Assume $f:K\rightarrow K$, is surjective and $K$ is a compact metric space and we have $d(f(x),f(y))\leq d(x,y)\, \forall x,y\in K$. How can I prove that $d(f(x),f(y))= d(x,y)\, \forall x,y\in K$?
Thank you for your help.
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What did you try? – egreg Apr 11 '14 at 15:15
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I tried assuming otherwise. I was thinking that $diam(f(K))$ could be proved to be less than $diam(K)$, but didn't get any good results. – User 1e5 Apr 11 '14 at 15:19
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I also know $f$ is equi-continuous. – User 1e5 Apr 11 '14 at 16:26