Suppose that to every map \(h: S^n \to S^n\) we have assigned an integer, denoted by \(\deg h\) and called the degree of \(h\), such that:
Homotopic maps have the same degree.
\(\deg(h \circ k) = (\deg h) \cdot (\deg k)\).
The identity map has degree 1, any constant map has degree 0, and the reflection map \(\rho(x_1, \ldots, x_{n+1}) = (x_1, \ldots, x_n, -x_{n+1})\) has degree \(-1\).
[One can construct such a function, using the tools of algebraic topology. Intuitively, \(\deg h\) measures how many times \(h\) wraps \(S^n\) about itself; the sign tells you whether \(h\) preserves orientation or not.] Prove the following:
There is no retraction \(r: B^{n+1} \to S^n\).
If \(h: S^n \to S^n\) has degree different from \((-1)^{n+1}\), then \(h\) has a fixed point. [Hint: Show that if \(h\) has no fixed point, then \(h\) is homotopic to the antipodal map \(a(x) = -x\).]
If \(h: S^n \to S^n\) has degree different from 1, then \(h\) maps some point \(x\) to its antipode \(-x\).
If \(S^n\) has a nonvanishing tangent vector field \(v\), then \(n\) is odd. [Hint: If \(v\) exists, show the identity map is homotopic to the antipodal map.]
Solution
First note that the antipodal map \(a(x)=-x\) is the composite of \(n+1\) coordinate reflections, each conjugate to the given reflection \(\rho\). Hence \[ \deg a=(-1)^{n+1}. \]
If \(r:B^{n+1}\to S^n\) were a retraction and \(i:S^n\hookrightarrow B^{n+1}\) the inclusion, then \(r\circ i=\operatorname{id}_{S^n}\). But \(i\) is nullhomotopic since the ball is contractible, so \(r\circ i\) is nullhomotopic. Thus \[ 1=\deg(\operatorname{id})=\deg(r\circ i)=0, \] a contradiction.
Suppose \(h\) has no fixed point. Define \[ H(x,t)=\frac{(1-t)h(x)-tx}{\|(1-t)h(x)-tx\|}. \] The denominator never vanishes: if it did, then \(h(x)=\frac{t}{1-t}x\); since both are unit vectors, this forces \(h(x)=x\), contrary to hypothesis. Thus \(H\) is a homotopy from \(h\) to the antipodal map. Hence a fixed-point-free map must satisfy \[ \deg h=\deg a=(-1)^{n+1}. \] The contrapositive proves (b).
Suppose \(h(x)\ne -x\) for every \(x\). Then \[ K(x,t)=\frac{(1-t)h(x)+tx}{\|(1-t)h(x)+tx\|} \] is well defined: a zero would force \(h(x)=-x\). Thus \(h\simeq\operatorname{id}\), so \(\deg h=1\). Contrapositively, if \(\deg h\ne1\), some \(x\) satisfies \(h(x)=-x\).
If \(v\) is a nonvanishing tangent vector field, normalize it so \(\|v(x)\|=1\). Since \(v(x)\perp x\), \[ F(x,t)=\cos(\pi t)x+\sin(\pi t)v(x) \] lies on \(S^n\) for all \(t\), and is a homotopy from the identity to the antipodal map. Therefore \[ 1=(-1)^{n+1}. \] Hence \(n+1\) is even, so \(n\) is odd.