# Reparameterisation of Curve as a Regular Curve (Topology)

There is a result that a curve or topological path can be reparameterized as a regular curve contained in the paper "Reparametrizations of continuous paths - Ulrich Fahrenberg and Martin Raussen" https://arxiv.org/pdf/0706.3560.pdf

For me the concepts are too advanced. Is there a simpler proof for this ?

The definitions that follow are taken from the paper (more or less).

A path is a continuous mapping $$p$$ from the closed unit interval $$I = [0, 1]$$ to a topological space $$X$$.
Excluding the case that the image of a path is a single point in X, then a path is regular if there is no closed interval $$[a, < b] \subset I$$ on which $$p$$ is constant.
A reparametrization $$\phi$$ is a non-decreasing surjective continuous map $$\phi: I \to I$$ with $$\phi(0) = 0; \phi(1) = 1$$.

Theorem: for any path $$p: I \to X$$ there is a regular path $$q: I \to X$$ and a reparametrization $$\phi$$ such that $$p = q \circ \phi$$.

It looks straightforward to prove this in the case that $$p$$ has a finite number of stop intervals (on which $$p$$ is constant) by cutting them out one by one and composing the corresponding $$\phi$$ functions. Clearly the number of stop intervals is countable, but how to deal with a countably infinite number of them ?

• Can you be explicit about what in the proof is too advanced for you? Without that, your potential answerers run the risk of writing something that is too advanced for you. – Lee Mosher Jun 6 at 15:57
• @LeeMosher. Thanks for the interest. I'm comfortable with basic point set topology, set theory and group theory, but terms like homotopy, compact open topology, category, group action and topological group are well outside my comfort zone. – Tom Collinge Jun 6 at 16:32

## 1 Answer

(*x) refer to footnotes added to the original answer...

You exclude the case that $$p : I \to X$$ is constant. Let $$\mathcal C$$ the set of all components (*a) of all preimages $$p^{-1}(x)$$ with $$x \in p(I)$$ and $$\mathcal S$$ the subset of all $$S \in \mathcal C$$ having more than than one point. Since $$S$$ is a connected subset of $$I$$, it is an interval. Since we require that $$S$$ has more than one point, it has length $$> 0$$ and we call it a stop interval of $$p$$. It may be an open, half-open or closed interval.

In the sequel we assume that all stop intervals are closed intervals. This is automatically satisfied if $$X$$ is a $$T_1$$-space (*b). Then all $$p^{-1}(x)$$ are closed in $$I$$ so that also all of its components are closed. If there exists a non-closed stop interval, some arguments below are no longer valid.

The set $$\mathcal S$$ is countable (*c) (either finite, including empty, or infinite). Let $$C = \bigcup_{S \in \mathcal S} S$$. Let $$\mathcal T$$ denote the set of components of $$I \setminus C$$. Each element of $$\mathcal T$$ is an interval - open, half-open or closed (which may be degenerate to point). (*d)

For each closed subinterval $$J = [c,d] \subset I$$ let $$\mathcal S_J = \{ S \cap J \mid S \in \mathcal S, S \cap J \ne \emptyset \} .$$ This is again a countable set of closed intervals. It is possible that $$S \cap J = \{c\},\{d\}$$ (degenerate intervals), but this does not matter. The number $$\lvert \mathcal S_J \rvert = \sum_{A \in \mathcal S_J} \lvert A \rvert$$ is well defined. Here $$\lvert A \rvert$$ denotes the length $$b - a$$ of the interval $$A = [a, b]$$. Obviously we always have $$\lvert \mathcal S_J \rvert \le \lvert J \rvert$$.

Define $$s : I \to I, s(t) = \lvert \mathcal S_{[0,t]} \rvert .(*e)$$ For $$t \le t'$$ we have $$s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert$$. To verify this, let $$S = [a,b] \in \mathcal S$$ such that $$S \cap [0,t'] \ne \emptyset$$. We have $$S \cap [0,t'] = (S \cap [0,t]) \cup (S \cap [t,t'])$$. If one these two intersections is empty, then the summand $$\lvert S \cap [0,t'] \rvert$$ of $$\lvert \mathcal S_{[0,t']} \rvert$$ occurs in exactly one of the sums $$\lvert \mathcal S_{[0,t]} \rvert$$, $$\lvert \mathcal S_{[0,t]} \rvert$$. If both intersections are nonempty, then $$\lvert S \cap [0,t'] \rvert = \lvert S \cap [0,t] \rvert + \lvert S \cap [t,t'] \rvert$$.

We conclude that

1. For $$t \le t'$$ we have $$s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert \le s(t) + (t' - t)$$.

2. $$s$$ is continuous because 1. implies $$\lvert s(t') - s(t) \rvert \le \lvert t' - t \rvert$$ for all $$t, t'$$.

