Math 215B: Solutions 3 (1) For this problem you may assume the classification of smooth one-dimensional manifolds: Any compact smooth one-dimensional manifold is diffeomorphic to a finite disjoint union of circles and closed intervals. Fix a basepoint x 0 S n, and suppose f, g : S n S n are smooth maps having x 0 as a regular value. Let #f 1 (x 0 ) (resp. #g 1 (x 0 )) denote the number of points in the zero-dimensional manifold f 1 (x 0 ) (resp. g 1 (x 0 )). (a) Suppose f and g are homotopic. Prove that #f 1 (x 0 ) = #g 1 (x 0 ) mod 2. (b) Let y 0 S n be another regular value for f. Prove that #f 1 (y 0 ) = #f 1 (x 0 ) mod 2. (c) Let [S n, S n ] denote the set of homotopy classes of continuous maps S n S n. Using parts (a) and (b) show that the map deg : [S n, S n ] Z/2 [f] [#f 1 (x)] mod Z/2 doesn t depend on the choice of representative of the class [f] (which is covered in part (a)), or on the choice or regular value x S n for that choice of representative (which is covered by part (b)). (d) Suppose that f : S n S n is null-homotopic. Prove that deg([f]) = 0. (e) Prove that deg([f g]) = deg([f]) deg([g]). (f) The sum of two elements [f], [g] [S n, S n ] is defined to be the homotopy class of the map given by the composite S n γ S n S n f g S n where γ is obtained by collapsing the equator to a point. Prove that deg([f] + [g] = deg([f]) + deg([g]). Solution. In order to apply analytic techniques to continuous maps, we will use the following smoothening theorem: Theorem 1 (cf. Hirsch, Ch. 5, Lemma 1.5). Let f : M N is a continuous map between C manifolds and suppose M is smooth outside a closed set Z M. Let U be an open set containing Z. Then there is a homotopy F t : M [0, 1] N, such that F 0 = f and F 1 is smooth and for all t we have F t = f outside U. 1
This is a straightforward generalization of the statement in Hirsch. In particular, taking Z = U = M shows that any continuous map between smooth manifolds is homotopic to a smooth one. If H t : M [0, 1] N is a homotopy between smooth maps f = H 0 and g = H 1, then applying the theorem to the homotopy f(m); 0 t 1/3, (m, t) H 3(t 1/3) (m); 1/3 t 2/3, g(m); 2/3 t 1, with Z = M [1/3, 2/3] and U = (1/4, 3/4) shows that they are homotopic through a smooth homotopy. We will also use Sard s theorem Theorem 2 (Hirsch, Ch. 3, Thm. 1.3). Let f : M N be a smooth map of C manifolds. Then the set of singular values of f has measure zero in N. Note that although N doesn t have a canonical measure, the euclidean structure on N allows us to talk about sets of measure zero. (a) By the above, there exists a smooth homotopy H t : M [0, 1] between f and g. The set of singular points of f is a closed subset of S n. Since the sphere is compact, its image is therefore closed, which is exactly the set of singular values of f. It follows that there exists an open ball x 0 B S n, such that every point in B is a regular value for both f and g. By Sard s theorem, there exists some y 0 B, which is a regular value for H t. Let γ be a smooth curve in B with endpoints x 0 and y 0 (that is, a 1-manifold with boundary). Then by the implicit function theorem with boundary (Hirsch, Ch. 1, Thm. 4.2), f 1 (γ) is a closed 1-manifold in S n with boundary contained in the preimages of {x 0 } and y 0. By compactness and the classification of 1-manifolds, the number of boundary points is finite and even. Thus #f 1 (x 0 ) = #f 1 (y 0 ) mod 2. By the same argument, #g 1 (x 0 ) = #g 1 (y 0 ) mod 2. Now y 0 is a regular point of the homotopy H t. Thus J := Ht 1 (y 0 ) is a neat 1-dimensional submanifold of S n [0, 1] and J = J S n [0, 1]. Thus # J = #H 1 t (y 0 ) S n {0, 1} = # ( f 1 (y 0 ) {0} g 1 (y 0 ) {1} ) = #f 1 (y 0 )+#g 1 (y 0 ). Since J is a one-dimensional manifold, this number is even. Thus we find #f 1 (y 0 ) = #g 1 (y 0 ) mod 2, and hence also #f 1 (x 0 ) = #g 1 (x 0 ) mod 2. (b) We can assume that x 0 y 0. Then embed the sphere so into R n+1 such that x 0 = e 1 and y 0 = e 2. Let r θ : R 2 R 2 be the positively oriented rotation by angle θ. Then R t := R t id R n 1 gives a smooth family of diffeomorphisms S n S n with R 0 = id S n and such that R π/2 maps x 0 to x 0. Thus f is homotopic to R π/2 f. In particular by (a): #f 1 (y 0 ) = #(R π/2 f) 1 (y 0 ) = #f 1 (x 0 ) mod 2. 2
(c) The map is defined by picking any smooth representative in the class of f, which is possible by the smoothening result mentioned above, and then choosing for x 0 any regular value of f. If g is another smooth representative, and y 0 a regular value of g, then by Sard s theorem, we can find a point z 0 on S n which is a regular value for both f and g. Then we have #f 1 (x 0 ) = #f 1 (z 0 ) = #g 1 (z 0 ) = g 1 (y 0 ) mod 2, where the first and last equality are due to part (b) and the middle equality is due to part (a). Thus the map is independent of any choices. (d) Suppose f is null-homotopic. Then the homotopy class of f contains a constant map form g p, for some p S n. In particular g is a smooth representative of the class [f]. Any q S n not equal