Math 751 Lecture Notes Week 7 Liban Mohamed 1 Tuesday, October 14 Definition 1.1. An isomorphism (or equivalence) of coverings is a homomorphism of coverings which is also a homeomorphism. Definition 1.2. If p : E ! B, p0 : E0 ! B are coverings, a homomorphism of coverings h :(E; p) ! (E0; p0) is a continuous map h : E ! E0 such that p0 ◦ h = p. 0 0 0 0 Theorem 1.1. Let p : E ! B, p : E ! B be coverings of B with p(e0) = p (e0) = b0 2 B. Then there is 0 0 0 0 0 0 an equivalence of coverings h : E ! E , h(e0) = e0 if and only if H = p∗(π1(E; e0)) and H = p∗(π1(E ; e0)) are equal: (E; e0) p h 0 0 0 p (E ; e0) (B; b0) 0 0 0 0 0 Proof. ): If h : E ! E is an equivalence with h(e0) = e0, then h∗(π1(E; e0)) = π1(E ; e0). Apply p∗ and, using p0 ◦ h = p, get H = H0. 0 0 0 (: Assume H = H . H ⊂ H implies that h :(E; e0) ! (E ; e0) exists by the lifting lemma, with 0 0 0 0 h(e0) = e0, p ◦ h = p. Reversing the roles of p and p , we get that H ⊂ H implies the existence of a lift 0 0 0 0 0 k :(E ; e0) ! (E; e0) of p with p ◦ k = p , k(e0) = e0. Consider the diagram: k◦h (E; e0) idE p p (E; e0) (B; b0) 0 p ◦ (k ◦ h) = (p ◦ h) ◦ k = p ◦ h = p. k ◦ h and idE are lifts of p that agree at e0 so, by uniqueness of lifts, 0 0 k ◦ h = idE. By similar reasoning on p , h ◦ k = idE. If p = p0, and equivalence of p interchanges points in the fiber over each b 2 B; is called an automorphism of (E; p), or a deck transformation. Definition 1.3. The deck transformations form a group under composition of maps, called the deck group of (E; p), and denoted D(E; p). Proposition 1.2. If h 2 D(E; p) so that h(x) = x for some x 2 E, then h = idE. 0 0 −1 0 Proposition 1.3. 9h 2 D(E; p) with h(e0) = e0 for e0; e0 2 p (b0) if and only if p∗π1(E; e0) = p∗π1(E; e0). 0 (E; e0) h p p (E; e0) (B; b0) Theorem 1.4 (Main Theorem). Two coverings (E; p) and (E0; p0) of B are equivalent if and only if for any 0 0 0 0 e0 2 E; e0 2 E with p(e0) = p (e0) = b0, we have that the corresponding subgroups p∗π1(E; e0) is conjugate 0 0 0 0 to H = p∗π1(E ; e0) 1 Definition 1.4. p : E ! B is called a universal covering map if E is simply connected; E is called a universal cover. Corollary. If a universal cover of B exists, it is unique up to equivalence of coverings, since the conjugacy class of the trivial subgroup is in any group has only one element. 0 00 Proof of Main Theorem. ): Assume we have an equivalence. If h : E ! E is an equivalence, let h(e0) = e0 . 00 0 00 00 0 −1 By the previous theorem, H = p∗π1(E; e0) equals H = p∗π1(E ; e0 ). By changing e0 to any e0 2 p (b0) via p0 ◦ h = p 0 p ◦ h(e0) = p(e0) 0 00 p (e0 ) = b0 00 0 0 0 0 0 get that H is conjugate to H = p∗π1(E ; e0). So H and H are conjugate. 0 0 0 0 (: IF H = P∗π1(E; e0) and H = p∗π1(E ; e0) are conjugate, use: Lemma 1.5. If p : E ! B is a covering, p(e0) = b0, and H = p∗π1(E; e0), given any K ⊂ π1(B; b0) −1 conjugate to H, there is an e1 2 p (b0) such that K = H1 = p∗π1(E; e1). −1 0 0 0 0 Using lemma, there is e1 2 p (b0) such that , p∗π1(E ; e0) = H = p∗π1(E; e1). By the lifting property, there is an equivalence h : E ! E0. −1 Proof of Lemma. K; H are conjugate in π1(B; b0), so there is a loop α at b0 in B such that H = [α]·K ·[α] . −1 Letα ~e0 be a lift to E of α under p, starting at e0, let e1 =α ~e0 (1). Then H = [p ◦ α~e0 ] · H1 · [p ◦ α~e0 ] . So K = H1. Example 1.1. Let B be the M¨obiusband, π1(B) ≡ Z. Conjugacty classes of subgroups are nZ for n 2 N. Even n yields an n-fold covering of B by the cylinder S1 × I with (z; I) 7! (zk;I). Odd n yields