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I am looking for the proof Prove of "any isometry S is a surjective mapping".

My attempt:

pick any two points $A, B$, consider their images $S(A) = A'$ , $S(B) = B'$ . To prove surjectivity, I need to find, for any point $X'$ in the plane, a point $X$ such that $S(X) = X'$ . I know from the definition of isometry that in this case we would have $XA = X'A'$ , $XB = X'B'$ .

Now my question is how to use this to construct possible candidates for X such that $S(X) = X'$.

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  • $\begingroup$ Isometries are onto by definition. Otherwise are called isometric embeddings. Ex. A line in a plane is an isometrica embedding. thus you cannot prove surjectivity by using just the isometry preperty. $\endgroup$
    – user126154
    Apr 7 '15 at 7:35
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It depends on your definition of an isometry. In my view, an isometry is not always surjective (it is, however, always injective, at least in the context of metric spaces). If you want to prove that it is always surjective, you have to either
- demand it in the definition, or
- restrict to a more specific context, such as square matrices.

You mention the "plane" in your question. This means that you are probably using the second case above. In this case, you could derive surjectivity from injectivity.

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