Show that antipodal points remain antipodal under any isometry of $S^2$ "The antipodal map of $S^2$ is the isometry $\bar m$ sending each point $P$ to its opposite point $-P$" (from textbook) Show that antipodal points remain antipodal under any isometry of $S^2$. In other words, for each isometry f of $S^2$, if $P$ and $P'$ are antipodal, then so are $f(P)$ and $f(P')$ I am unsure how I can define the $P$ and $P'$ so that they are opposites of each other in the antipodal map, and am I supposed to apply the traditional isometries (translation, reflection and rotation) to the points? I don't understand this problem.

If $P\in S^2$, $-P$ is the only point $P'$ of $S^2$ such that $d(P,P')=2$. So, since isometries preserve distance and since $d(P,-P)=2$, $d\left(\overline m(P),\overline m(-P)\right)=2$ and therefore $\overline m(P)=\overline m(-P)$.