prove maps $-f$ and $g$ are homotopic.

Question

Let $f,g:X \rightarrow S^2$ be continuous, prove that if $f(x) \neq g(x)$ for each $x \in X$, then $g$ is homotopic to the map $X \rightarrow S^2:x \mapsto -f(x)$.

So I am trying to find a homotopy between $f$ and $g$, I tried $$\frac{(1-t)g - tf}{\vert\vert (1-t)g - tf \vert \vert}$$ and thought $f(x) \neq g(x)$ as a condition to make sure the bottom doesn't equal to 0, but not quite. Help?

Answer 1

Your map is correct.

The points $f(x)$ and $-g(x)$ cannot be antipodal because $f(x) \neq g(x)$, so the line through $f(x)$ and $-g(x)$ does not cross zero. The set $\{(1 - t)g(x) - tf(x) \ | \ t \in \mathbb{R} \} = \{g(x) - t(f(x) - -g(x)) \ | \ t \in \mathbb{R}\}$ is precisely that line, so it cannot be zero for any $t$.