The definition of equivalent paths is as follows :

Two paths $f: [a,b] \rightarrow \mathbb{R^n} $ and $g: [c,d] \rightarrow \mathbb{R^n} $ are equivalent if there exist a $C^{1}$ bijection $\phi: [a,b]\rightarrow [c,d]$ such that $\phi'(t) > 0 $ for all $t \in [a,b]$ and $f = g \circ \phi.$

And I was asked to show that those 2 paths:

$f(t): [0, 2\pi] \rightarrow \mathbb{R^2}$ and $h(t): [0, 4\pi] \rightarrow \mathbb{R^2}$. Where $f(t) = \Big( \cos(t), \sin(t) \Big)$ and $h(t) =\Big( \cos(t), \sin(t) \Big) $

Are not equivalent.

My attempt :

Assume towards contradiction that they are equivalent, then there is a $C^{1}$ bijection $\phi : [0, 2 \pi] \rightarrow [0, 4 \pi] $, such that, $\phi'(t) > 0, \forall t \in [0, 2\pi]$ and $ f = h \circ \phi.$

Which means that $(0,1) = f(0) = (h \circ \phi)(0) = h(\phi(0))= (\cos(\phi (0)), \sin (\phi (0)))$,

which means that $\phi(0) = \phi (2 \pi) = 0 $, contradicting that $\phi $ is an increasing function. hence our first assumption was wrong.

Is my argument correct?

  • $\begingroup$ Why is $\phi(0) = 0$ a contradiction? $\endgroup$ – Dayton Apr 18 at 9:21
  • $\begingroup$ @Dayton21 we will have that $\phi (0) = \phi (2 \pi)$ =0, contracting that $\phi$ is an increasing function. $\endgroup$ – Intuition Apr 18 at 9:24
  • $\begingroup$ Yes exactly, but the statement "$\phi(0) = 0$ or $2\pi$" does not contradict anything. However $\phi(0) = \phi(2\pi)$ does $\endgroup$ – Dayton Apr 18 at 9:28
  • $\begingroup$ I am sorry .... I will correct it ..... but my argument is correct? @Dayton21 $\endgroup$ – Intuition Apr 18 at 9:29
  • 1
    $\begingroup$ yeah but the options are $\phi(0) \in \{0,2\pi, 4\pi\}$ and $\phi(2\pi) \in \{0, 2\pi,4\pi \}$ so why is it necessary that $\phi(0) = \phi(2\pi)$? $\endgroup$ – Dayton Apr 18 at 9:43

Suppose that there there is a $C^1$ bijection $ϕ:[0,2π]→[0,4π]$, such that, $ϕ′(t)>0,∀t∈[0,2π]$ and $f=h∘ϕ.$

Then we have

$ \cos(t)= \cos( \phi(t))$ and $ \sin(t)= \sin( \phi(t))$ for all $t∈[0,2π]$. We take derivatives and get

$\cos(t)= \cos(\phi(t)) \phi'(t)$ , thus $ \cos(t)=\cos(t) \phi'(t)$ for all $t∈[0,2π]$.

We conclude that $\phi'(t)=1$ for all $t∈[0,2π]$. Hence $\phi(t)=t+c$. Since $\phi(0)=0$. we have $c=0.$

But then we derive $ 4 \pi= \phi( 2 \pi)=2 \pi$, a contradiction.

  • $\begingroup$ So is my argument above correct or no? $\endgroup$ – Intuition Apr 18 at 9:36
  • $\begingroup$ math.stackexchange.com/questions/3192108/… Could you look at this if you have time please? $\endgroup$ – Intuition Apr 18 at 9:36
  • $\begingroup$ How do you arrive at the identity in you last line? $\endgroup$ – Intuition Apr 18 at 10:13
  • $\begingroup$ it sems like you divided by 0 -because $\pi/2$ is included in our interval- in your first conclusion in the second line from below...... or what did you do? $\endgroup$ – Intuition Apr 18 at 10:33

(Somewhat) alternative hint: If $f,h$ are equivalent, then for each $v\in \Bbb R^n$, the sets $f^{-1}(v)$ and $h^{-1}(v)$ have equal cardinality. $f$ assumes all values once or twice, $h$ does it twice or three times.

  • $\begingroup$ yes I know this argument but I wanted to solve it through the definition given. $\endgroup$ – Intuition Apr 18 at 9:31
  • $\begingroup$ Could you please look at this question if you have math.stackexchange.com/questions/3192108/… ? $\endgroup$ – Intuition Apr 18 at 9:32
  • $\begingroup$ why is $f$ and $g$ invertible ? $\endgroup$ – Intuition Apr 18 at 9:43
  • $\begingroup$ or you mean the inverse image? $\endgroup$ – Intuition Apr 18 at 10:13

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