My question is similar to another on this site: Two maps are homotopic if never antipodal
Let $f:S^n\rightarrow S^n$ be a continuous map such that $f(x)\neq -f(-x)$ for all $x$, I want to define a homotopy from $f(x)$ to a map $g(x)$ satisfying $g(x)=g(-x)$ for all $x$. By the assumption, there must exist a unique geodesic $\gamma_x:[0,1]\rightarrow S^n$ such that $\gamma_x(0)=f(-x)$ and $\gamma_x(1)=f(x)$ . I will then define a 'homotopy' $H:S^n\times [0,1] \rightarrow S^n$ by:
$H(x;t):=\gamma_x(t)$
The answers on the aforementioned thread agrees that this should be a homotopy, however does not deal with the proof. I wish to prove it rigorously. Since projecting onto the second coordinate give the geodesic which is continuous, it only remains to show that projecting onto the first coordinate is continuous. i.e:
$x\mapsto \gamma_x(t_0)$ is a continuous map for all $t_0$. Is there an elegant method which is escaping me in showing this?
And one other question, if we assume that $f$ is a smooth map (in terms of smooth manifolds), will $x\mapsto \gamma_x(t_0)$ be smooth as well for all $t_0\in [0,1]$?
If you want to do it rigorously, you have to make precise what you understand by a "geodesic" $\gamma$. On the sphere it is an embedding $\gamma : I \to S^n$ whose image is contained in a great circle, but do you accept any parametrization or do you want that the speed is constant? Note that you cannot expect that the speed is constant $1$ if $\gamma$ lives on $I = [0,1]$.
In my opinon there is no use to require constant speed in your question, it would only make things more technical.
If you accept this, then you see JHance's answer to Showing that two maps of the sphere are homotopic if their values are never antipodal yields a geodesic $H(x,-) : I \to S^n$ from $f(x)$ to $g(x)$(not only for $n=2$, but for all $n$). If $f$ is smooth, then so is $H$.