Questions: Degree Theory for Maps of Spheres

Question 1

What is the degree of the antipodal map a: S^n → S^n defined by a(x) = -x?

• A. 1
• B. -1
• C. (-1)^n
• D. (-1)^{n+1}

Answer: (D)

The antipodal map is the composition of (n+1) reflections, each reflecting one coordinate of R^{n+1}. Each reflection has degree -1 (it reverses the orientation of one axis). Therefore deg(a) = (-1)^{n+1}. For S^1 (n=1): deg = (-1)^2 = 1, which is correct since rotation by π is homotopic to the identity on S^1 (you can continuously rotate). Wait — actually the antipodal map on S^1 sends (cos θ, sin θ) → (-cos θ, -sin θ) = rotation by π, which has degree 1. For S^2 (n=2): deg = (-1)^3 = -1, which is correct since the antipodal map reverses orientation of S^2.

Question 2

If f: S^n → S^n has degree 0, then f is homotopic to a constant map.

• T. True
• F. False

Answer: True

A theorem of Hopf states that two continuous maps S^n → S^n are homotopic if and only if they have the same degree. Since a constant map has degree 0 (the induced map on H_n sends [S^n] to 0), any map of degree 0 is homotopic to a constant. Combined with the Hurewicz theorem (π_n(S^n) ≅ H_n(S^n) ≅ Z), this says the homotopy class of f is completely determined by its degree — the single integer deg(f) is a complete homotopy invariant for self-maps of spheres.

Question 3

If f: S^n → S^n has no fixed points (f(x) ≠ x for all x), what can you conclude about deg(f)?

• A. deg(f) = 0
• B. deg(f) = 1
• C. deg(f) = (-1)^{n+1} (same as the antipodal map)
• D. deg(f) = -1

Answer: (C)

If f has no fixed points, then f(x) ≠ x for all x, so the line segment from f(x) to -x never passes through the origin (since f(x) and x are distinct unit vectors, f(x) ≠ x implies f(x) and -x are not antipodal in a way that creates the origin issue — actually the correct statement is: t·f(x) + (1-t)·(-x) ≠ 0 because if it were zero, f(x) = x). Therefore H(x,t) = (tf(x) + (1-t)(-x))/|tf(x) + (1-t)(-x)| is a valid homotopy from -x to f(x), so f is homotopic to the antipodal map a(x) = -x, giving deg(f) = (-1)^{n+1}.

Question 4

Explain why deg(g ∘ f) = deg(g) · deg(f) for maps f, g: S^n → S^n.

Model answer: By functoriality of homology: (g ∘ f)_* = g_* ∘ f_*. The induced map f_*: H_n(S^n) → H_n(S^n) is multiplication by deg(f), and g_* is multiplication by deg(g). Their composition is multiplication by deg(g) · deg(f). Since (g ∘ f)_* is multiplication by deg(g ∘ f), we get deg(g ∘ f) = deg(g) · deg(f). This multiplicativity makes the degree a homomorphism from the monoid of self-maps of S^n (under composition) to the integers (under multiplication).

This multiplicativity has strong consequences. For instance, if f is a homeomorphism, then f ∘ f^{-1} = id has degree 1, so deg(f) · deg(f^{-1}) = 1, forcing deg(f) = ±1. Homeomorphisms have degree ±1. If deg(f) ≠ 0, then f is surjective (it covers S^n with nonzero 'algebraic multiplicity'). If |deg(f)| > 1, then f maps every point of S^n to at least one other point as well — it 'wraps' the sphere multiple times.

Question 5

For n = 1, the degree of a map f: S^1 → S^1 coincides with the classical winding number.

• T. True
• F. False

Answer: True

For S^1, the degree is the induced map on H_1(S^1) ≅ Z. The fundamental class [S^1] is the generator corresponding to one counterclockwise traversal. The map f wraps S^1 around itself some integer number of times, and this integer — the winding number — is exactly deg(f). The homological definition of degree generalizes the winding number from circles to arbitrary-dimensional spheres, providing a unified framework for counting 'how many times a map wraps around its target.'