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?

It seems that my answer to Reparameterisation of Curve as a Regular Curve (Topology) also answers your question in the affirmative. In the referenced question arbitrary paths $p : [0,1] \to X$ are considered and it is shown that there exists a reparametrization $\phi$ (which is a non-decreasing surjective continuous map $\phi: [0,1] \to [0,1]$ with $\phi(0)=0, \phi(1)=1$) such that $p = q \circ \phi$ with a path $q : [0,1] \to X$ which is not constant on any closed subinterval $[a,b]$, $0 \le a < b \le 1$, of $I =[0,1]$.
Applying this to your $f : I \to \mathbb R^2$, we get $\phi$ and $g : I \to \mathbb R^2$ such that $f =  g \circ \phi$ and  $g$ is not constant on any closed subinterval of $I$. Now assume that $g$ is not injective. Then $g(a) = g(b)$ for suitable $0 < a < b < 1$. Choose $a', b'$ such that $\phi(a') = a, \phi(b') = b$. Clearly $0 < a' < b' < 1$ and $\phi([a',b']) = [a,b]$ since $\phi$ is non-decreasing. Then $f(a') = f(b')$ and $f(x) = c$ for $x \in [a',b']$. Hence $g(y) = c$ for $y = \phi(x) \in \phi([a',b']) = [a,b]$, thus $g$ is constant on $[a,b]$ which is a contradiction.
