On local isometry of mappings Apropos discussion at this link: locally isometric is not a symmetric relation.
My instructor told me that a counterexample to the symmetric relation here could be actually even simpler than the complete hyperbolic manifold. For example, a map from a circle to real line or from a torus $\mathbb{T}^2$ to $\mathbb{R}^2$. Is this true? If yes, then how do I go about proving this?
-R.
 A: Since $S^1$ is compact, any map $f$ of the circle into $\mathbb{R}$ must have a compact connected image.  Thus the image must be a closed interval (possibly degenerating to a point).  
Now consider a point $x$ on the circle s.t. $f(x)$ is the least upper bound of the image $f(S^1)$.  Removing $x$ from a small connected neighborhood $\mathcal{O}$ (any open arc containing $x$ but properly contained in $S^1$ will do) will separate it into two connected components.
But the same is not true of removing $f(x)$ from $f(\mathcal{O})$.  Because $f(x)$ is the upper endpoint of $f(S^1)$, $f(\mathcal{O})$ (a subinterval of $f(S^1)$) remains connected with the removal of $f(x)$.
Thus $f$ is not a homeomorphism on a neighborhood of $x$, and certainly not a local isometry there.
On the other hand a uniform wrapping of the real line around the circle, $g:\mathbb{R}\to S^1$, preserving arc length, is a local isometry.
Added:
Here's a more "metric" way of proving no local isometry $f$ maps $S^1$ continuously into $\mathbb{R}$. Suppose one exists, for the sake of contradiction.
Let $\epsilon: S^1 \to \mathbb{R}^+$ denote the largest radius $\epsilon(x)$ for each point $x\in S^1$ such that on the metric ball $\mathscr{B}(x,\epsilon(x))$ centered on $x$, the map $f$ is an isometry.  By assumption $\epsilon(x) \gt 0$ at each point, and $\epsilon$ is easily seen to be continuous.  Since $S^1$ is compact, there exists positive greatest lower bound $\epsilon_0$, the minimum of $\epsilon(x)$, such that:
$$ \epsilon(x) \ge \epsilon_0 \gt 0 \;\;\; \forall x \in S^1 $$
For any point $x_0$ in $S^1$, consider that $f$ restricted to the ball (arc) $\mathscr{B}(x_0,\epsilon_0)$ is an isometry and thus has as image the real interval $(f(x_0)-\epsilon_0,f(x_0)+\epsilon_0)$.  The same argument applies to the endpoints of the arc $\mathscr{B}(x_0,\epsilon_0)$, so that those neighborhoods are mapped to intervals $(f(x_0)-2\epsilon_0,f(x_0))$ and $(f(x_0),f(x_0)+2\epsilon_0)$ in one order or the other.
Piecing these neighborhood restrictions of $f$ together tells us $f$ is an isometry on $(f(x_0)-2\epsilon_0,f(x_0)+2\epsilon_0)$.  Since $x_0$ was arbitrary, this shows every point in $S^1$ has a metric ball of radius $2\epsilon_0$ on which $f$ is an isometry.  But this contradicts the minimality of $\epsilon_0$ for $\epsilon:S^1 \to \mathbb{R}^+$.
Therefore no local isometry $f$ exists.
