Math 751 Lecture Notes Week 3

Transcription

1 Math 751 Lecture Notes Week 3 Setember 25, Fundamental grou of a circle Theorem 1. Let φ : Z π 1 (S 1 ) be given by n [ω n ], where ω n : I S 1 R 2 is the loo ω n (s) = (cos(2πns), sin(2πns)). Then φ is an isomorhism. Proof. Let : R S 1, s (cos(2πs), sin(2πs)). Then 1 (1, 0) = Z. Embed i : R R 3 as a helix, s (cos(2πs), sin(2πs), s). Then = r 12 i where r 12 (x, y, z) = (x, y). Let ω n : I R be given by s ns. Note that ω n (0) = 0 and ω n (1) = n. Also, ω n = ω n, so φ(n) = [ ω n ]. In fact, φ(n) = [ f] for any ath f : I R from 0 to n. Indeed ω n and f are homotoic in R via the homotoy (1 t) ω n + t f. So ω n f. Define the translation τ m : R R by τ m (x) = x + m, and notice that ω m is a ath from 0 to m and τ m ( ω n ) is a ath from m to n + m; their concatenation is thus a ath from 0 to n + m. We have: φ(m + n) = [ ω n+m ] = [ ( ω m τ m ( ω n ))] = [( ω m ) ( τ m ( ω n )] = [ω m ω n ] = [ω m ] [ω n ] = φ(m)φ(n), so φ is a homomorhism. To rove that φ is bijective, we need two lemmas. Lemma 1.1 (ath lifting). For every f : I S 1 with f(0) = x 0 S 1 and for any x 0 1 (x 0 ), there is a unique f : I R such that f = f and f(0) = x 0.! f R f x 0 I S 1 f 0 x 0 f Lemma 1.2 (homotoy lifting). For every homotoy f t : I S 1 with f t (0) = x 0 S 1 and for any x 0 1 (x 0 ), there is a unique homotoy f t : I R such that f t = f and f t (0) = x 0. 1

2 In our case, we use x 0 = (0, 1) and x 0 = 0. Let f : I S 1 be a loo at (1, 0). By ath lifting, this means there is an f : I R such that f = f and f(0) = 0 Z. Say f(1) = n Z, so f is a ath in R from 0 to n. Then φ(n) = [ f] = [f]. Since f was arbitrary, φ must be surjective. Now suose φ(m) = φ(n) for some m and n from Z. So [ω m ] = [ω n ], or ω m ω n. So there is a homotoy f t with f 0 = ω m and f 1 = ω n. By homotoy lifting, there is a f t : I R such that f t = f t and f t (0) = 0. But f t (1) is indeendent of t, so f 0 (1) = f 1 (1). Now f 0 and ω m are both lifts to R of f 0 = ω m which start at 0. By the uniqueness of ath lifting, this means f 0 = ω m. In articular, f 0 (1) = ω m (1) = m. Similarly, f 1 (1) = ω n (1) = n. So m = n. Both lifting lemmas that we used in the above roof result from a single general lifting lemma which we will rove here. Lemma 1.3 (lifting). Let Y be a connected sace. Given F : Y I S 1 and F : Y {0} R which lifts F Y {0} to R, there is a unique lift F : Y I R of all of F which restricts to the given lift on Y {0}. R F! F F Y {0} Y I S 1 Proof. First we define F locally, that is, on N I for some neighborhood N of a given y 0 Y. Then we show the uniqueness of F on sets of the form {y 0 } I. This uniquely defines F on all of Y I. (Ste i) There is an oen cover {U α } α of S 1 so that for each α, 1 (U α ) = β (Ũβ), where each Ũβ is an oen interval in R that satisfies (Ũβ) = U α and such that restricts to a homeomorhism between Ũβ and U α. For all airs (y 0, t) Y I, let α be such that F (y 0, t) U α. F is continuous, so there is a neighborhood N (a t, b t ) of (y 0, t) so that F (N t (a t, b t )) U α. Since {y 0 } I is comact, it can be covered by finitely many such N t (a t, b t ). We can choose a single neighborhood N of y 0 and a artition of I given by 0 = t 0 < t 1 < < t m = 1 so that for each i there is an α i with F (N [t i, t i+1 ]) U αi. Assume (for induction) that F has been defined on N [0, t i ], starting with the given lift on N {0} for i = 0. We can extend it to N [t i, t i+1 ] as follows. Recall that since F (N [t i, t i+1 ]) U αi, we have F (N {t i }) Ũβ i. Define F on N [t i, t i+1 ] by F = ( 1 Uαi : U αi Ũβ i ) F. (Ste ii) Now we show uniqueness for the case when Y is a single oint. Choose a artition of I by 0 = t 0 < t 1 < < t m = 1 so that for all i there is an oen U αi that comletely contains F ([t i, t i+1 ]). Assume we have F and F, two lifts of F : I S 1. We have that F (0) = F (0), since we are choosing a secific starting oint x 0 R. For induction, suose F and F coincide on [0, t i ]. F is 2

