3. Covering and Fibration

Covering and Lifting

Definition 3.1. Let be in . A trivialization of over an open set is a homeomorphism over , i.e. , the following diagram commutes is called locally trivial if there exists an open cover of such that has a trivialization over each open . Such is called a fiber bundle, is called the fiber and is called the base. We denote it bywhere there is no ambiguity from the context. If we can find a trivialization of over the whole , then is homeomorphic to and we say is a trivial fiber bundle.

Example 3.2. The projection mapis a trivial fiber bundle with fiber .

Example 3.3. A real vector bundle of rank over a manifold is a fiber bundle with fiber .

Example 3.4. We identify as the unit sphere in parametrized byThere is a natural -action on given byThis action is free, and the orbit space can be identified with the -dim complex projective space Then the projection map is a fiber bundle with fiber . It is a nontrivial fact that they are not trivial fiber bundles. The case when gives the Hopf fibratioinwhich is particularly interesting.

In this case, the projection sends to . In polar coordinates, we havefor . For a fixed , we obtain a torus in .

When identifying with the compatification of (or considering the stereographic projection ), we can visualize (Figure 1) the foliation of by these tori , where degenerates to the unit circle on the -plane of and degenerates to the -axis. Each -fiber is a slope 1 simple closed curve on one of the tori , and the image of the projection is exactly the compatification of any half planeof (for a fixed ).
AT Hopf.svg
Figure 1. A visualization of Hopf fibration

Moreover, the disjoint union of any two fibers over two distinct points of is in fact the so called Hopf link, shown in Figure 2. Such the link is nontrivial, the Hopf fibration is not a trivial fiber bundle.

Figure 2. Hopf link

Definition 3.5. A covering (space) is a locally trivial map with discrete fiber . A covering map which is a trivial fiber bundle is also called a trivial covering. If we would like to specify the fiber, we call it a -covering. If the fiber has points, we also call it a -fold covering.

Figure 3. Trivialization (left) and covering (right)
Example 3.6. The map is a -covering.
Figure 4. The -covering of

If , then

Example 3.7. The map is an -fold covering, for any .

Example 3.8. The map , , is not a covering (why?). But

the map , , is an -fold covering, where and .

the map , is a -covering.

Example 3.9 (From Hatcher [??]). The figure-8has two coverings as follows (the left is a -fold (or double) covering and the right is a -fold covering).
The 4-regular tree is its universal cover (a covering which is simply connected), see Figure 5.
Figure 5. 4-regular tree

Example 3.10. Recall that the number of holes (genus) and the number of boundary components determine the homeomorphism type of a compact oriented topological surface. Let be a surface of genus and with boundary components.

admits a -fold covering from , cf. Figure 6.

In general, admits a -fold covering from .

Figure 6. A 7-covering

Example 3.11. Denote by the real projective space of dimension Let be the -sphere. Then there is a natural double cover .

Example 3.12 (Branched double cover). Figure 7 shows a branched double cover of a disk
Figure 7. The Birman-Hilden double cover via the twisted surface
Namely, when deleting those (red) points (denoted by ) from both and , we obtain a -fold covering:with a bijection on . This can be used to show the homeomorphism in Figure ?? (it is a special case of Figure 7 for . In general, it has genus with boundary components.)
Figure 8. The normal view of the branched double cover of the punctured disk

Definition 3.13. Let . A lifting of along is a map such that

Lemma 3.14. Let be a covering. LetThen is both open and closed.

Proof. Exercise.

Theorem 3.15 (Uniqueness of lifting). Let be a covering. Let be two liftings of . Suppose is connected and agree somewhere. Then .

Proof. Let be defined in Lemma 3.14. Consider the map . By assumption, we have . Moreover, Lemma 3.14 implies that is both open and closed. Since is connected, we find which is equivalent to .

Fibration

Definition 3.16. A map is said to have the homotopy lifting property (HLP) with respect to if for any maps and such that , there exists a lifting of along such that , i.e., the following diagram is commutative

Definition 3.17. A map is called a fibration (or Hurewicz fibration) if has HLP for any space.

Theorem 3.18. A covering is a fibration.

Proof. Let be the data as in Definition 3.16. We only need to show the existence of for some neighbourhood of any given point .In fact, for any two such neighbourhoods and with , we have and agree at some point on and hence agree everywhere in by the uniqueness of lifting (Theorem 3.15). Thus glue to give the required lifting .

Since is compact, given we can find a neighbourhood and a partitionsuch that has a trivialization over open sets . Now we construct the lifting on , for , by induction on .

For , the lifting on to one of the sheets of is determined by :

Assume that we have constructed on for some . Now, the lifting on to one of the sheets of is determined by , which can be glued to the lifting on by the uniqueness of lifting again. This finishes the inductive step.

We obtain a lifting of on as required.

Corollary 3.19. Let be a covering. Then for any path and such that , there exists a unique path which lifts and .

Proof. Apply HLP to .

Corollary 3.20. Let be a covering. Then is a faithful functor. In particular, the induced map is injective.

