Suppose $(X,d)$ is a metric space with the nearest point property and $a,b \in X$ with $a \ne b$. Suppose there is a path of finite length in $X$ from $a$ to $b$ and let $m$ be the infimum of the lengths of all paths from $a$ to $b$.Then, by Lipschitz reparametrization, there exists a path $g:[0,1] \rightarrow X$ from $a$ to $b$ that satisfies $lth\_t(g) = tm ~\forall~t \in [0,1].$ and $g$ is lipschitz with length $m. lth\_t(g)$ represents the length of the function $g$ upto a point $t$.


Then is the function $g$ always injective? 


I am kind of stuck here and can't think of any counterexamples as well. Any inputs will be appreciated, thanks!



Is the function $g$ always injective where $g$ is obtained by lipschitz re-parametrization
It is injective because if $g(s)=g(t)$, $s<t$, then you can remove the interval $[s,t]$ from the domain of definition of $g$ and make the curve shorter. Then you rescale the domain to be $[0,1]$. Rescaling does not change the length of the curve. See also Lemma 3.10 in my notes linked to the answer to another related question: [does there always exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and ..?](https://mathoverflow.net/questions/295446/does-there-always-exists-a-path-g0-1-rightarrow-x-from-f0-to-f1-th?rq=1)


