In Thurston's and Levy's "Three dimensional Geometry and Topology, page 6, they define the induced metric on the torus from the euclidian metric of its covering space, the plane.
Specifically, for any point $x\in T$, one can pick a neighborhood $U$ small enough such that its preimage under the covering map $\rho$ is the disjoint union of $U_i$ homeomorphic to $U$. What I want to ask is why do we have to shrink $U$ even further so that the diameter of the $U_i$ will be less than the distance between any two of them before projecting the metric (i.e. declaring $\rho$ to be an isometry between some $U_i$ and $U$).
For instance cover the circle with the line via $\rho: t \mapsto e^{2\pi i t}$. Then $U = S^1 - \{1\}$ (the circle minus the east pole) is evenly covered by the disjoint union of open intervals $\cup_n (n, n+1)$. Now choose $\epsilon >0$ small. You don't want to define the distance between $e^{2 \pi i \epsilon}$ and $e^{2 \pi i (-\epsilon)}$ to be $1-2\epsilon$ (which is the distance between their pre images in any single $(n, n+1)$) because they are actually very close together on the circle.