No, this is not always possible.
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Let $M$ be the manifold shown above, i.e. an open square subset of $R^2$, with three triangles removed, four parallelograms appended, and three identifications made.
We can realize this to have the following geodesics:
- A B C
- D E F
- G H I
- A D G
- B E H
- C F I
- C E G
- A I E
- B G F
Then any two of these geodesics intersect at most once, in the indicated points. Furthermore, by making the identified pieces sufficiently narrow, and perhaps by cutting away other pieces, we can ensure that there is only one geodesic between each of these pairs of points. Finally, by exhaustive search, there is no way to order these nine points that makes the system acyclic.
By contrast, any proper subsystem of these geodesics does have an ordering that makes the system acyclic, and in that sense this is a minimal counterexample.
Update: A simple proof of the central combinatorial fact.
Assume wlog that G < E. Since BGF, we have either
- (1) B < G < E or (2) F < G < E
In case 1, by BEH, GHI, AIE we have: B < G < E < H < I < A
But then ABC and CEG would put C in contradictory places.
In case 2, by CEG, CFI, AIE we have: A < I < F < G < E < C
But then ADG and DEF would put D in contradictory places.
In either case, no ordering can satisfy all the constraints.
