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!

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.