Proof. Let be two paths and . Let . Suppose and we need to show that .

Let be a homotopy. Consider the following commutative diagram with the lifting by HLPThen the uniqueness of lifting implies . Thus, .

Transport functor

Let be a covering. Let be a path in from to . It defines a mapwhere is a lift of with initial condition .

Figure 9. The transportation

Assume in . HLP implies that . We find a well-defined map:

This leads to the following definition (check the functor property!).

Definition 3.21. The following datadefine a functor, called the transport functor. In particular, we have a well-defined mapHere consists of all set isomorphisms in for .

Example 3.22. Consider the covering mapConsider the following path representing an element of

Start with any point in the fiber, lifts to a map to We find . Therefore is

Proposition 3.23. Let be a covering, be path connected. Let . Then the action of on is transitive, whose stabilizer at is . In other words,as a coset space, i.e. we have the following short exact sequence

Proof. For any point , let be a path in and . Then . This shows the surjectivity of .

HLP implies that is injective and we can view as a subgroup of . By definition, for , we have , i.e. . On the other hand, if , then the lift of is a loop, i.e. . Therefore, . This implies , which finishes the proof.

Lifting Criterion

Theorem 3.24 (Lifting Criterion). Let be a covering. Consider a continuous map , where is path connected and locally path connected. Let such that . Then there exists a lift of with if and only if

Proof. If such exists, thenConversely, letand consider the following commutative diagramThe projection is also a covering. We have an induced commutative diagram of functorswhich induces natural group homomorphismsLet . The condition says that stabilizes . By Proposition 3.23, this implies we have a group isomorphismSince is locally path connected, is also locally path connected. Then path connected components and connected components of coincide. Let be the (path) connected component of containing , then implies that is a covering with fiber a single point, hence a homeomorphism. Its inverse defines a continuous map whose composition with gives .

-principal covering

Definition 3.25. Let be a discrete group. A continuous action is called properly discontinuous if for any , there exists an open neighborhood of such thatWe define the orbit space where for any .

Proposition 3.26. Assume acts properly discontinuously on , then the quotient map is a covering.

Proof. For any , let be the neighbourhood satisfying . Thenis a disjoint union of open sets. Thus, is locally trivial with discrete fiber , hence a covering.

Definition 3.27. A left (right) -principal covering is a covering with a left (right) properly discontinuous -action on over such that the induced map is a homeomorphism.

Example 3.28. is a -principal covering for the action .

Example 3.29. is a -principal covering.

Proposition 3.30. Let be a -principal covering. Then the transport is -equivariant, i.e.,

Proof. Let and . Then for some . If we apply the transformation to the path , we find another lift of but with endpoints and . ThereforeIt follows that . This proves the proposition.

Figure 10. Transport commutes with -action

Theorem 3.31. Let be a -principal covering, path connected, . Then we have an exact sequence of groupsIn other words, is a normal subgroup of and .

Proof. Let . The previous proposition implies that -action and -action on commute. It induces a -action on . Consider its stabilizer at and two projections is an isomorphism and is an epimorphism with .

Apply this theorem to the covering , we find a group isomorphismwhich is called the degree map. An example of degree map is

Applications

Definition 3.32. Let be an inclusion. A continuous map is called a retraction if . It is called a deformation retraction if furthermore we have a homotopy . We say is a (deformation) retract of if such a (deformation) retraction exists.

Proposition 3.33. If is a retract, then is injective.

Proof. Let such that . Then the compositionis the identity. It follows that is injective.

Corollary 3.34. Let be the unit disk in . Then its boundary is not a retract of .

Proof. Since is contractible, we have . But . Then the corollary follows from the proposition above.

Theorem 3.35 (Brouwer fixed point Theorem). Let . Then there exists such that .

Proof. Assume has no fixed point. Let be the ray starting from pointing toward . Thenis a retraction of . Contradiction.

Theorem 3.36 (Fundamental Theorem of Algebra). Let be a polynomial with . Then there exists such that .

Proof. Assume has no root in . Define a homotopy of maps from to By construction, and . But they are homotopic hence representing the same element in . Contradiction.

Proposition 3.37 (Antipode). Let be an antipode-preserving map, i.e. . Then is odd. In particular, is NOT null homotopic.

Proof. Let be a lifting of for the covering map . ThenSince is antipode-preserving, for some . So which implies is odd.

Example 3.38. Let be the antipode map defined by . Then .

Theorem 3.39 (Borsuk-Ulam). Let . Then there exists such that .

Proof. Assume . DefineLet be the upper hemi-sphere of . It defines a homotopy between constant map and , hence . On the other hand, is antipode-preserving: , hence is odd. Contradiction.

Theorem 3.40 (Ham Sandwich Theorem). Let be two bounded regions of positive areas in . Then there exists a line which cuts each into half of equal areas.

Proof. Let .

Given , let be the plane passing the origin and perpendicular to the unit vector . Let Define the continuous mapBy Borsuk-Ulam, such that . The intersection gives the required line since