Are all injective arcs the images of an injective interval? Given a continous $f:[0,1] \to \mathbb{R}^2$ call it an injective arc if:


*

*$f(0) \neq f(1)$

*$f(a) = f(b) =c $ for $a<b$ implies that $f([a,b]) = {c}$.
Given such $f$ we also have an ordering on $Im(f)$ by taking the one of $[0,1]$.
Is it true that for any such $f$ there is an injective continous $g:[0,1] \to \mathbb{R}^2$ that is a homeomorphism on $Im(f)$ and preserves ordering?

Here is a sketch, but I haven't checked it yet. I can expand tomorrow if you'd like. You can build such a $g$ from the $f$.
Firstly, since your definition of injective arc implies that for any $x \in \mathbb{R}^2$ is either empty (if it is not in the image), a point, or an interval. Consider the quotient of $[0,1]$ by collapsing $f^{-1}(x) = [a,b]$ to $a$ and denote this map $\pi$. By the universal property of the quotient, this gives you a function $\tilde{f}$ s.t. $f = \tilde{f} \circ \pi$. Now $\tilde{f}$ is injective and has the same image. It is also continuous and since $\mathbb{R}^2$ is Hausdorff and the constructed quotient is compact, we have that $\tilde{f}$ is homeo. Now it is left to show that $\tilde{f}$ induces the same ordering, which can be checkec peacewise in between each collapse. 
