it follows that agrees with on .
Now suppose that and that the set is nonempty. We then find that the point must lie in and it also lies in because is convex. The maps and both agree with at , so they agree with each other. The restricted map is small, so we conclude that . Because of this consistency property, we see that the maps can be combined to give a map . As the sets are all open, an open patching argument shows that is continuous. As each map is a lift of , we see that is a lift of . As agrees with on , we see that agrees with on .
It is also easy to see that is contained in the domain of , so we can define to be the restriction of to . This is a continuous lift of extending and therefore satisfying as required.
As is assumed to be compact, it must be bounded. We therefore have for sufficiently large , and this completes the proof. ∎