# 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.

• Hint: For the usual metrics on the sphere, $P$ and $P'$ are antipodal iff the distance between them is ... Commented Mar 6, 2018 at 7:06
• double the radius? Commented Mar 6, 2018 at 7:08
• Completing Robert Israel's comment: ... the maximum allowed. In other terms, if $a,b$ are the endpoints of a diameter of $X$ and $\varphi$ is an isometry on $X$, $\varphi(a),\varphi(b)$ are the endpoints of a diameter of $\varphi(X)$. Commented Mar 6, 2018 at 13:24

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 $\overline m(P)=\overline m(-P)$.
• theres a second part to this question where I need to show that $\bar mf=f\bar m$ for any isometry. Can you help me understand what it's even asking by reversing the order? Commented Mar 6, 2018 at 7:20