an n-fold covering of B by the the M¨obiusband under the same map. 2 Thursday, October 16 Universal Covering Spaces Q: When does a (path connected, locally path connected) space B have a universal cover? Theorem 2.1. B has a universal cover if and only if B is locally simply connected. Definition 2.1. B is semi-locally simply connected if, for any b 2 B, there is a neighborhood of b such that the inclusion i : Ub ,! B induces a trivial homomrophism i∗ : π1(Ub; b) ! π1(B; b). Example 2.1. If B is simply-connected, then B is semi-locally simply connected Example 2.2. The Hawaiian earring has no universal cover, since it is not semi-locally simply connected. 1 The Hawaiian earring is the space X = [nCn, where Cn is the circle of center (1=n; 0) and radius n . Claim: If U is any neighborhood of the origin, then i∗ : π1(U; 0) ! π1(X; 0) is nontrivial. Proof of Claim: Given n, there is a retraction r : X ! Cn, r : Ci 7! 0 when i 6= n and r : Cn 7! n, rjCn = idCn . Choose n large enough that Cn ⊂ U. j j∗ Cn X π1(Cn; 0) X k k i ∗ i∗ U U j∗ = i∗k∗ is injective since r∗ ◦ j∗ = idZ, so i∗ can't be trivial. 2 Proof of Theorem 2.1, ) Let p : E ! B be a universal covering map, p(e0) = b0. Let U be an evenly covered neighborhood of b0 ~ −1 ~ and let U be the slice of p (U) containing e0. Let f be a loop in U at b0. pjU~ : U ! U is a homeomorphism, ~ ~ ~ ~ ~ so f lifts to a loop, f, in U at e0. Hence f is a loop in E at e0, so there is a homotopy, F from f to conste0 . ~ ~ Thus p ◦ F is a homotpy in B from p ◦ f = f to constb0 . Theorem 2.2. Let B be path connected, locally path connected and semi-locally simply connected. Let b0 2 B −1 and H ⊂ π1(B; b0). Then there is a covering p : E ! B and e0 2 p (b0) such that p∗π1(E; e0) = H. Sketch of proof. Let P be the set of all paths in B starting at b0. Define an equivalence relation on P by α ∼ β if α(1) = β(1) and [α ∗ β¯] 2 H. Let α# be the equivalence class of α 2 P . Let E = fα#jα 2 P g. Define p : E ! B, p : α# 7! α(1). p is surjective since B is path-connected. Topologize E so that p becomes a covering map. Definition 2.2. Let X be a topological space, G ⊂ Hom(X). G acts on X: G × X ! X by (g; x) 7! g(x). Let [x] = fgxjg 2 Gg, the orbit of x. Let the orbit space X=G := f[x]jx 2 Xg. Definition 2.3. G acts freely on X if whenever g · x = x for some x 2 X, g = eG. Definition 2.4. G acts properly discontinuously on X if 8x 2 X, There is a neighborhood Ux 3 x such that gUx \ Ux = ; for all g 6= eG.() gUx \ hUx = ; if h 6= g 2 G). Proposition 2.3. If X is Hausdorff and G is a finite group of homeomorphisms of X acting freely on X, the action of G is properly discontinuous (hence π : X ! X=G is a covering by following theorem). Theorem 2.4. Let X be path-connected, locally path-connected and G ⊂ Hom(X). Then π : X ! X=G is a covering if and only if the action of G is properly discontinuous. Moreover, if this is the case, the deck group of this covering is G and the covering is regular, i.e. π∗π(X; x) C π1(X=G; π(x)), for all x 2 X). Definition 2.5. p : E ! B is a regular cover if p∗π1(E; e) C π1(B; p(e)). Proposition 2.5. p : E ! B is regular if and only if D(E; p) acts transitively on fibers. That is, 8e1; e2 2 −1 p , 9h 2 D(E; p) such that h(e1) = e2. −1 Proof. ( If e1; e2 2 p (b) with h 2 D(e; p) such that h(e1) = e2, then p∗π1(E; e1) = p∗π1(E; e2) ∼ p∗π1(E; e1), so p∗π1(E; e1) C π1(B; b) ∼ Theorem 2.6. If p : E ! B is a cover with H = p∗π1(E; e) ⊂ π1(B; p(e), then D(E; p) = N(H)=H, where −1 N(H) = fg 2 π1(B; p(e))jgHg = Hg.