N-dimensional sphere and degree problem Let $f,g:\mathbb{S}^{n}\rightarrow\mathbb{S}^{n}$ be two differential mappings such that $\langle f(x),g(x)\rangle\neq 0$ for all $x\in\mathbb{S}^{n}$. Prove that $deg(f) = \pm deg(g)$. Here it is worth mentioning that $\mathbb{S}^{n} = \{\textbf{x}\in\mathbb{R}^{n+1}; |\textbf{x}| = 1\}$. Moreover, the notation $deg(f)$ denotes the degree of the application $f$. I've tried to solve it but did not succeed so far. Thank you in advance.
 A: I take $S^n$ to be the unit sphere in $\Bbb R^{n + 1}$.  Thus
$\vert y \vert = 1 \tag 0$
for all $y \in S^n$.
Since for all $x \in S^n$, 
$\langle f(x), g(x) \rangle \ne 0, \tag 1$
we can assume that $\langle f(x), g(x) \rangle$ does not change sign on $S^n$; so suppose we assume to begin that
$\langle f(x), g(x) \rangle > 0. \tag 2$
For $t \in [0. 1]$, we define
$L(x, t) = (1 - t) f(x) + t g(x); \tag 3$
then
$\vert L(x, t) \vert^2 = \langle L(x, t), L(x, t) \rangle = \langle (1 - t) f(x) + t g(x), (1 - t) f(x) + t g(x) \rangle$
$= (1 - t)^2 \langle f(x), f(x) \rangle + 2t(1 - t) \langle f(x), g(x) \rangle + t^2 \langle g(x), g(x) \rangle$
$= (1 - t)^2 \vert f(x) \vert^2 + 2t(1 - t)\langle f(x), g(x) \rangle + t^2 \vert g(x) \vert^2 ; \tag 4$
by virtue of (0) and (2), we thus have
$\vert L(x, t) \vert^2 = (1 - t)^2 + 2t(1 - t) \langle f(x), g(x) \rangle + t^2 \ge (1 - t)^2 + t^2 = 2t^2 - 2t + 1; \tag 5$
since
$\dfrac{d(2t^2 - 2t + 1)}{dt} = 4t - 2, \tag 6$
which is zero for
$t = \dfrac{1}{2}, \tag 7$
$2t^2 - 2t + 1$ takes its minimum value of $1/2$ at $t = 1/2$; thus we have that
$\vert L(x, t) \vert^2 \ge \dfrac{1}{2}, \tag 8$
so
$\vert L(x, t) \vert \ge \dfrac{1}{\sqrt 2}; \tag 9$
in the light of (9) we see that
$H(x, t) = \dfrac{L(x, t)}{\vert L(x, t) \vert} \tag{10}$
is well-defined for all $x \in S^n$ and $t \in [0, 1]$; also
$\vert H(x, t) \vert = \dfrac{\vert L(x, t) \vert}{\vert L(x, t) \vert} = 1, \tag{11}$
$H(x, 0) = f(x), \tag{12}$
$H(x, 1) = g(x); \tag{13}$
we see via (11)-(13) that $H(x, t)$ is in fact a continuous homotopy 'twixt
$f(x)$ and $g(x)$ in $S^n$; thus 
$\deg f = \deg g. \tag{14}$
In the event that 
$\langle f(x), g(x) \rangle < 0, \tag{15}$
we set
$L(x, t) = (1 - t)(-f(x)) + tg(x); \tag{16}$
then in a manner similar to the above we find once again that
$\vert L(x, t) \vert^2 = (1 - t)^2 -2t(1 - t)\langle f(x), g(x) \rangle + t^2 \ge 2t^2 - 2t + 1; \tag{17}$
again we have
$\vert L(x, t) \vert \ge \dfrac{1}{\sqrt 2}; \tag{18}$
so again we may set
$H(x, t) = \dfrac{L(x, t)}{\vert L(x, t) \vert} \tag{19}$
and now we obtain a homotopy 'twixt $-f(x)$ and $g(x)$; thus
$\deg g = - \deg f. \tag{20}$
As long as (1) binds, we have
$\deg g = \pm \deg f, \tag{21}$
which was to be demonstrated.
A: I think Lewis's answer has a small mistake, taking $f(x)=x,\ g(x)=-x$ i.e. Identity map and an antipodal map on $S^n$. Now we have $(f(x),g(x))=(x,-x)\equiv -1\ on\ S^n$. However we know that: $\deg (f)=1,\ \deg(g)=(-1)^{n+1}$. It seems to be an error to say $\deg f=-\deg g$, due to it should be dependent on dimision.
Now check out that from the homotopic Lewis constructed above, we obtain $$
\deg (-f)=\deg g,\ \forall (f,g)<0\ on\ S^n$$
Denote the antipodal map as $\alpha:S^n\rightarrow S^n,\ x\mapsto -x$, then we have:$$
\deg(-f)=\deg(\alpha\circ f)=\deg\alpha\cdot\deg f=(-1)^{n+1}\deg f
$$
Hence the result maintain the same but a little bit different when $(f,g)<0\ on\ S^n$.
