Bijective local isometry to global isometry Suppose that I have a bijective local isometry $f: X \rightarrow Y$ where $X$ and $Y$ are length spaces. Can I show that $f$ is a global isometry? My thought is to consider a path $\gamma$ from $x$ to $y$ in $X$ and look at a subdivision $(t_1, \ldots, t_n)$ of $\gamma$. Then map the subdivision $(t_1, \ldots t_n)$ to a subdivision of $f(\gamma)$ in $Y$. For a sufficiently fine subdivision each consecutive pair of points should be in the same chart on $Y$. Then the sum of the lengths should be the same in the two spaces so $f$ is an isometry. Can anyone help me formalize this a bit?
 A: Your argument can be firmed up to show that $f$ preserves the lengths of curves.  This in turn implies that $f$ is nonexpanding, i.e. $d'(f(x_1),f(x_2))\leq d(x_1,x_2)$. But without knowing that $f^{-1}$ is continuous, it's conceivable that there's some path in $Y$ from $f(x_1)$ to $f(x_2)$ which is not the image of a path in $X$ and has a shorter length than all such paths. That would stop $f$ from being an isometry. We can show that a bijective local isometry is a global isometry if and only if $f^{-1}$ is continuous.
By length space, I assume you mean a metric space whose induced arc-length metric agrees with the original metric (which, in particular, implies that the space is path-connected). And by local isometry, I assume you mean a map $f:(X,d) \to (Y,d')$ such that every point $x \in X$ has a neighborhood $U$ in which $d'(f(x_1),f(x_2))=d(x_1,x_2)$ for all $x_1,x_2 \in U$.
Claim 1. A local isometry between length spaces preserves the lengths of curves.
Proof. Let $\gamma:[0,1] \to X$ be a path from $x_0=\gamma(0)$ to $x_1=\gamma(1)$. Given $\epsilon>0$, we can find a subdivision $t_0<\cdots<t_n$ such that $$\sum_{i =1}^n d'(f(\gamma(t_i)),f(\gamma(t_{i+1})))<L_{d'}(f\circ \gamma)+\epsilon.$$
We can cover the image of $\gamma$ in $X$ with open sets on which $f$ is an isometry. This image is compact, so we can extract a finite subcover. It's straightforward to see that we can add terms to the subdivision $t_0<\cdots<t_n$ until each interval $[t_i,t_{i +1}]$ is mapped into one of the open sets on which $f$ is an isometry. (For example, pull back the finite cover and use the Lebesgue Number Lemma.) By adding even more terms, we ensure that 
 $$\sum_{i =1}^n d(\gamma(t_i),\gamma(t_{i+1}))<L_{d}( \gamma)+\epsilon.$$
Now we have 
\begin{align*}
L_d(\gamma)&< \epsilon+\sum_{i =1}^n d(\gamma(t_i),\gamma(t_{i+1}))\\
&=\epsilon+\sum_{i =1}^n d'(f(\gamma(t_i)),f(\gamma(t_{i+1})))\\
&\leq L_{d'}(f\circ \gamma)+\epsilon.
\end{align*}
We also have
\begin{align*}
L_{d}(\gamma) &>\sum_{i =1}^n d(\gamma(t_i),\gamma(t_{i +1}))-\epsilon\\
&=\sum_{i =1}^n d'(f(\gamma(t_i)),f(\gamma( t_{i +1})))-\epsilon\\
&\geq L_{d'}(f\circ \gamma)-\epsilon,
\end{align*}
as desired.
Claim 2. If a bijective local isometry between length spaces has a continuous inverse, then it is a global isometry.
Proof. Since $f^{-1}$ is continuous, each neighborhood $U$ on which $f$ is an isometry is carried to an open set $(f^{-1})^{-1}(U)=f(U)$. Then it is easy to see that $f^{-1}$ is an isometry on $f(U)$. It follows that $f^{-1}$ is also a local isometry, hence preserves the lengths of curves. This means $f$ respects the arc-length metric. And since the arc-length metric agrees with the original metric, we conclude that $f$ is a global isometry. $\square$
