Topology (H) Lecture 22 Lecturer: Zuoqin Wang Time: May 27, 2021 CLASSIFICATION OF COVERING SPACES 1. The universal covering { The universal covering. f In PSet11-1-2 we have seen that p∗ : π1(X; x~0) ! π1(X; x0) is injective. So the f f subgroup p∗(π1(X; x~0)) in π1(X; x0) is isomorphic to π1(X; x~0). The main theme today is to establish a \one-to-one correspondence" (in reverse order) between subgroups of π1(X; x0) and covering spaces over X. We first start with the smallest subgroup of π1(X; x0), namely the subgroup feg, which will give us the \largest" covering space. f f Definition 1.1. We say a covering space X is a universal covering if π1(X) ' feg. Example 1.2. • R is a universal covering space of S1, while S1 is not. • S2 is a universal covering space of S2, R2 is a universal covering space of T2, 2 2 1 the unit disc D is a universal covering space of Σ2 = T #T (shown below) . n • Sn is a universal covering space of RP (n ≥ 2). S3 is a universal covering space of L(p; q). SU(2) is a universal covering space of SO(3)2. • The universal covering of S1 _ S1 is the Cayley graph of ha; bi shown below.3 1 According to the uniformization theorem, D is a universal covering space of Σn for all n ≥ 2. 2In general the spin group Spin(n) is a universal covering of SO(n) for n ≥ 3. 3In general the Cayley graph of the free group on n generators is a universal covering of S1 _···_S1. 1 2 CLASSIFICATION OF COVERING SPACES { The existence of universal covering space. Given a topological space X, we want to construct a universal covering. It turns out that it is not always possible to do so. To see this let's suppose p : Xf ! X is a universal covering. Then for any x 2 X, there is a neighborhood U of x and an open Ü f Ü set U in X such that pjUe : U ! U is a homeomorphism. Now each loop γ in U can Ü −1 f be lifted to a loopγ ~ in U by the map (pjUe) . Since π1(X) = feg, the lifted loopγ ~ is null-homotopic in Xf. Composing the null-homotopy with the projection map p, we conclude that γ must be null-homotopic in X. In other words, for any x there is a neighborhood U of x such that any loop in U is null-homotopic in X. So the group homomorphism i∗ : π1(U; x) ! π1(X; x) induced by the inclusion map is a trivial homomorphism. (Roughly speaking, X can't have arbitrarily small \holes".) Definition 1.3. We say a topological space X is semi-locally simply connected 4 if for each x 2 X, there is a neighborhood U of x such that the group homomorphism i∗ : π1(U; x) ! π1(X; x) induced by the inclusion map i : U,! X is trivial. Example 1.4. In PSet 3-2-3, we have seen the Hawai- ian earring X, i.e. a shrinking wedge of circles as showed in the picture. It is path-connected, locally path-connected, but NOT semi-locally simply connect- ed, and thus admits no universal covering. On the other hand, the cone CX over the Hawaiian earring is contractible and thus is semi-locally simply connected, but it is not locally simply connected. Theorem 1.5. Suppose X is path-connected, locally path-connected. Then X admits a universal covering space Xf if and only if X is semi-locally simply connected. Proof. We have seen that the semi-locally simply connectedness is necessary. It re- mains to prove the converse. Idea. Given S1, how do you construct the topological space R? Any point in R is the end point of a lifted path (with fixed starting point) which corresponds to a path in S1 (with fixed starting point). Two lifted paths have the same end point if and only if the projected paths in S1 are path homotopic. So: points in R are in one-to-one correspondence with path homotopy classes of paths in S1 (with fixed starting point)! But, how do we define a topology on the space of homotopy classes of paths? We can simply take the compact-open