# Showing that 2 paths are not equivalent.

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?

• Why is $\phi(0) = 0$ a contradiction? – Dayton Apr 18 at 9:21
• @Dayton21 we will have that $\phi (0) = \phi (2 \pi)$ =0, contracting that $\phi$ is an increasing function. – Intuition Apr 18 at 9:24
• Yes exactly, but the statement "$\phi(0) = 0$ or $2\pi$" does not contradict anything. However $\phi(0) = \phi(2\pi)$ does – Dayton Apr 18 at 9:28
• I am sorry .... I will correct it ..... but my argument is correct? @Dayton21 – Intuition Apr 18 at 9:29
• 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 $\phi(0) = \phi(2\pi)$? – 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.

• So is my argument above correct or no? – Intuition Apr 18 at 9:36
• math.stackexchange.com/questions/3192108/… Could you look at this if you have time please? – Intuition Apr 18 at 9:36
• How do you arrive at the identity in you last line? – Intuition Apr 18 at 10:13
• 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? – 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.

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