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Let $(M,d)$ be a metric space. An isometry is a distance preserving map. A constant displacement map is a function $f$ such that $d(x,f(x)=d(y,f(x))$ for all $x$ and $y$. I know that not all isometries are bijective. But are all maps that are both constant displacement and distance-preserving also bijective? I would like to see a proof or a counterexample. I have tried working on this problem but I have not succeeded in proving anything.

dfeuer
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user107952
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  • Can you find a non-bijective isometric constant displacement map from $\Bbb R_+$ to $\Bbb R_+$? – dfeuer Nov 23 '13 at 10:43
  • In addition to dfeuer's remark, there are noninjective constant displacement maps from real line union one point to itself. – Moishe Kohan Nov 23 '13 at 16:03
  • @studiosu, the OP was asking specifically about constant displacement maps that are also isometries. An isometry in a metric space (as opposed to a pseudometric space) is always injective. – dfeuer Nov 24 '13 at 02:18
  • Oops, I did not see the distance preserving condition. – Moishe Kohan Nov 24 '13 at 06:46

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No. The map $x\mapsto x+1$ on the space $M=[0,\infty)$ preserves distances, and is constant-displacement, but is not onto. (Following dfeuer's hint).

user111000
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