A bijective arcwise isometry is an isometry? Let $X,Y$ be length spaces. Suppose $\, f:X \to Y$ is an arcwise isometry, i.e $L_X(\gamma) = L_Y(f \circ \gamma)$ for every continuous path $\gamma:I \to X$. (In particular, $f$ takes non-recitifiable paths to non-recitifiable paths).
In addition, assume $f$ is a bijection.
Is it true that $f$ is an isometry?

The naive approach would be to say
$$ d_Y(f(p),f(q))= \inf \{ L_Y(\beta): \beta \, \, \text{is a path from } f(p) \text{ to } f(q)\} \stackrel{(*)}{=}\inf \{ L_X(\alpha): \alpha \, \, \text{is a path from } p \text{ to } q\}=d_X(p,q),$$
so the answer is positive. However, there is a problem with this argument:
Equality $(*)$ relies upon the assumption we have a "length-preserving" bijection between the two sets of paths, namely $\alpha \to f \circ \alpha$.
The problem is that we do not know $f^{-1}$ is continuous*, so for a path $\beta$ in $Y$, $f^{-1} \circ \beta$ is not necessarily a path in $X$ (paths must be continuous by definition). This causes an obstacle in showing surjectivity: If $\beta$ is a path in $Y$, we would like to say $\beta=f(f^{-1}(\beta))$, but why is $f^{-1}(\beta)$ a legitimate path?

*We do know $f$ is continuous, in fact it's $1$-Lipschitz.

Hint: consider the age old counterexample $f : [0,2\pi) \to S^1$ given by $f(t) = (\cos(t),\sin(t))$.
