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If $X$ and $Y$ are isometric spaces, does the mapping between them $f:X\to Y$ have to be bijective? I feel only injectivity is required to satisfy the relation $$d_y(f(a),f(b))=d_x(a,b)$$

and surjectivity is not needed.

Even this proof of the completion theorem seems to imply $f$ need only be injective. However, my textbook seems to insist $f$ should be bijective. I don't know if surjectivity is a condition that is useful later on.

Thanks in advance!

  • Hello, you're of course right that it only need be injective. Consider an inclusion of a subspace. That said, it is a fact that any isometry of a compact metric space with itself is surjective. Maybe that's what they refer to? – Alex Youcis Aug 12 '13 at 09:49
  • Yes but what if $X$ and $Y$ are different? –  Aug 12 '13 at 09:52
  • I just gave you an example where this is untrue if X and Y are unequal. Include any proper subspace into the full space. – Alex Youcis Aug 12 '13 at 09:53
  • A map $f \colon X \to Y$ with the property $d_Y(f(a),f(b)) = d_X(a,b)$ for all $a,,b\in X$ is an isometric embedding of $X$ in $Y$. Occasionally also called an isometry from $X$ into $Y$. Two metric spaces $X$, $Y$ are called isometric, if there is a bijective isometric map between them. Such a bijective isometry is also sometimes called an isometry between the two spaces. The relation "isometric" between metric spaces is (and should be) an equivalence relation. Note that when dealing with completions, it is supposed that the space to be completed is dense in its completion, therefore – Daniel Fischer Aug 12 '13 at 10:06
  • an isometry from one completion $Y$ of $X$ into another $Z$, that respects the embeddings of $X$ into $Y$ resp. $Z$ automatically is surjective, hence bijective. – Daniel Fischer Aug 12 '13 at 10:08

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For some people (I personally don't adhere to this definition), the statement '$X$ and $Y$ are isometric spaces' is not equivalent to 'there is an isometry $f\colon X\rightarrow Y$'. This comes down to whether or not you impose the extra condition that isometries are bijective. I would personally call such a map which wasn't necessarily surjective an isometric embedding and then if the map also happens to be surjective I would call it an isometry, so that isometries became the class of isomorphisms in the category of metric spaces. Many authors adopt this position.

However, many others adopt the definition that an isometry* (I'll use a * to distinguish the two definitions) only needs to preserve distance, and so if $f\colon X\rightarrow Y$ is an isometry* then the space $X$ and $f(X)$ are isometric metric spaces. However, $X$ and $Y$ are only isometric if $f(X)=Y$, that is $f$ is surjective.

The moral of the story is that an author should make clear which definition they are using, and a reader should check to see which definition the author is using.

Dan Rust
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