6

Suppose $X$ is a Hausdorff topological space (or metric space if you like) and $f:[0,1]\to X$ is any non-constant path. It could be that $f$ is constant on a closed interval $[a,b]$, and it is possible to collapse this interval to a point in $[0,1]$ and get a path (by the universal property of quotient spaces) which has "fewer" intervals where it is constant.

However, it's possible that $f$ is constant on infinitely many closed intervals. I'd like to be able to replace $f$ with a path $f'$ that is not constant on any interval.

Is there always a monotone and onto function $m:[0,1]\to [0,1]$ and path $f':[0,1]\to X$ which is not constant on any interval such that $f'\circ m=f$? It seems like the obvious thing to do would be to collapse all intervals on which $f$ is constant to points, but it is not obvious to me that the resulting quotient is always $[0,1]$. Is there some result in the theory of Peano spaces which makes this really easy?

J.K.T.
  • 1,536
  • Under some conditions on the ambient space, there should be a notion of arc length, by which you can reparametrize the entire path (without trying to fix one interval at a time). – Greg Martin Apr 11 '14 at 20:18
  • I seem to have posted the same question and got a good answer here https://math.stackexchange.com/q/3317511 which formalizes your intuition for the solution. – Tom Collinge Aug 12 '19 at 10:35

1 Answers1

1

The Bing Shrinking Criterion is what you want. Your case is a particularly easy one and has its own proof: for any quotient map $[0,1] \to J$, if each decomposition element is a point or a closed interval then $J$ is homeomorphic to $[0,1]$.

Lee Mosher
  • 120,280