Given two continuous function $f,g:S^1 \to S^1$ such that $f(x) \neq g(x) \forall x\in S^1$, I have to prove that $f$ and $g$ are homotopic.
So, I have to find a homotopy $$F: S^1 \times [0,1] \longrightarrow S^1 \\(x,0) \mapsto f(x) \\(x,1) \mapsto g(x)$$
Is it right to choose $$F: S^1 \times [0,1] \longrightarrow S^1 \\ (x,t) \mapsto \frac{tf(x)+(1-t)g(x)}{\vert\vert{tf(x)+(1-t)g(x)}\vert\vert} ?$$ In this case I don't understand why the two functions need to be different $\forall x$.
Have you any idea?
Thanks to the previous comments, the homotopy provided in the question is not correct because the segment in the denominator passes through zero.
So I could use the homotopy $$F: S^1 \times [0,1] \longrightarrow S^1 \\ (x,t) \mapsto \frac{f(x)(1+e^{i\pi t})-g(x)(e^{i\pi t}-1)}{\parallel f(x)(1+e^{i\pi t})-g(x)(e^{i\pi t}-1)\parallel}.$$