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 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.

I know this question sounds a lot like this question, 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.

  • $\begingroup$ How does your question (3) differ from the question(s) in the post that you link to? $\endgroup$
    – Yemon Choi
    Aug 21, 2011 at 4:22
  • $\begingroup$ In the post I link to, they are asking the same question but for monomorphisms, rather than embeddings. $\endgroup$ Aug 21, 2011 at 4:37
  • $\begingroup$ But for more general "concrete" categories, what do you mean by an "embedding"? Something like an extremal monomorphism? $\endgroup$
    – Yemon Choi
    Aug 21, 2011 at 4:55
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    $\begingroup$ I am thinking of the definition here in (en.wikipedia.org/wiki/Embedding#Category_theory) which I believe is the same as in Adámek, Herrlich, and Strecker (The Joy of CATs). Briefly, to be an embedding, a morphism has to be an injective function wrt the underlying sets and an initial morphism. $\endgroup$ Aug 21, 2011 at 5:17
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    $\begingroup$ @Gerald, done, thanks for the suggestion. @Martin, unless I am totally mixed up, in (3) the OP and most of the answerers are talking about "monomorphisms" whereas I'm talking about "embeddings". $\endgroup$ Aug 21, 2011 at 14:46

3 Answers 3


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)).

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    $\begingroup$ A slightly stronger statement: a non-expanding (i.e. 1-Lipschitz) map $f:X\to X$ on a compact metric space is an isometry if and only if it is surjective. $\endgroup$ Aug 21, 2011 at 17:34
  • $\begingroup$ @PietroMajer May I ask for some references or proof outline for your statement. I worked out the direction "isometry to subjective", but have no idea to prove the other way around. Thanks in advance $\endgroup$
    – hmeng
    Oct 13, 2021 at 23:06
  • $\begingroup$ I do not have a reference handy... So maybe I'll post a proof here $\endgroup$ Oct 14, 2021 at 18:36
  • $\begingroup$ The statement mentioned by Pietro Majer is Theorem 1.6.15 in A course in metric geometry of Burago, Burago, and Ivanov. $\endgroup$ Mar 24, 2023 at 16:23

The answer to (2) is "no". Take, for example, $X$ to consist of a countable number of copies of $[0,1]^2$ with the standard metric, where every two points in different components have distance 1000 from one another.

Then take $Y$ to be $X$, but with some stuff cut out of one of the components so it looks like an annulus or a letter of the alphabet or some other cute shape.

$X$ and $Y$ are clearly not isomorphic, because $Y$ has a weird component and $X$ doesn't. But there's clearly embeddings from each to the other, as you can embed the weird component in a normal one, and the cardinality of components is the same in $X$ and $Y$, whether you count the weird component or not.

  • $\begingroup$ Thanks. Is it true for compact metric spaces? $\endgroup$ Aug 21, 2011 at 14:27

Let me present a counter-example to question (2).

Let $\ X:=\mathbb Z_{\ge 0}\ $ be the set of all non-negative integers in their natural metrics. Let $\ Y:=X\setminus\{1\}.\ $ Then inclusion $\ Y\subseteq X\ $ is one isometric embedding, and $\ x\mapsto x+2\ $ is an isometric embedding in the opposite direction.

Spaces $\ X\ $ and $\ Y\ $ are not isometric because $\ Y\ $ has a point, namely $\ 0\in Y,\ $ which is not distant from any other point of $\ Y\ $ by $\ 1,\ $ while $\ X\ $ does not have any such point (each point $\ x\in X\ $ is distant form $\ x+1\in X\ $ by $\ 1).$


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