I am reading the book A course in metric geometry by Burago, Burago and Ivanov. I have some difficulties with an exercise 3.4.6 on page 78. The exercise is the following: Let a group $G$ act by isometries on a length space $X$. Assume that all the orbits are discrete. Prove that the projection map $p:X\rightarrow X/G$ is a local isometry at a point $x$ if and only if $x$ is not fixed by a nonidentity element of $G$.
I could prove one direction, namely that if $x$ is not fixed by a nonidentity element of $G$ then $p$ is a local isometry. For the other direction I have no idea how to do. Does anyone have an idea how to do it, or a hint?
Thanks, Maurice
I think the statement is wrong: Consider the subset $X$ of $\mathbb R^2$ that is the union of the two coordinate axes equipped with the induced intrinsic metric; $$X = \{(x,y) \in \mathbb R^2\ | x = 0\} \cup \{(x,y) \in \mathbb R^2\ | y = 0\}.$$ This is clearly a length space. Now consider the action of $\mathbb Z_2$ on $X$ via reflecting at the $y$-axis (the one coming from the involution $(x,y) \mapsto (-x,y)$). Then the action on $X$ is effective, isometric and the $y$-axis equals the fixed point set of the action. It follows easily that $p : X \to X/\mathbb Z_2$ is a local isometry at every point of the $y$-axis different from $(0,0)$.
Edit: It is not hard to see, that the statement is true for, say, manifolds.