topology on the space 4As usual, X is called locally simply connected if any point has a simply connected neighborhood, i.e. a neighborhood U so that any loop in U is null-homotopic in U. This is stronger than semi-locally simply connected. CLASSIFICATION OF COVERING SPACES 3 of paths, and then take the quotient topology on the space of homotopy classes. But for our purpose below, we will take an alternative way which is more direct: we can show that \locally" the projection map is bijective. Thus by choosing a base of S1 consisting of small open sets (which form a base of the topology), we can define the \lifting" of these open sets to be open in the space of homotopy classes and thus get a base of a topology. That is what we need. Step 1. Construction of the set Xf. Let X be a path-connected, locally path-connected and semi-locally simply con- nected topological space with base point x0. We define f X = f[γ]p j γ is a path in X staring at x0g: By construction, there is a natural projection map from Xf to X, given by f p : X ! X; [γ]p ! γ(1): Since X is path-connected, p is surjective. Step 2. p is \locally" bijective. For any path γ : [0; 1] ! X with γ(0) = x0, let U ⊂ X be an open neighborhood of γ(1) which is path-connected and such that any loop in U is null-homotopic in X. Define U[γ]p = f[γ∗λ]p j λ is a path in U with λ(0) = γ(1)g: Note that U[γ]p is well-defined, i.e. depends only on the path homotopy class [γ]p instead of on the path γ. Moreover, we have 0 0 (>) If [γ ]p 2 U[γ]p ; then U[γ]p = U[γ ]p : 0 0 0 To see this we suppose γ = γ ∗λ, then any element in U[γ ]p has the form [γ ∗λ∗λ ]p which is still an element in U[γ]p , and conversely any element in U[γ]p has the form 0 0 ¯ 0 [γ∗λ ]p = [γ∗λ∗λ∗λ ]p which lies in U[γ ]p . It is not hard to see that p = pj : U ! U is bijective: U;[γ]p U[γ]p [γ]p • Since U is path-connected, pU;[γ]p (U[γ]p ) = U: • Since any loop in U is null-homotopic in X, different choices of λ from γ(1) to a given point x 2 U are homotopic in X and thus the map p is injective. Step 3. Define a topology on Xf. To define a topology on Xf, we first consider the following collection U = fU ⊂ X j U is open, path-connected and any loop in U is null-homotopic in Xg of open sets in X. Since X is locally path-connected and semi-locally simply connected, U is a basis of the topology on X. 4 CLASSIFICATION OF COVERING SPACES Claim: the collection Ü U := fU[γ]p j U 2 U; γ is a path in X from x0 to a point in Ug form a basis for a topology on Xf. f 00 Reason: Obviously the union of such sets is X. Now suppose [γ ]p 2 00 0 U[γ]p \ V[γ ]p . Then γ (1) 2 U \ V . Since U is a basis of the given topology on X, there is a set W 2 U such that γ00(1) 2 W ⊂ U \ V . 00 0 00 According to (>), we have U[γ]p = U[γ ]p and V[γ ]p = V[γ ]p . It follows 00 00 00 00 0 [γ ]p 2 W[γ ]p ⊂ U[γ ]p \ V[γ ]p = U[γ]p \ V[γ ]p : Step 4. p is a covering map. We have seen in Step 2 that pU;[γ]p : U[γ]p ! U is bijective. In fact, pU;[γ]p also gives 0 a bijection between subsets V[γ ]p ⊂ U[γ]p and V 2 U satisfying V ⊂ U: in one direction −1 we have p (V 0 ) = V , and in the other direction we have p (V ) \ U = V 0 , U;[γ]p [γ ]p U;[γ]p [γ]p [γ ]p 0 0 where [γ ]p 2 U[γ]p is a path homotopy class such that γ has end point in V . It follows pU;[γ]p is a homeomorphism with respect to the topology constructed in Step 3. f Since p is continuous on each open set U[γ]p , p : XS! X is continuous. It is a −1 covering map since for each U 2 U, we have p (U) = [γ]p U[γ]p , which is a disjoint union in view of (>). Step 5. Xf is simply connected. f f Let's take [cx0 ]p to be our base point in X. For any [γ]p 2 X, if we let γt(s) = γ(st), f f f then [γt]p 2 X and f(t) := [γt]p is a path in X connecting [cx0 ]p and [γ]p.