Let $(X,d)$ and $(Y,p)$ be metric spaces and $\pi : Y\rightarrow X$ be a locally isometric covering map. Let $g:X\rightarrow X$, $f: Y \rightarrow Y$ be homeomorphisms such that $\pi\circ f = g\circ \pi$. Further Suppose that $X$ is compact and there exists $\delta_{0} >0$ such that for each $y\in Y$ and $0<\delta <\delta_{0}$, the open set $U_{\delta}(y)$ is connected and $\pi : \pi^{-1}(\pi(U_{\delta}(y)))\rightarrow U_{\delta}(\pi(y))$ is an isometry. Then show that for every pair $x_{1}\neq x_{2}\in X$, there exist $y_{1}, y_{2}\in Y$ with $\pi(y_{1})=x_{1}$ and $\pi(y_{2})=x_{2}$ and $p(y_{1}, y_{2})= d(x_{1}, x_{2})$.
(Locally isometric covering map: Let $(X,d)$ and $(Y,p)$ be metric spaces and let $\pi : Y\rightarrow X$ be a continuous map then $\pi$ is called a locally isometric covering map if for each $x\in X$, there exists a neighborhood $U(x)$ of $x$ such that $\pi^{-1}(U(x)) = \cup_{\alpha}U_{\alpha}$ where $\lbrace U_{\alpha}\rbrace$ is a pairwise disjoint family of open sets and $\pi|_{U_{\alpha}} : U_{\alpha}\rightarrow U(x)$ is an isometry for each $\alpha$.)