to p is a regular value for g, so deg f = deg g = #g 1 (q) = 0 mod 2. (e) Replace f and g by smooth representatives of their homotopy classes. This doesn t change the homotopy class of their composition, as can be easily checked. Now suppose that g is not surjective. Then it is null-homotopic, as can be seen by composing with a deformation retraction of the punctured sphere onto a point. Thus we can assume that g is constant, so f g is constant, thus deg([f g]) = 0 = deg([f]) deg([g]) in this case. So assume that g is surjective. Let z 0 S n be a regular point of f g. Then for every point x (f g) 1 (z 0 ) the differential D x (f g) = D g(x) f D x g is surjective, hence an isomorphism. Thus D g(x) f and D x g are isomorphisms. It follows that every y f 1 (z 0 ) is a regular point for g and, since g is surjective, that z 0 is a regular point of f. Therefore (f g) 1 (z 0 ) = y f 1 (z 0)g 1 (y). Since all preimages are taken of regular points, by taking cardinalities of this equality of sets, we obtain the result, (f) We take some care to exhibit a smooth representative of [f] + [g]. Say we want to construct S n S n by connecting two copies of the sphere along a chosen point p. In order to obtain a continuous map f g, one needs to alter f or g by a homotopy so that f(p) = g(p), say by composing one with a rotation of the image sphere. One can check that the homotopy class of f g is independent of any such variation and of the choice of p. So suppose f(p) = g(p) =: q. We can construct a map ϕ, which collapses a small closed ball B 0 around q to q, and is the identity of S n outside a slightly bigger open ball B 1 and such that ϕ is homotopic to the identity. Thus [f] = [ϕ f] and [g] = [ϕ g], and we can assume that f and g are constant in a neighborhood of p. Now model S n S n as the union of the spheres with radius 1/2 around the centers (1/2, 0,..., 0) and ( 1/2, 0,..., 0) in R n+1, with γ the map from S n leaving the first coordinate constant and projecting radially onto S n S n in the other coordinates. This is smooth and a bijection from the sphere with the equator 3
removed to S n S n with the connecting point removed. Moreover, since we chose f and g to be constant around p, we see that f g is smooth, since around the gluing point it is the constant map. Now choose any point y 0 S n which is a regular value for all of f, g and f g γ. In particular y 0 f 0 (p). Then #γ 1 ((f g) 1 (y 0 )) = #(f g) 1 (y 0 ) = #f 1 (y 0 ) + #g 1 (y 0 ). (3) Hirsch, p.118 Exercise 5: Let M V be a closed neat submanifold of codimension k. Then there is a map f : (V, M) (S k, p) such that p is a regular value and f 1 (p) = M, if and only if M has a trivial normal bundle. Solution. The solution uses the tubular neighborhood theorem: Theorem 3 (Hirsch, Ch. 4, Thm. 6.2). Let M V be a neat submanifold. Then M has a tubular neighborhood in V. In fact, it follows that one can always find a normal tubular neighborhood. Thus there exists an embedding g : ν M V, which restricts to id M on the zero section. Now we assume that there exists an f as described. Then we obtain the morphism of vector bundles Df : T V T S k, which restricts to Df M : T M V T p S k. The assumption that p is a regular value is equivalent to this map being fiberwise surjective. Since f M is constant with value p, the tangent bundle of M is in the kernel of Df M. Thus Df M descends to a vector bundle epimorphism ν M T p S k, since the bundles have the same rank k, it is fiberwise bijective. Thus the map pulls back to an isomorphism ν M f M (T ps k ). Since T p S k {p} R k is trivial, this shows that ν M is trivializable. Now assume conversely that there exists an isomorphism of vector bundles ν M M R k. By the tubular neighborhood theorem, there exists a map g : M R k V, which is a diffeomorphism onto an open neighborhood of M in V. Let p, q S k be antipodal points. Stereographic projection gives a smooth embedding h : R k S n with φ(0) = p, which is a diffeomorphism onto S k \ {q}. Let B 1 B 1 B 2 small open balls around q. Then there exists a smooth map ϕ : S k S k, which is the identity on S k \ B 2 and with ϕ(b 1 ) = {q}. Let h := ϕ h. This has p as a regular value with h 1 (p) = {0} and there is some constant K, s.t. h(x) = q whenever x K. Let U 1 := g(m R k ) V and Z 1 = V \ U 1. Since Z 1 and M are disjoint closed sets of V, there exists some smooth function ψ : V [0, 1] which is constant equal to zero on M and constant equal to 1 on Z 1. Let Z 2 := ψ 1 ([0, 1/3]) and U 2 = V \ Z 2. Then Z 2 is a closed neighborhood of M in V, and Z 2 U 1, so V = U 1 U 2. In particular, we can find a smooth map λ : M R >0, such that m M : g({m} B λ (0)) Z 2. 4
Now we define f : V S k ϕ h(cxλ(m) 1 ) if v U 1, v = g(m, x), v q if v U 2. Here we use the diffeomorphism U 1 M R k. By construction, the two assignments agree on U 1 U 2. Thus f is a smooth function on V, with f 1 ({p}) = M. To conclude that p is a regular value, it is enough to note that for fixed m M the map R k S k, x ϕ h(cxλ(m) 1 ) is submersive at 0. 5