What I want to prove: Let $f,g:S^n\to S^n$ be continuous. If there is no point $x\in S^n$ with $f(x)=-g(x)$ then $f\sim g$, then .
I am using this as a lemma to prove a slightly bigger result: Namely, For $n\in \mathbb{N}$, if $f:S^n\to S^n$ has no fixed points, then $f\sim T(x)=-x$ for all $x\in S^n$.
If you believe the above result, then it can be proven as follows:
Suppose there is no fixed point, and for the purposes of contradiction, $f\nsim T(x)$. Then $\exists x\in S^n$ such that $f(x)=-T(x)=x$ which means $x$ is a fixed point $\rightarrow \leftarrow$. Therefore, $f\sim T(x)$.
The problem is I am not even sure if the lemma is true. I believe it geometrically for $S^2$ but I don't know how to argue without using the language of analysis--I want to prove this without using a metric.
My attempt: I want to argue indirectly-- that is, suppose $f\sim g$ and for some $x\in S^n$, $f(x)=-g(x)$. In $\mathbb{R}^{n+1}$, if $f\sim g$, then for $t\in [0,1]$, $h(t)=tf(x)+(t-1)g(x)$. This is indeed a homotopty. If for some $x$, $f(x)=-g(x)$, then $-tg(x)+tg(x)-g(x)=-g(x)$ which is certianly not $g(x)$ at time $t$. Am I cheating when I use the linear homotopy as it goes outside of $S^n$?