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 ?
Best Answer
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. $H(x,t) = H_1(x,t)/\|H_1(x,t)\|$.