Math 751 Week 6 Notes

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1 Math 751 Week 6 Notes Joe Timmerman October 26, October 7 Definition 1.1. A map p: E B is called a covering if 1. P is continuous and onto. 2. For all b B, there exists an open neighborhood U of b which is evenly covered, i.e., p 1 (U) = α V α, where the v α are disjoint and open, and p Vα : V α U is a homeomophism. Example p: R S 1, t e 2πit 2. id X : X X 3. p: X {1,..., n} X, (x, k) x 4. p: S 1 S 1, z z n 5. p: S n RP n, x [x] = {±x} 6. p: C C, z e x 7. Products of covering maps: If p i : E i B i, i = 1, 2 are coverings, then p 1 p 2 : E 1 E 2 B 1 B 2 is a covering. 1. Being a covering implies the map is open and locally a homeomor- Remark 1.3. phism. 2. Not any local homeomorphism is a covering. 3. p 1 (b) is discrete, for all b B (by disjointness.) Definition 1.4. Let p 1 : E 1 B, p 2 : E 2 B be two coverings. We say p 1 and p 2 are equivalent if there exists a homemorphism f : E 1 E 2 such that p 2 f = p 1. Note: This is an equivalence relation. Problem: Find all coverings of a space B (up to equivalence.) 1

2 Lemma 1.5. If p: E B is a covering, B 0 B, and E 0 := p 1 (B 0 ), then p E0 : E 0 B 0 is a covering. Example 1.6. Let p: R 2 T 2 be a covering. Overlay the integer lattice on R 2, and identity each square with a torus in the usual way. Let p 0 = (1, 0) S 1, and let B 0 = S 1 {p 0 } {p 0 } S 1. Then p 1 (B 0 ) = R Z ζ R (so it gets rid of the inside of the squares.) Theorem 1.7 (Path lifting property). Let P : E B be a covering, b 0 B, and e 0 p 1 (b 0 ). If γ : I B is a path in B starting at b 0, then there is a unique lift γ e0 : I E such that γ e0 (0) = e 0. The proof of this theorem follows from the previous lemma. Theorem 1.8 (Homotopy lifting property). Let F : I I B be a homotopy with b 0 := F (0, 0). Then there is a unique lift F : I I E of F such that F (0, 0) = e 0. Corollary 1.9. If γ 1, γ 2 are paths in B which are homotopic by some F, then γ 1 (0) = γ 2 (0) = b 0, then ( γ 1 ) e0 F ( γ 2 ) e0. In particular, these lifts have the same endpoints: ( γ 1 ) e0 (1) = ( γ 2 ) e0 (1). Definition Let b 0 B. For e 0 p 1 (b 0 ), define Φ e0 : π 1 (B, b 0 ) p 1 (b 0 ) [γ] γ e0 (1) Note that by corollary 1.9, this map is well-defined. Theorem Let Φ e0 be defined as above. Then Φ e0 is onto if E is path-connected, and it is injective if E is simply connected. Proof. First suppose that E is path-connected. Let e 1 p 1 (b 0 ), and let δ be a path in E from e 0 to e 1. Then p δ is a path in B. Further, γ := p δ : I B is a loop in B with base point b 0. Then δ is a lift of γ starting at e 0. Then we have φ e0 ([γ]) = γ e0 (1) = δ(1) = δ 1, so φ e0 is surjective. Note that the equality γ e0 (1) = δ(1) comes from the uniqueness of lifts. Now suppose E is simply connected. Let γ 1, γ 2 be loops in B with base point b 0 such that φ e0 ([γ 1 ]) = φ e0 ([γ 2 ]) = e 1. By definition, this means that ( γ 1 ) e0 (1) = ( γ 2 ) e0 (1). To show that φ e0 is injective, we must show γ 1 γ 2. Since E is simply connected, there is a unique homotopy class of paths from e 0 to e 1, so ( γ 1 ) e0 ( γ 2 ) e0 by some homotopy F. This gives a homotopy p F : I I B from p ( γ 1 ) e0 = γ 1 to p ( γ 2 ) e0 = γ 2, which shows that φ e0 is injective. Example Let p: S n RP n be a covering. For n 2, S n is path-connected and simply connected. Then by theorem 1.11, Φ e0 : π 1 (RP n, b 0 ) (p 1 (b 0 ) is a bijection. Since #p 1 (b 0 ) = 2, it must be that π 1 (RP n, b 0 ) = Z/2Z. 2

