TITLE: Pursuit-Evasion on a Manifold
QUESTION [6 upvotes]: 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? (By diameter here 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!

REPLY [2 votes]: I don't have enough reputation to comment. So I apologize that I write this as an answer. I don't understand the answer of Anton Petrunin. According to it, two pursuers are enough to capture an evader in the sphere. But as Barry Cipra said in the comments, actually the opposite is true: the evader is always able to escape (just consider the great circle through the two pursuers and the hemisphere defined by it containing the evader. The evader just have to go so close to the center of this hemisphere as possible. The pursuers won't be able to reach him.)
On the other hand, I actually like to think of two antipodal points in the sphere as a convex subset (convex meaning that any two points in the subset at distance $<\pi$ are connected by a geodesic). Thus the analogy with the cited paper won't apply (one needs three or more pursuers).