I know pursuit-evasion has been studied in many contexts, including on a manifold (e.g., Melikyan, "Geometry of Pursuit-Evasion Games on Two-Dimensional Manifolds"), but I have not seen this version:

There is a pursuer, a point $p_t$ at time $t$, while the evader is at $v_t$. They start at $p_0$ and $v_0$. For each $t \in \mathbb{N}$, the evader jumps from $v_t$ to $v_{t+1}$, and then the pursuer jumps from $p_t$ to $p_{t+1}$. Both jumps are at most a unit distance, measured by shortest path on the manifold $\cal{M}$. Both pursuer and evader know the other's location at all times. The evader is captured if $p_t = v_t$ at some time $t$.

Let $\cal{M}$ be a sphere. Then the pursuer might fail to capture the evader, who can just run away on a great circle. But what if there are two pursuers, $p_t$ and $q_t$? Specifically:

Q1. For $\cal{M}$ a sphere, can an evader always be captured by two pursuers initially at the north and south poles?

There is a general principle for $\cal{M}=\mathbb{R}^n$:
it suffices for the evader to be inside the convex hull of
the pursuers (e.g.,
Kopparty, Ravishankar, "A framework for pursuit-evasion games in $R^n$,"
*Information Proc. Lett.*, 2005).
It seems a natural extension that two antipodal pursuers on a sphere
suffice.

Q2. For $\cal{M}$ a surface homeomorphic to a sphere embedded in $\mathbb{R^3}$, with metric inherited from the Euclidean metric in $\mathbb{R^3}$, will placing $p_0$ and $q_0$ at points realizing the diameter of $\cal{M}$ suffice to capture any evader? (Bydiameterhere I mean that the shortest path from $p_0$ to $q_0$ on the surface is maximum over all pairs of points on $\cal{M}$.)

This is much less clear to me. A counterexample would be interesting.
But if the answer to Q2 is **Yes**,
I would be interested to learn if there are closed,
bounded surfaces of higher genus
for which just two well-placed pursuers suffice.

The literature on pursuit-evasion is vast, and I suspect the answers to my questions are known. Thanks for suggestions or pointers!

thenthe pursuers jump from $p_t,q_t$ to $p_{t+1},q_{t+1}$? Otherwise randomness on the part of the evader would (with probability $1$) prevent $p_t=v_t$ and $q_t=v_t.$ – Aaron Meyerowitz May 22 '12 at 9:12