I have three questions about when you can show there is an isometry between metric spaces. 


(1) If there is an injective non-expanding map from $X$ to $Y$ and an injective non-expanding map from $Y$ to $X$, are $X$ and $Y$ isometric?


I think the answer must be no, just let $X=[0,1]$ and $Y=[0,1/2]$ with the Euclidean metric on each and let the morphisms just shrink each of the intervals by a 1/2. But $X$ and $Y$ are not isometric as metric spaces. The only reason I ask is that [this question](https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold) seems to imply that this is true for compact metric spaces. So maybe I am just missing something.


(2) If there is an isometric embedding from $X$ to $Y$ and an isometric embedding from $Y$ to $X$ is it true that $X$ and $Y$ are isometric?


Here by an isometric embedding I mean a map that preserves the metric. 


(3) If the answer to (2) is yes, is there something to be said about which concrete categories this result holds for, with respect to embeddings?


Here I am taking the definition of concrete categories and embeddings from [Adámek, Herrlich, Strecker](http://katmat.math.uni-bremen.de/acc/acc.pdf).


I know this question sounds a lot like [this question](https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold), but unless I am confused, they are talking about injective maps (monomorphisms) which make sense in any category, whereas I am talking about embeddings which are only defined for concrete categories.


EDIT: Edited to remove jargon and make clearer.


Thanks very much for any information.



Isometries between metric spaces
As already observed by James, the answer to (2) is negative in general. However, the answer is positive if we assume that $X$ (or $Y$) is compact. I will show in a moment how this can be deduced from the following claim:


Let $f\colon X\to X$ be a distance-preserving map of a metric space into itself. If $X$ is compact, then $f$ is an isometry (i.e. it is surjective, its injectivity being obvious).


Let me first prove the claim: set $Y=f(X)$ and suppose that $Y\neq X$. Pick a point $x\_0\in X\setminus Y$. Since $X$ is compact, $Y$ is also compact, so $x\_0$ has positive distance, say $\epsilon$, from $Y$. Now let $x\_n=f^n(x)$. For every $i\leq j$ we have 
$d(x\_i,x\_j)=d(f^i(x\_0),f^i(f^{j-i}(x\_0)))=d(x\_0,f^{j-i}(x\_0))\geq d(x\_0,Y)=\epsilon$. Therefore, no subsequence of $\{x\_n\}$ can be a Cauchy sequence, and this contradicts the compactness of $X$, thus proving the claim.


Coming back to question (2) (and in some sense (3)), if $h\colon X\to Y$ and $g\colon Y\to X$ are embeddings, and $X$ is compact, then by the claim $g\circ h$ is an isomorphism of $X$. In particular, $g$ is surjective, so it is an isomorphism between $Y$ and $X$.


Therefore, even if (2) does not hold in general, it holds in the category of compact metric spaces (this gives a partial answer to (3)).


