# Proof Critique - Hatcher Exercise 2.2.3

Let $f: S^n \to S^n$ be a map of degree zero. Show that there exist points $x,y \in S^n$ with $f(x) = x$ and $f(y) = -y$. Use this to show that if $F$ is a continuous vector field defined on the unit ball $D^n \subset \mathbb{R}^n$ such that $F(x) \neq 0$ for all $x$, then there exists a point on $\partial D^n$ where $F$ points radially outward and another point on $\partial D^n$ where $F$ points radially inward.

Suppose that $f: S^n \to S^n$ is a map of degree zero and $f(x) \neq x$ for all $x \in S^n$. Therefore $f$ has no fixed points. Consequently the degree of $f$ is $(-1)^{n+1}$. This contradicts the assumed degree of $f$.

Similarly, suppose that $f(y) \neq -y$ for all $y \in S^n$. Then $-f$ has no fixed points. Hence $\deg -f = (-1)^{n+1}$. Observe however that $$(-1)^{n+1} = \deg -f = \deg -\text{id} \cdot \deg f = 0.$$ The contradiction yields the desired result.

Now suppose that $F$ is a continuous vector field defined on the unit ball $D^n \subset \mathbb{R}^n$ such that $F(x) \neq 0$ for all $x$. By considering the inclusion map $i : D^n \hookrightarrow \mathbb{R}^n$ the map $$f_t(x) = (1-t)F(x) + ti(x)$$ establishes a homotopy between $F$ and the inclusion map. Since the inclusion map is not surjective, the degree of $i$ is zero. The homotopy assures us that $F$ has degree zero. Moreover, since $F(x) \neq 0$ for all $x$, we can normalize $F(x)$ such that $$\hat{F}(x) = \frac{F(x)}{\| F(x) \|} \in \partial D^n \cong S^{n-1}.$$ Suppose that $F(x) \neq x$ for all $x$. Then by letting $G : S^{n-1} \to S^{n-1}$ be the restriction of $F$ to $S^{n-1}$, it follows that $G$ satisfies the relation $G(x) \neq x$ for all $x \in \partial D^n$. Hence, $\deg G = (-1)^{n+1}$; but $$\deg F \ \vert \deg G,$$ since $G$ is simply a restriction of $F$. This yields the desired contradiction. An analogous argument yields that $F(x) = -x$ for some $x \in \partial D^n$.

• We can talk about degree of a map between spheres of any dimension, as long as the domain and codomain spheres have the same dimension. Sep 27, 2016 at 6:01
• @BalarkaSen Thanks, removing my suspension argument, is the rest of the proof ok? I've edited my submission.
– user319128
Sep 27, 2016 at 6:05

First of all, notice that the degree is only defined for maps $S^n\to S^n$. The inclusion $i: D^n\to \mathbb{R}^n$ does not satisfy this property. If your argument were true, any map defined on $S^n$ being homotopic to the inclusion would have degree zero. Observing that $G$ factors through $S^n\hookrightarrow D^n\stackrel{F/|F|}{\rightarrow} S^n$ and using $\tilde{H}_n(D^n)=0$ gives $G_*=0$, i.e., $G$ has degree zero. Also your conclusion is wrong. By the preceeding argument, that a degree zero map has a fix point and point which is only reflected under that map, we have $G(x)=x$ and $G(y)=-y$ for some $x,y\in S^n$. This yields $F(x)=|F(x)|x|| x$ and $F(y)=-|F(y)|y || -y$ as desired.