Let $(X,d)$ and $(X',d')$ be metric spaces. Suppose that there exist isometric embeddings $f:(X,d)\to(X',d')$ and $g:(X',d')\to(X,d)$. Then:
- Is it true that $(X,d)$ and $(X',d')$ are isometric?
If yes:
- Can we construct such an isometry from $f$ and $g$?
And if not:
- What is a counterexample?
I would suppose that the answer to these questions is known. Here is the only thing I've figured out so far: under the assumptions above $g\circ f: (X,d) \to (X,d)$ is an isometric embedding. But an endo-isometric embedding is not automatically surjective, take as a counterexample the map $\Bbb{N}\to\Bbb{N}$ given by $n\mapsto n+1$, and so it's not an isometry. If the answer to the question above is yes, than one cannot proceed exactly this way.
Consider $X={\mathbb N}$ and $Y={\mathbb N} - \{2\}$ equipped with the natural metrics (given by $d(n.m)=|n-m|$).