Lift of isometries of spherical space forms

Question

If we have an isometry between two spherical space forms, then it is said that it lifts to an isometry of the sphere. Why is that?

Answer 1

I'll be assuming your spheres are at least two dimensional. (If they are one dimensional, it's tautologically true because sphereical space forms of 1 dimension are just circles themselves).

Call the two spherical space forms $X$ and $Y$. Then we have universal covering maps (which are local isometries) $\pi_1:S^n\rightarrow X$ and $\pi_2:S^n\rightarrow Y$.

Suppose $f:X\rightarrow Y$ is an isometry. Consider the map $S^n\rightarrow Y$ given by $f\circ \pi_1$. Since $S^n$ is simply connected, this induces the $0$ map on fundamental groups. By the universal lifting property of covering maps, there is a function $g:S^n\rightarrow S^n$ with $f\circ \pi_1 = \pi_2 \circ g$. I claim that $g$ is an isometry.

First, to see $g$ is a diffeomorphism, use the same argument on $f^{-1} \circ \pi_2$ to get a map $h:S^n\rightarrow S^n$. By fiddling with basepoints a bit, one can then show that $h = g^{-1}$. Thus, $g$ is a diffeomorphism.

Now, since $f$, $\pi_1$, and $\pi_2$ are all local isometries, on any neighborhood where $\pi_2$ is $1-1$, we have $g^\ast = \pi_1^\ast f^\ast (\pi_2^{\ast})^{-1}$, so is a composition of metric preserving maps, so it preserves the metric as well.