Parametrisation of curve

Question

If there exist $f: [a,b]\rightarrow R^2$ 1-1 C^1 map for some $a<b$.

And if there exist $g:[c,d]\rightarrow Im(f)$ C^1 map, will $g$ be 1-1 map?

such that $g(c) = f(a)$ and $g(d) = f(b)$

Answer 1

Not necessarily.

Example. Take any one-to-one $C^1$-function $f:[-1,1]\to\Bbb R^2$. Then the $C^1$-function

$$g:[0,3\pi]\to\Bbb R^2,\quad g(t)= f(-\cos t)$$

starts at $g(0)=f(-1)$ and ends at $g(3\pi)=f(1)$ but is not one-to-one (because it is not injective).

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I think the problem gets much harder when we claim that $g$ is injective (no self-intersection) and we only have to show that it is surjective (onto).

Answer 2

let $f:[0,1] \to \mathbb R^2$ be defined by $f(t)=(t,t)$. Then $f$ is $C^1$ and $1-1$ .

Now consider $g:[-1,1] \to Im(f)$ , defined by $g(t)=(t^2,t^2)$. $g$ is $C^1$ but not $1-1$ .