3 Example Let p: R S 1, t e 2πit. Since R is both simply connected and patheconnected, theorem 1.11 tell us that φ e0 : π 1 (S 1, b 0 ) Z is a bijection. To show that the groups are isomorphic, we now need to show that φ e0 is a homomorphism. Let γ, δ π 1 (S 1, b 0 ), and let γ 0, δ 0 be their lifts in R. Let γ 0 (1) = n Z, δ 0 (1) = m Z. By definition, φ e0 ([g]) = n, φ e0 ([δ]) = m. Claim φ e0 ([g] [δ]) = n + m (i.e., it is a homomorphism.) Proof. We have φ e0 ([δ] [γ]) = φ e0 ([δ γ]) = (γ δ) 0 (1) = ( γ 0 δ )(1) = δ (1) = n + m Now set δ (t) = n + δ 0 (t) so that δ (0) = n, δ (1) = n + m. Thus, φ e0 and therefore an isomorphism. is a homomorphism Proposition If p: E B is a covering and B is path-connected, and b 0, b 1 B, there there is a bijection p 1 (b 0 ) p 1 (b 1 ). Proof. Let γ be a path in B from b 0 to b 1 (which exists because B is path-connected.) Define the bijection f γ : p 1 (b 0 ) p 1 (b 1 ) by e 0 γ e0 (1). It has inverse (f γ ) 1 = f γ. 2 October 9 Proposition 2.1. Let E be path connected, p: E B a covering, and p(e 0 ) = b 0. Then p : π 1 (E, e 0 ) π 1 (B, b 0 ) is injective. Further, if e 0 is changed to some other e 1 p 1 (b 0 ), then the images of p are conjugate in π 1 (B, b 0 ). Proof. Let p ([γ 1 ]) = p ([γ 2 ]). Then P γ 1 p γ 2 by some homotopy F. By homotopy lifting, we have that ( p γ 1 ) e0 ( p γ 2 ) e0, which implies that γ 1 γ 2, by the uniqueness of lifts. Thus, p is injective. Now let e 1 be a different point in the fiber of p over b 0. Define H 1 = P π 1 (E, e 0 ), H 2 = p π 1 (E, e 1 ). We want to show these are conjugate. First let δ be a path in E from e 0 e 1. Then the following diagram commutes: p π 1 (E, e 0 ) π 1 (B, b 0 ) δ # (p δ) # π 1 (E, e 1 ) π 1 (B, b 0 ) Note that δ # is an isomorphism. So we have H 1 = (p δ)# H 2, by conjugation with [p δ]. Theorem 2.2. Let E be path-connected, p: E B a covering map, and e 0 p 1 (b 0 ). Let H := p π 1 (E, e 0 ) π 1 (B, b 0 ). Then: 3

