Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 25J (Differential Geometry)

The homotopy lemma says that homotopic smooth maps between compact connected manifolds have the same mod-two degree, computed as the parity of the inverse image of a regular value. The homogeneity lemma says that any two points of a connected smooth manifold are related by a diffeomorphism isotopic to the identity.

Given regular values $p,q$ of $f$, choose such a diffeomorphism $h$ with $h(p)=q$. Then $h\circ f$ is homotopic to $f$, while $(h\circ f{)}^{-1}(q)=f^{-1}(p)$. The homotopy lemma proves that the two inverse-image counts have equal parity.

For smooth Brouwer, suppose a smooth self-map of a ball had no fixed point. Following the ray from $f(x)$ through $x$ to the boundary constructs a smooth retraction of the ball onto its sphere. Its restriction to the sphere is the identity, but it is also null-homotopic through the ball. The identity has odd mod-two degree and a constant map has even degree at a different regular value, contradicting the homotopy lemma. Thus a fixed point exists.

Orthogonality and $\Vert \tilde{\varphi }(x)\Vert =\Vert x\Vert$ giveso ${\psi }_{\epsilon }$ maps each boundary sphere to itself. AlsoOn compact $A$, this derivative is uniformly close to the identity; for sufficiently small $\epsilon$ its determinant is positive and it is locally invertible. The map is properly homotopic to the identity and has degree one, so the local diffeomorphism has one sheet and is a global diffeomorphism $A\to A$.

Apply change of variables to the orientation-preserving diffeomorphism ${\psi }_{\epsilon }:A\to A$:Rearranging gives

The left side in part (c) is a polynomial in $\epsilon$ of degree at most $n+1$. If $n$ is even, $(n+1)/2$ is a half-integer, sois not a polynomial and cannot agree with the left side on an interval. This contradiction shows that an everywhere nonzero tangent unit vector field $\varphi$ cannot exist on an even-dimensional sphere.