Let $f: S^{n} \rightarrow S^{n}$ and $g: S^{n} \rightarrow S^{n}$ be two continuous map on $S^{n}$. Suppose that, for all $x \in S^{n}$, we have $$f(x)\cdot g(x)=0\,.$$ Then show that $f \simeq g$.
For the definition of two homotopic functions, see here.
For information, the operator $\cdot$ is the usual dot product in $\mathbb{R}^{n+1}$. Also $S^n\subseteq\mathbb{R}^{n+1}$ is the unit sphere with center at the origin.
For $n=1$, the claim is not difficult to prove. If $f(x)\cdot g(x)=0$, then $f(x)\perp g(x)$. Treating $\mathbb{R}^2$ as $\mathbb{C}$. Therefore, $f(x)=e^{\pm i\pi/2} g(x)$. Because $f$ and $g$ are continuous, we must have $$f(x)=e^{si\pi/2}g(x)$$ for all $x\in S^1$, where $s=\pm 1$ is fixed. In this case, we have a homotopy: $$H(x,t)=e^{si\pi t/2}g(x)$$ where $t\in[0,1]$ and $x\in S^1$. Clearly, $H(x,0)=g(x)$ and $H(x,1)=f(x)$ for all $x\in S^1$.
How do we prove the statement for a general $n$? Thanks in advance for sharing your knowledge.
Presumably, the operator $\cdot$ is the standard inner product on $\mathbb{R}^{n+1}$, and $$S^n:=\big\{x\in\mathbb{R}^{n+1}\,\big|\,\|x\|=1\big\}\,,$$ where $\|\_\|$ is the norm induced by the inner product $\cdot$. Let $$h(t,x):=(1-t)\,f(x)+t\,g(x)\,,$$ where $t\in[0,1]$ and $x\in S^n$. Because $f(x)\cdot g(x)=0$ for all $x\in S^n$, we conclude that $h$ is a nonvanishing continuous map from $[0,1]\times S^n$ to $\mathbb{R}^{n+1}$. Define $$H(t,x):=\frac{h(t,x)}{\big\|h(t,x)\big\|}$$ for all $t\in[0,1]$ and $x\in S^n$. Then, $H:[0,1]\times S^n\to S^n$ is a continuous map such that $H(0,\_)=f$ and $H(1,\_)=g$. Consequently, $H$ is a homotopy from $f$ to $g$.
Remark. It suffices to know that $f(x)$ and $g(x)$ are linearly independent for every $x\in S^n$. If this assumption is met, then $h(t,x)\neq 0$ for any $x\in S^n$ and $t\in[0,1]$. Therefore, the same argument works.