Showing that 2 curves are not equivalent.

Question

I want to verify that the 2 following curves are not equivalent:

$c_1: [0, 2\pi) \rightarrow \mathbb{R^2}$ and $c_2: [0, 2\pi) \rightarrow \mathbb{R^2}$. Where $c_1(t) = \Big( \cos(t), \sin(t) \Big)$ and $c_2(t) =\Big( \cos(t + \pi), \sin(t + \pi) \Big) $

Here is the solution : Let's suppose that $c_1$ ~ $c_2$

Then $\exists$ a diffeomorhism $\phi $ s.t $c_2 = c_1 \circ \phi$

We have $\Big( -1, 0 \Big) = c_2 (0) = (c_1 \circ \phi)(0) = \Big( \cos ( \phi (0)), \sin(\phi(0)) \Big)$

so $\phi(0) = \pi$ because $\phi \in [0, 2\pi)$

Since $\phi$ is injective and continuous, it is monotone.

• If it's increasing the $0$ is not in the image of $\phi$

• If it's decreasing then $3\pi/2$ is not in the image of $\phi$

$\phi$ is not bijective $\implies \phi $ is not a diffeomorphism $\implies$ $c_1$ and $c_2$ are not equivalent.

I don't understand why If it's increasing the $0$ is not in the image of $\phi$ and why If it's decreasing then $3\pi/2$ is not in the image of $\phi$.

Answer 1

$\phi$ increasing and injective implies $\phi(t) > \phi(0)$ for all $t >0$. You have $\phi(0) = \pi$ and so if $\phi(t) = 0$ then $t<0$ (contradiction since $t \in [0,2\pi)$), otherwise $t>0$ and $\phi(t)<\phi(0)$. Apply this same logic to the other condition.