When I read the book "Algebra Topology-A First Course", I find a problem. It is on the Page 97, Exercise (16.15).
Problem We define $f$, $g\colon \mathbb{S}^{n}\to\mathbb{S}^{n}$ to be orthogonal at a point $x\in\mathbb{S^{n-1}}$, if the inner product $\langle f(x),g(x)\rangle =0$. Prove if $|\deg(f)|\not=|\deg(g)|$, then $f,g$ are orthogonal at some $x$.
The basic thought of mine is to make an orthogonal function of $f$, which we denote $f^{\perp}$. And then we just need to prove $f^{\perp}$ and $g$ have the fixed point or the antipode, which is to say $f^{\perp}(x)=g(x)$ or $f^{\perp}(x)=-g(x)$.
The first step is make an orthogonal function of $f$.
Lemma 1 $f$ have an orthogonal function $\Longleftrightarrow$ there exists a $g$ which has no fixed point and antipode with $f$. That is to say, $f$ and $g$ are linearly independent.
Proof By Gram-Schmidt Orthogonal Process.However, like the nowhere vanishing vector field is only on the odd dimensional sphere. We can find a linearly independent $g$ only on odd dimensional case.
The next step is prove it is true for odd dimensional case.
Lemma 2 $f,g$ have no fixed point $\Longrightarrow$ $f\simeq{ag}$ and $a$ is the antipodal map.
Proof Easy to Prove.So in odd dimensional case, if $f^{\perp}$ and $g$ has no fixed point, then we will have $\deg(f)=\deg(f^{\perp})=(-1)^{n+1}\deg(g)$. This is contradiction to the assumption.
So my question is how to prove the even dimensional case?
If $f$ and $g$ are nowhere orthogonal, we have that the function $\left<f,g\right>$ has fixed sign, suppose it is positive. If it is negative, you can compose $g$ with the antipodal map: the degree may only change by a sign, not affecting the hypothesis. Now, for $t\in [0,1]$ you have the homotopy between $f$ and $g$
$$\frac{(1-t)f+tg}{\left\|(1-t)f+tg\right\|}$$
The fact that $\left<f,g\right>>0$ ensures you that the homotopy is well defined (you may have some problems if $f(x)=-g(x)$ for some $x$, but then $\left<f(x),g(x)\right><0$), hence $\operatorname{deg}f=\operatorname{deg}g$, contradiction. If $n$ is odd, the antipodal map has degree $+1$, hence it's enough to know that $\operatorname{deg}f\neq\operatorname{deg}g$ to get a contradiction.