4 a A closed path γ in B based at b 0 lifts to a loop in E at e 0 iff [γ] H. b φ e0 : H\π 1 (B, b 0 ) p 1 (b 0 ), [γ] γ e0 (1) is a bijection. In particular #p 1 (b 0 ) = [π 1 (B, b 0 ): p π 1 (E, e 0 )]. Proof of (b). First show that φ e0 is well-defined, i.e., if [δ] H, then φ e0 ([δ[ [γ]) = φ e0 ([γ]). We have φ e0 ([γ] [δ]) = φ e0 ([δ γ]) = ( δ γ) e0 (1) = ( δ e0 γ δe0 (1) )(1) = γ δe0 (1) By part (a), since [δ] H, we have that δ e0 (1) = e 0. Thus, γ δe0 (1) = φ e 0 ([γ]), so it s well defined. From last class we know that φ e0 is onto, so it remains to show that it s injective. Suppose that φ e0 ([γ 1 ]) = φ e0 ([γ 2 ]). By definition, this means that ( γ 1 ) e0 (1) = ( γ 2 ) e0 (1). Thus, ( γ 1 ) e0 ( γ 2 ) e0 is a loop in E based at e 0, which is in turn a lift of γ 1 γ 2. By (a), [γ 1 γ 2 ] H. Finally, [γ 1 ] = [γ 1 γ 2 γ 2 ] = [γ 1 γ 2 ] [γ 2 ]. Note that [γ 1 γ 2 ] H, so γ 1, γ 2 are equivalent in the set of cosets. Thus, the function is injective. Theorem 2.3 (Lifting Lemma). Let E, B, Y be path-connected and locally path-connected (i.e., for all x X and for all neighborhoods U x of x, there exists a V k which is path connected-connected, contains x, and is contained in U x.) Let p: E B be a cover, b 0 B, e 0 p 1 (b 0 ), and f : Y B such that f(y 0 ) = b 0. Then there exists a lift f : Y E such that f(y 0 ) = e 0, p f = f iff f (π 1 (Y, y 0 ) p π 1 (E, e 0 ). E, e 0 f p f Y, y 0 B, b 0 Proof. The direction is clear. Let y Y. How should we define f(y)? Let α be a path iny from y 0 to y 1. Then f γ is a path in B starting at b 0. Define f(y) := ( f α)(1). We have (p f)(y) = p ( f α) e0 (1) = f α(1) = f(y). Thus, f is actually a lift. Now we need to show f is well-defined (i.e., independent of α). If β is another path iny from y 0 to y, then α β π 1 (Y, y 0 ), so f (α β) f π 1 (Y, y 0 ). Now by assumption, we have f π 1 (Y, y 0 ) p π 1 (E, e 0 ). This means that ( f (α β)) e0 is a loop at e 0. Then we have ( f α) e0 ( f β) ( f α) e0 = ( f α) (f β) = (f α) (f β) This means that ( f α) e0 = ( f β) e0, which is what we wanted to show. Now we need to show that f is continuous. Let y Y, and let U be a path connected neighborhood of f 1 (y 1 ) B (which exists by the locally-connected assumption). Let V be the slice in p 1 (U) which contains f(y 1 ). By the continuity of f, there is some pathconnected neighborhood of y, say W, in Y such that f(w ) U. Then f = (p V ) 1 f W is continuous. 4

5 Corollary 2.4. If Y is simply connected, then such a lift always exists. Proposition 2.5. (Lift uniqueness) If Y is connected and f 1, f 2 : Y E are two lifts as in the previous theorem, then f 1 = f 2. Proof. Let A = {y : f 1 (y) = f 2 (y)} =. We will show A = Y by showing that A is both open and closed. Let y Y, and let U be an evenly covered neighborhood of f(y) in B. Then we have p 1 (u) = α Ũ α such that p Ũα : Ũα U is a homeomorphism. Let Ũ1, Ũ2 be the Ũα containing f 1 (y) and f 2 (y), respectively. Note that the f i are continuous, so there is a neighborhood N of y such that f 1 (N) Ũ1 and f 2 (N) Ũ2. If f 1 (y) f 2 (y), then Ũ 1 Ũ2, so Ũ1 Ũ2 =. This means that f 1 f 2 on N, so A is closed (as the complement of an open set). On the other hand, if f 1 (y) = f 2 (y), then Ũ1 = Ũ2, which implies that f 1 = f 2 on N (since p f 1 = p f 2 = f, and p is injective on Ũ1 = Ũ2). Thus, A is open, which proves the proposition. Exercise 1. Any continuous map RP 2 S 1 is null-homotopic. 5