Two paths f1:[a1,b1] and f2:[a2,b2] are called equivalent if there exists a strictly isotone surjective function g:[a1,b1]-->[a2,b2] s.t. f1=f2 composed g.
Prove that this relation is indeed an equivalence relation on the set of all curves in a fixed metric space X.
Have to show it's reflexive, symmetric, and transitive.
I'm guessing since you have a question about topology that you possess the techniques to work out the mathematical details I am going to omit.
Here's a crude outline showing how this can be done:
(let C1, C2 and C3 be curves in X)
Obviously a path is equivalent to itself (let g be the identity function), so the relation is reflexive.
Suppose C1 related to C2 by some mapping as described. The mapping is surjective so we know every point C2 is going to point back somewhere in C1. Since the map is strictly isotone, then no two points in C2 can point back to the same point in C1. So the inverse of the original map is strictly isotone and surjective, and the relation is therefore symmetric.
Show transitivity using almost exactly the same logic as above. (Suppose C1 relates to C2 and C2 relates to C3, and show that C1 then relates to C3).
Do that and you're done!
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