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I got this problem:

Find examples of continuous functions $f:[0,1]\to[0,1]$ that satisfy $\forall x\in[0,1], f(f(x))=f(x)$ other than $f(x)=x$.

I proved that $f([0,1])=\{x\in[0,1]|f(x)=x\}$, But I couldn't find an example of a function other than $f(x)=x$ that satisfies the conditions.

Thanks on any examples.

MathNerd
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1 Answers1

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The condition $f(f(x)) = f(x)$ is equivalent to $f(y)= y$ for all $y \in $ image of $f$.

Take an $f$ that is not surjective: $f(x) = x$ for $x\le 1/2$ and $f(x) = 1/2$ for $x\ge 1/2$.

orangeskid
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    Any retraction of $[0,1]$ onto a closed sub-interval. – orangeskid Sep 28 '14 at 10:08
  • Nice answer, by the way can you think of a subjection that satisfies the conditions? – MathNerd Sep 28 '14 at 10:12
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    $f$ is surjection means image of $f = [0,1]$. But we have the condition $f(y) = y$ for $y$ in image. Therefore if $f$ surjection then $f(x) = x$ for all $x \in [0,1]$. – orangeskid Sep 28 '14 at 10:19
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    Any retraction of $[0,1]$ onto a closed sub-interval. Pick the image $[\alpha ,\beta] \subset [0,1]$. On $[\alpha ,\beta]$ $f$ is the identity. Then flap the arms $[0,\alpha]$ and $[\beta, 1]$ into $[\alpha, \beta]$ (lots of freedom here). – orangeskid Sep 28 '14 at 10:19
  • No worries. Try sketching some graphs of such functions. – orangeskid Sep 28 '14 at 10:27
  • @Mathematics Man: It will equal the image of $f$, $\ f([0,1])$, so it will be an interval (closed). – orangeskid Feb 19 '21 at 00:38