3 continuous and [t i, t i+1 ] is connected, so F ([t i, t i+1 ]) is connected. Thus there is a unique Ũβ i that comletely contains F ([t i, t i+1 ]). Similarly, there is a unique Ũ β i F ([t i, t i+1 ]). F i (Ũβ i ) R I [t i, t i+1 ] U αi S 1 F Suose Ũβ i Ũ β i. But F (t i ) = F (t i ) by the induction hyothesis, and {Ũβ i } are disjoint or equal, so Ũβ i = Ũ β i. Also, Ũβi is a homeomorhism, so is injective on Ũβ i and F = F. Hence F = F on [t i, t i+1 ]. (Ste iii) The lifts F constructed on the sets N I in (Ste i) are unique by (Ste ii) on each segment {y} I, so two such lifts must agree on their overlas. This means, by gluing, that we get a well-defined lift F : Y I R. F is continuous since it is on each N I. Finally, F is unique by (Ste ii). To see how ath lifting comes from this result, let Y be a single oint and ath lifting trivially follows. To get to homotoy lifting, let Y = I. However, we aren t given a lift F : I {0} R of F : I I S 1. Given a homotoy F : I I S 1, define f t (s) = F (s, t). There is a unique lift F : I {0} R obtained by the ath-lifting lemma alied to f 0 : I S 1. By the general lifting lemma, there is a unique F : I I R; ft (s) = F (s, t) is a homotoy of aths lifting the homotoy f t, since F {0} I and F {1} I are lifts of constant aths, and by uniqueness, they are also contstant aths. 2 Fundamental Theorem of Algebra Theorem 2. Let f(z) = z n +a 1 z n 1 + +a n 1 z+a n be a comlex olynomial. Then f has a comlex root, i.e., f(z) = 0 has a solution in C. Proof. If a n = 0, then z = 0 is a solution. So assume a n is nonzero. Define F (z, t) = z n +t(a 1 z n 1 + +a n ). Clearly F is continuous. F (z, 0) = z n = n (z) and F (z, 1) = f(z). Restrict F to a circle C r with radius r, with C r := {z z = r}. We can see that for large enough r, F is nonzero: ) ( ) ( F (z, t) z n t a 1 z n a n = r (1 t n a1 r + + a ) n r n > 0 So for large r, F is a homotoy C r I C = C\0 from f to n. 3

4 Assume now that f never vanishes. Define G(z, t) = f(tz). Notice that G(z, 0) = f(0) = a n and G(z, 1) = f(z). Restricting z to C r, G is a homotoy C r I C from f to the constant ma e an. By transitivity we have n (z) = z n and the constant ma are homotoic as mas C r C. We thus obtain the following commutative diagram Z = π 1 (C r, r) (e an ) ( n ) Z = π 1 (C, a n ) Z = π 1 (C, r n ) δ where δ is a ath in C from r n to a n. Let 1 denote the generator of π 1 (C r, r), a loo at r going around the whole circle. Then since (e an ) has to be trivial, (e an ) (1) = 0, and since the diagram is commutative and δ is an isomorhism, δ ( n ) must be trivial, so ( n ) (1) = 0. But this contradicts the fact that ( n ) (1) = n 1. Hence our assumtion that f never vanishes is false, so f must have a comlex root. 3 Seifert-Van Kamen Theorem We would like to comute π 1 (X) in terms of {π 1 (U α )} for {U α } α a nice oen cover of X. The next theorem shows how to do this. Theorem 3 (Seifert-Van Kamen). Let X be a ath connected sace, and X = U V where U, V, and U V are oen ath connected subsets of X. Fix x 0 U V and consider the inclusion mas: U V i j U V α β U V Then π 1 (X, x 0 ) = π 1 (U, x 0 ) π 1 (V, x 0 ) N i (ξ)j (ξ) 1 ξ π 1 (U V, x 0 ) where is the free roduct and N x is the normal subgrou generated by x. Corollary 3.1. If U V is simly connected, then π 1 (U V ) = π 1 (U) π 1 (V ). Corollary 3.2. The union of two simly connected saces is simly connected, rovided their intersection is nonemty and ath connected. Examle 1. Let X = S n for n 2. Define U to be the comlement of north ole, and V to be the comlement of the south ole. Then U, V, and U V are all oen and ath connected, and U and V are contractible. So, by the last corollary, S n is simly connected. 4

5 Definition 1. Given two saces X and Y with distinguished oints x 0 and y 0 resectively, the join of X and Y is defined by: X Y := X Y / x 0 y 0 Examle 2. Let X n = n i=1 S1 be a join of n circles at a single oint (called a bouquet of n circles). Then π 1 (X n ) = Z n, that is, a free roduct of n coies of Z. Proof. For n = 1 we get a single circle, so the result is clear. For induction, suose we have shown π 1 (X n 1 ) = Z (n 1). Let x 0 be the join oint of n circles. For each i choose i x 0 to be a oint on the i-th circle. Let U = X n \{ n } n 1 i=1 S 1 = X n 1 and V = X n \{ 1,, n 1 } S 1. Then U V {x 0 }, so by the Seifert-Van Kamen theorem (articularly Corollary 3.1), π 1 (X n ) = π 1 (U) π 1 (V ) = π 1 (U) Z, and by the induction hyothesis, this gives Z (n 1) Z = Z (n). Examle 3. Let X = R 2 \{x 1,, x n }. Then X deformation retracts to a bouquet of n circles, one going around each x i. So π 1 (X) = Z n. Examle 4. Let X = R 3 \{coordinate axes} S 2 \{6 oints} = R 2 \{5 oints}, where was by x x x and = is by stereograhic rojection. Then it is clear that π 1 (X) = Z 5. Examle 5. Let X = S 2 {equatorial disk D 2 }, so S 2 D 2 = {the equator}. Take U = X\{north ole} S 2, and take V = X\{south ole} S 2, so U V D 2. By Corollary 3.2, X is simly connected. Examle 6. Let X = S 2 {north-south diameter}. Let P be a oint on the diameter (neither of the oles). Let Q be a oint on the shere (neither of the oles). Choose U = X\{P } S 2 and V = X\{Q} S 1. Notice that U V S 2 \{Q} = R 2. Since U V is simly connected, by Corollary 3.1 π 1 (X) = π 1 (U) π 1 (V ) = 0 Z = Z. 5