Showing that two maps of the sphere are homotopic if their values are never antipodal

Question

Let $X$ be a topological space and let $S^2 \subset \mathbb{R^3}$ be the unit sphere with the metric $d$ inherited from $\mathbb{R^3}$. Show that if $f,g:X\to S^2$ are continuous maps such that $d(f(x),g(x))<2$ for all $x\in X$, then $f$ and $g$ are homotopic.

I am not sure if the following is rigorous enough.

My try

$d(f(x),g(x))<2$ for all $x\in X$, implies that $f(x)$ and $g(x)$ are not antipodal for all $x \in X$, hence there is a unique shortest path (geodesic) between them, denote this path by $C_x$. $C_x$ can be thought of as an arc with central angle $\theta_x<\pi$.

for each $x \in X$, there is a copy of complex plane $\mathbb{C_x}$ which contains the arc $C_x$, hence we can define a map $H:X\times I \to S^2$ by $$H(x,t)=f(x)e^{it\theta_x}$$ If in $\mathbb{C_x}$ we can join $f(x)$ to $g(x)$ by traversing the angle $\theta_x$ in the counter clock wise direction along $C_x$, otherwise define $$H(x,t)=f(x)e^{-it\theta_x}$$

$H$ is the required homotopy between $f(x)$ and $g(x)$.

Is this an acceptable proof ? is there an easier way to show this ?

Answer 1

Your answer is correct, but there's a quicker and more rigorous way to write it such that it is obviously continuous. Due to the fact that this rests in $\mathbb{R}^3$ there is a quicker way. In $\mathbb{R}^3$ the "obvious" homotopy is the one by convex combinations:

$$H_1: X \times I \to S^2$$

$$H_1(x,t) = tf(x) + (1-t)g(x)$$

Now we clearly can't just do this, as it rarely lies in the sphere. But our condition tells us the points $f(x),g(x)$ for the same $x$ are never antipodal. Thus the line $H_1(x,-)$ doesn't pass through the origin. So we simply project onto the sphere. $H(x,t) = H_1(x,t)/\|H_1(x,t)\|$.

Answer 2

Your argument looks good to me. Here's an alternative approach (essentially changing only the speed at which points move during the homotopy), which you might or might not regard as simpler. First, work in the solid ball $B^3$ instead of its boundary $S^2$. Here you can move $f(x)$ to $g(x)$ at constant speed along a straight line segment through the solid ball. (This is just the usual argument for convex sets being homotopically trivial.) In your case, since $f(x)$ and $g(x)$ are never antipodal, these line segments don't go through the center $O$ of the ball. So you actually have a homotopy in $B^3-\{O\}$. But this punctured ball can be retracted to $S^2$ by pushing everything radially outward from $O$. That sends your homotopy to a homotopy in $S^2$.