Metric on the unit circle Let $d$ be a metric on the unit circle $S^1$ which defines the usual topology. Let $D = \sup_{x, y \in S^1} d(x,y)$ be the diameter for $S^1$ under this metric.
Is it true that the map $d : S^1 \times S^1 \to [0,D]$ is an open map?
Thank you!
 A: The map is not always open. Here is a simple counter-example.
The unit circle is homeomorphic to the following figure.

This figure is endowed with the euclidean metric of the plane, that we then transport to a metric on the unit circle using the homeomorphism.
Now consider the two points in red on the figure (or more exactly, their image on the unit circle by the homeomorphism). Let us call them $x$ and $y$. The distance will send any small neighborhood of $(x,y)$ to a set of the form $(\delta, \delta']$, with $\delta' < D$. 
Consequently, this map is not open.
A: Yes, it is. A basis for the product $S_1\times S_1$ is given by the product of open sets $U_1\times U_2$, where $U_i$ is an open ball in $S_1$.
The function maps pairs of such open sets to the set:
$$d(U_1, U_2) = \{x\in \Bbb{R} | \inf_{(x_1, x_2)\in (U_1, U_2)} d(x_1, x_2) < x < \sup_{(x_1, x_2)\in (U_1, U_2)} d(x_1, x_2)\} $$
And you can see that the only such sets where the image is a closed interval (or half-closed) are those that partially overlap or such that they include points that are opposite to each other. In those cases, the image includes the points $0$ or $D$. Intervals that include those points are open in $[0, D]$ in the subspace topology.
Since the function maps basis sets to open sets, it is an open map.
