Let $(M,d)$ be a metric space. A constant displacement map is a function $f$ from $M$ to $M$ such that $d(x,f(x))=d(y,f(y))$. My question is this: Is the composition of two bijective constant displacement maps also a constant displacement map? And if not, is the composition of two isometric bijective constant displacement map also constant displacement?
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Hint: Consider the case of the (standard) discrete metric. What are isometries of such space? What are constant displacement maps? Now, construct counter examples. – Moishe Kohan Nov 25 '13 at 20:09
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@studiosus Thanks for the hint! I got the answer now. – user107952 Nov 26 '13 at 03:41
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Since the answer is already known to the OP (thanks to the hint by studiosus), I post it here. Let $M=\{1,2,3,4,5\}$ with the discrete metric ($d(x,y)=1$ unless $x=y$). The permutation $(12)(345)$ is a bijective isometry and a constant displacement map. But its square is the cycle $(354)$, which is not a constant displacement map.
The above metric space can be realized as a subset of $\mathbb R^4$ with the standard metric. I wonder if there are counterexamples that are subsets of $\mathbb R^d$, $d\le 3$.
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