Math 550. Homework 6. Solutions

Math 550. Homework 6. Solutions Definition 1. Two continuous mappings f , g : X → Y are homotopic if there is a continuous mapping H : X × [0, 1] → Y such that H(x, 0) = f (x) and H(x, 1) = g(x) for all x in X. Problem 1. Let C and C0 be circles. (i) Prove that if a continuous mapping F : C → C0 is not surjective, then deg F = 0. (ii) Find an example of a continuous mapping F : C → C0 that is surjective but has deg F = 0. Solution. (i) If F is not surjective, then there is a point P in C0 not contained in F(C). It follows that F(C) is contained in a sector (the complement of the ray issuing from the center of C0 and passing through P). As we have shown before, this implies that deg F = 0. (ii) Assume that C = C0 are the unit circle. Use polar coordinates (cost, sint), 0 ≤ t ≤ 2π and define F by  (cos 2t, sin 2t), 0≤t ≤π F(cost, sint) = (cos 2t, − sin 2t), π ≤ t ≤ 2π. This mapping F wraps the upper semicircle and the lower semicircle each onto the full circle, but in opposite directions. The degree of F is the winding number of the path γ(t) = F(cost, sint) around the origin. We can write γ as a sum of two paths γ = γ1 + γ2 , where γ1 (t) = (cos 2t, sin 2t), 0≤t ≤π and γ2 (t) = (cos 2t, − sin 2t), π ≤ t ≤ 2π Then W (γ, 0) = W (γ1 , 0) +W (γ2 , 0) = 1 − 1 = 0. Problem 2. (i) Prove that two continuous mappings F, G : C → C0 are homotopic if and only if they have the same degree. (ii) Conclude that a continuous mapping F : S1 → S1 has degree n if and only if F is homotopic to the map z 7→ zn of S1 onto itself. Solution. (i) This is equivalent to saying that two continuous closed paths in γ, δ : [0, 1] → R2 \ {0} are homotopic is and only if W (γ, 0) = W (δ, 0). This was done in Homework 5. (ii) You also showed in Homework 5 that the mapping z 7→ zn of the unit circle has degree n. Thus (ii) follows at once from (i). Problem 3. (i) Let F, F 0 : X → Y and G, G0 : Y → Z be continuous mappings. Prove that if F is homotopic to F 0 , and G is homotopic to G0 , then the composite G ◦ F is homotopic to G0 ◦ F 0 . 1 (ii) Let F, G be continuous mappings from the unit circle S1 into itself. Prove that

deg(G ◦ F) = deg F · deg G . Solution. (i) If H : X × [0, 1] → Y is a homotopy from F to F 0 and K : Y × [0, 1] → Z is a homotopy from G to G0 (as in the definition above), then define Γ : X × [0, 1] → Z by Γ(x, s) = K(H(x, s), s). Then Γ is continuous because H and K are both continuous, Γ(x, 0) = K(H(x, 0), 0) = K(F(x), 0) = G(F(x)) = G ◦ F(x), and Γ(x, 0) = K(H(x, 1), 1) = K(F 0 (x), 1) = G0 (F 0 (x)) = G0 ◦ F 0 (x). (ii) If deg F = n, then F is homotopic to z 7→ zn , and if deg G = m, then G is homotopic to z 7→ zm . By (i), the composite G ◦ F is homotopic to the composition of the nth power map and the mth power map, that is to z 7→ znm , which has degree nm. Problem 4. Suppose that F is a continuous mapping from the positive octant {(x, y, z) | x ≥ 0, y ≥ 0, z ≥ 0} to itself. Show that there is a unit vector P in this octant, and a nonnegative number λ, such that F(P) = λP. Solution. The positive octant intersects the unit sphere in a triangular sector T . Define a mapping G form T into T by G(P) = F(P)/|F(P)|. Because T is homeomorphic to a disk (it is a challenging exercise for you to describe one such homeomorphism), there is a point P in T such that G(P) = P. Then F(P) = λP with λ = |F(P)|, as desired. Problem 5. Let f : C → C0 be a continuous mapping between circles. (i) Prove that if f (P∗ ) = f (P) for all P, then the degree of f is even. (ii) Prove that if f (P∗ ) 6= f (P) for all P, then f is surjective. Solution. There is no loss of generality if we assume that C = C0 = S1 is the unit circle. (Recall that the antipode of a point P in the unit circle is P∗ = −P. (i) The degree of f is the winding number W (γ, 0) of the path γ(t) = F(cost, sint) (0 ≤ t ≤ 2π) around the origin. Let δ1 be the path δ1 (t) = F(cost, sint) (0 ≤ t ≤ π) and δ2 be the path δ2 (t) = F(cos(t +π), sin(t +π)) (0 ≤ t ≤ π). Then δ2 (t) = F(− cost, − sint) = F(cost, sint) = δ1 (t), so W (δ2 , 0) = W (δ1 , 0). But W (γ, 0) = W (δ1 , 0) +W (δ2 , 0) = 2W (δ1 , 0), an even number. (ii) If f (P) 6= f (P∗ ) for all P in C, define g(P) = f (P) − f (P∗) . | f (P) − f (P∗)| Then g(P∗ ) = −g(P), so that deg(g) is odd; in particular deg(g) 6= 0. Now g is homotopic to f . Indeed, because for every point P in the unit circle the point f (P) is in the unit circle and f (P) 6= f (P∗ ), the vector f (P) − s f (P∗) 6= 0. Therefore H(P, s) = f (P) − s f (P∗ ) | f (P) − s f (P∗)| defines a continuous mapping H : S1 × [0, 1] → S1 that satisfies H(P, 1) = g(P) and H(P, 0) = f (P)/| f (P)| = f (P) for all P in S1 . By Problem 2, the mappings f and g have the same degree, and so deg( f ) 6= 0 also. By Problem 1, f must be surjective. 2