Is arc length function a bijection? Let $\gamma\colon [a,b] \to \mathbb{R}^2$, with $a<b \in \mathbb{R}$, be a simple curve (i.e. continuous and injective). The length of $\gamma$ is defined as 
$$
\sup \left\{{\sum_{i=1}^{n} \left\| \gamma(t_i)-\gamma(t_{i-1}) \right\| : n\in \mathbb{N} ,\, a=t_0 < t_1 < \dots < t_n= b} \right\}
$$
Suppose $\gamma$ has finite length $L$ and define a function $\Lambda\colon[a,b] \to [0,L]$ that gives the length of $\gamma\restriction_{[a,t]}$. I think I have proved that $\Lambda$ is strictly increasing, and thus injective. But I can't show it is surjective (I'd like it to be a bijection).
It suffices to show that it is continuous, but I don't know how to deal with continuity of this sort of function... Indeed, I'm not even sure it is a bijection, I guessed, but I may be wrong. My question: is it a bijection? If so, could you give me a hint on how to prove it, please?
 A: You have proven that under the assumptions made about $\gamma$ the function
$$\Lambda:\quad[a,b]\to[0,L],\quad t\mapsto {\rm length}\bigl(\gamma\restriction[a,t]\,\bigr)$$
is well defined and (strictly) increasing. It remains to prove that $\Lambda$ is continuous. Fix a $\tau\in\>]a,b]$. The limit $$\lim_{t\to\tau-}\Lambda(t)=:\lambda\leq\Lambda(\tau)$$ exists. We have to prove that in fact $\lambda=\Lambda(\tau)$.
Given an $\epsilon>0$ there is a $\tau'<\tau$ with $$\bigl|\gamma(t)-\gamma(\tau)\bigr|<\epsilon\qquad(\tau'<t\leq\tau)\ ,$$ and a partition
$$a=t_0<t_1<\ldots<t_{N-1}<t_N=\tau$$
with $t_{N-1}>\tau'$ such that
$$\sum_{k=1}^N \bigl|\gamma(t_k)-\gamma(t_{k-1})\bigr|\geq L(\tau)-\epsilon\ .$$
It follows that
$$\Lambda(t_{N-1})\geq\sum_{k=1}^{N-1} \bigl|\gamma(t_k)-\gamma(t_{k-1})\bigr|\geq L(\tau)-\epsilon-\bigl|\gamma(\tau)-\gamma(t_{N-1})\bigr|>\Lambda(\tau)-2\epsilon\ ,$$
hence $$\Lambda(t)>\Lambda(\tau)-2\epsilon\qquad(t_{N-1}<t\leq\tau)\ .$$
In the same way, but "working from above", one proves that $\lim_{t\to\tau+}\Lambda(t)=\Lambda(\tau)$ for all $\tau\in[a,b[\>$.
