Math 205A: Differential Topology, Homework 6
Ian Coley
November 17, 2013
Problem 2.4.9.
Suppose f : X → S k is smooth, where X is compact and 0 < dim X < k. Then for all closed
Z ⊂ S k of dimension complementary to X, I2 (X, Z) = 0.
Solution.
By earlier considerations, since f is not surjective, there exists p ∈
/ f (X) ∩ Z. Then by
k
removing the point {p}, we may use stereographic projection onto R . If π is this projection,
we have I2 (X, Z) = I2 (π(X), π(Z)) = I2 (π ◦ f, π(Z)). By Exercise 2.4.5, since π(S k \ {p}) =
Rk is contractible and π(Z) is closed and of complementary dimension to (π ◦ f )(X), we
have I2 (π ◦ f, Z) = 0. Therefore I2 (X, Z) = 0 as required, and we are done.
Problem 2.4.10.
Prove that S 2 and the torus are not diffeomorphic.
Solution.
By earlier considerations, S 2 is simply connected. As such, any map from S 1 to S 2 is
contractible to a point. In particular, if γ1 , γ2 are two loops in S 2 , then I2 (γ1 , γ2 ) = 0 since
these loops are homotopic to two distinct points.
Let T 2 = S 1 × S 1 be the torus. Then there is a smooth map γ1 : S 1 → T 2 by x 7→ (x, q)
for some fixed q ∈ S 1 . Similarly there is γ2 : S 1 → T 2 such that x 7→ (p, x) for a fixed p ∈ S 1 .
These two loops intersect at exactly one point, namely (p, q) ∈ T 2 . Therefore S 2 ∼
6= T 2 since
every two loops in S 2 must have intersection number 0 mod 2.
Problem 2.4.11.
Suppose that f : X → Y has deg2 f 6= 0. Prove that f is onto.
Solution.
Suppose that f : X → Y is not a surjective map, where X is compact, Y is connected, and
dim X = dim Y . Then there exists y ∈ Y outside the image of f . Then f t {y} vacuously
and I2 (f, {y}) = 0. By degree theory, since Y is connected, I2 (f, {y}) is constant everywhere
and is equal to the degree, i.e. deg2 f = 0. Since we have proved the contrapositive, we are
done.
Problem 2.4.14.
Two compact sub manifolds X and Z in Y are cobordant if there exists a compact manifold
with boundary W ⊂ Y × I such that ∂W = X × {0} × Z × {1}. Show that if X may be
deformed into Z, then X and Z are cobordant.
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Solution.
If X can be deformed into Z, then there exists a function f : X × I → Y such that each
f (−, t) = ft is an embedding with f0 (x) = x is the inclusion of X into Y and f1 (x) = z
is the embedding onto Z. We can extend this map to a map F : X × I → Y × I where
F (x, t) = (ft (x), t). Then the image of F , say W , is a compact manifold in Y × I such that
∂W is the image of F at t = 0 and t = 1, namely X × {0} ∪ Z × {1} as required. This
completes the proof.
Problem 2.4.15.
Prove that if X and Z are cobordant in Y , then for every compact manifold C in Y with
dimension complementary to X and Z, I2 (X, C) = I2 (Z, C).
Solution.
Let W ⊂ Y × I be the compact manifold whose boundary is X × {0} ∪ Z × {1}. Consider
the projection π : Y × I → Y and i : W ,→ Y × I. Let π ◦ i = f : W → Y . Then
since ∂f : ∂W → Y is a smooth map with the extension aforementioned extension to
W , by the Boundary Theorem we have I2 (∂W, C) = 0 = I2 (X, C) + I2 (Z, C). Therefore
I2 (X, C) = I2 (Z, C).
Problem 2.4.18.
Suppose that Z is a compact submanifold of Y with dim Z =
globally definable by independent functions, then I2 (Z, Z) = 0.
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2
dim Y . Prove that if Z is
Solution.
By Exercise 2.3.20, N (Z; Y ) is trivial, so N (Z; Y ) = Z × Rk . Then by the Tubular Neighbourhood Theorem, there exists a neighbourhood Z ⊂ U in Y homeomorphic to a neighbourhood V of Z in N (Z; Y ). In V , Z × {0} is homotopic to a small deformation Z × {v},
so I2 (Z × {0}, Z × {0}) = I2 (Z × {0}, Z × {v}) = 0. Therefore in U , Z is homotopic to some
Z 0 disjoint from Z under the homeomorphism. Therefore I2 (Z, Z) = I2 (Z, Z 0 ) = 0. This
completes the proof.
Problem 2.4.19.
Show that the central circle X in the open Möbius band has mod 2 intersection number
I2 (X, X) = 1.
Solution.
Based on Figure 2.18 in the book, there is a clear smooth deformation from X to X 0 in I × I
from a constant function to a smooth bump function. Under the gluing of the ends, this is
still a smooth deformation. Therefore I2 (X, X) = I2 (X, X 0 ) = 1.
Problem 2.6.1.
Show that the Borsuk-Ulam theorem is equivalent to the following assertion: if f : S k → S k
carries antipodal points to antipodal points, then deg2 f = 1.
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Solution.
Recall that W2 (f, z) = deg2 u where
u(x) =
f (x) − z
.
|f (x) − z|
Suppose first that the Borsuk-Ulman theorem holds. Then f : S k → S k carrying antipodal
points to antipodal points is a map of the form f : S k → Rk+1 where f (−x) = −f (x). Then
(x)
W2 (f, 0) = 1, i.e. deg2 u = 1 where u(x) = |ff (x)|
. But since we are on the circle, |f (x)| = 1
everywhere, so u(x) = f (x). Therefore deg2 f = 1.
Conversely, suppose that the assertion holds. Let f : S k → Rk+1 be a smooth map such
that −f (x) = f (−x) and f (x) 6= 0 for any x ∈ S k . Then let u(x) be as above, satisfying
the premises of the assertion. Therefore deg2 u = 1. Since W2 (f, 0) = deg2 u = 1, so we are
done.
Problem 2.6.2.
Prove that any map f : S 1 → S 1 mapping antipodal points to antipodal points has deg2 f = 1
by a direct computation with exercise 2.4.8.
Solution.
Exercise 2.4.8 states that this f corresponds to a smooth map g : R → R such that
f (cos t, sin t) = (cos g(t), sin g(t)). We know that g(t + 2π) = g(t) + 2πq for some integer q. Since f takes antipodal points to antipodal points, we know g(0) = (2n + 1)π so that
f (1, 0) = (−1, 0). Since we may take this to the general formula of g, i.e. g(t+2π) = g(t)+2πq
where g(t) is defined on [0, 2π), where we must have q = 2n + 1. Since deg2 f = q (mod 2),
we have deg2 f = 1. This completes the proof.
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