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Let's roll the sphere, keeping the contact point on a line of constant lattitude $\phi$. (Here $\phi \in (0, \pi)$, with $\pi/2$ meaning the equator.) The path $b$ is an arc of a circle. The arc has radius $\tan \phi$, and sweeps out the angle $2 \pi \cos \phi$. So the distance from one end of the arc to the other is $2 (\tan \phi) \sin\left( \pi \cos \phi \right) = 2 O(\phi) \sin (\pi - O(\phi^2)) = O(\phi^3)$. So we see that the displacement goes to $0$ as $O(\phi^3)$, as desired.

Let's roll the sphere, keeping the contact point on a line of constant latitude $\phi$. (Here $\phi \in (0, \pi)$, with $\pi/2$ meaning the equator.) The path $b$ is an arc of a circle. The arc has radius $\tan \phi$, and sweeps out the angle $2 \pi \cos \phi$. So the distance from one end of the arc to the other is $2 (\tan \phi) \sin\left( \pi \cos \phi \right) = 2 O(\phi) \sin (\pi - O(\phi^2)) = O(\phi^3)$. So we see that the displacement goes to $0$ as $O(\phi^3)$, as desired.

rolling a ball http://www.math.lsa.umich.edu/%7Espeyer/Closing.png

The fallacy http://www.math.lsa.umich.edu/%7Espeyer/Hairpins.png

False Proof: Identify my contractible path $\gamma$ with a square of side length $1$. Subdivide it into $N^2$ little squares $s_1$, $s_2$, ..., $s_{N^2}$. Let $p$ be a corner of the big square. Let $\delta_i$ be a path which goes from $p$ to $s_i$, circles $s_i$, and goes back to $p$. Let $\delta$ be the concatenation of the $\delta_i$'s. If you choose the ordering of the $s_i$'s correctly (boustrophedonically, to be specific), and choose the right $\delta_i$'s, then $\delta$ is simply $\gamma$ with a whole lot of backtracking put in.

False Proof: Identify my contractible path $\gamma$ with a square of side length $1$. Subdivide it into $N^2$ little squares $s_1$, $s_2$, ..., $s_{N^2}$. Let $p$ be a corner of the big square. Let $\delta_i$ be a path which goes from $p$ to $s_i$, circles $s_i$, and goes back to $p$. Let $\delta$ be the concatenation of the $\delta_i$'s. If you choose the ordering of the $s_i$'s correctly and choose the right $\delta_i$'s, then $\delta$ is simply $\gamma$ with a whole lot of backtracking put in.