3. The restriction of $$s$$ to any $$S = [a,b] \in \mathcal S$$ has the form $$s(t) = s(a) + \lvert \mathcal S_{[a,t]} \rvert = s(a) + \lvert [a,t]\rvert = s(a) + (t-a)$$.

4. The restriction of $$s$$ to any $$T \in \mathcal T$$ is constant (with some value $$c_T$$). To see this, let $$t,t' \in T$$ with $$t \le t'$$. Then $$s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert = s(t)$$ because $$[t,t'] \subset T \subset I \setminus C$$ does not intersect any $$S \in \mathcal S$$.

Define $$r(t) = t - s(t)$$. This is a continuous function such that $$r(t) \ge 0$$ and $$r(0) = 0$$. The function $$r$$ is non-decreasing because for $$t \le t'$$ we have $$r(t') - r(t) = t' - s(t') - (t - s(t)) = (t' - t) - (s(t') - s(t)) \ge 0$$ (see 1.). Since $$r(1) = 1 - s(1)$$, we regard $$r$$ as continuous non-decreasing surjection $$r : I \to [0,1-s(1)]$$.

Moreover, for $$t\in S = [a,b] \in \mathcal S$$ we have $$r(t) = t - s(a) - (t-a) = a - s(a)$$ and for $$t \in T \in \mathcal T$$ we have $$r(t) = t - c_T$$. This means that the stop intervals of $$r$$ are the same as those of $$p$$. Hence $$p = p' \circ r$$ with a unique function $$p' : [0,1-s(1)] \to X$$. Because $$I$$ is compact, $$r$$ is a quotient map and $$p'$$ is continuous (*f, *g, *h). Stretching $$[0,1-s(1)]$$ to $$I$$ yields the desired result. Note that $$s(1) )= 1$$ is impossible because in that case $$p$$ would be constant.

Let us finally see where the argument breaks down if there exists a non-closed stop interval $$S$$. Then $$S \subsetneqq \overline{S} = [a, b]$$ and 3. holds on $$\overline{S}$$. Hence $$r$$ is constant on $$\overline{S}$$ (this follows also from continuity). Thus $$r$$ and $$p$$ do not have the same stop intervals. As an example consider any non-constant map $$f : I \to \{0,1\}$$, where $$\{0,1\}$$ has the trivial topology, such that $$p(x) = 0$$ for $$x < 1/2$$ and $$p(x) = 1$$ for $$x \ge 1/2$$. Then $$p$$ has stop intervals $$[0,1/2)$$ and $$[1/2,1]$$, but $$r$$ has stop interval $$I$$.

Remark:

One could also use the definition $$s(t) = \int_0^t \chi_C(x)dx$$ where $$\chi_C$$ is the characteristic function of the subset $$C \subset I$$ (i.e. $$\chi_C(x) = 1$$ for $$x \in C$$, $$\chi_C(x) = 0$$ for $$x \notin C$$). However, a proof that $$\chi_C$$ is integrable is needed..

Clarifications and references ...

(*a) implicit in the definition of "components" is that they are connected.
(*b) A space is $$T_1$$ if and only iff every singleton point set is closed.
(*c) $$\mathcal S$$ must be countable in order for the sum of positive lengths of $$S \in \mathcal S$$ to be finite.
(*d) Although each $$S$$ is closed the (countably) infinite union need not be closed - hence the possible different type of interval in $$\mathcal T$$.
(*e) $$s(t)$$ is then the total "stopped length" up to the point $$t$$.
(*f) Munkres - Topology, p.135 definition of quotient map: a continuous surjective closed map is a quotient map.
(*g) https://math.stackexchange.com/q/548598 - continuous map from compact space to Hausdorff is closed.
(*h) Munkres - Topology, p.142 Theorem 22.2 - commutivity diagram: existence and continuity of $$p'$$

• Thanks, that's really helpful. I took a while to get to grips with some of the points and added my own notes (and corrected a couple of typos) in an edit - too long for a comment, hope you don't mind. I'll wait 'til later to accept in case you want to change anything. – Tom Collinge Aug 9 at 12:45
• I added a remark concerning $s$. This was my final edit. – Paul Frost Aug 9 at 14:39
• I think $C$ is Borel measurable being the countable union of closed intervals ? – Tom Collinge Aug 9 at 15:15
• Ah, you are right if we use the Lebesgue integral which is absolutely legitimate. But I think the characteristic function is also Riemann integrable. – Paul Frost Aug 9 at 22:02
• I think in this case it would be. Since $C$ is the countable union of closed intervals, the discontinuities of its characteristic function should be countable, therefore Lebesgue measure zero, therefore $\chi_C(x)$ would be Riemann integrable. – Tom Collinge Aug 10 at 6:57