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If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?

Thanks!

2 Answers 2

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We have that $f(f(x))=f(f(y))$ implies $f(f(f(x)))=f(f(f(y)))$ so $f(x)=f(y)$. For any $a,b\in \mathbb R$ we have some $x,y\in \mathbb R$ such that $f(x)=a,f(y)=b$ since $f$ is onto. Thus $f(a)=f(b)\implies f(f(x))=f(f(y))\implies f(x)=f(y)\implies a=b$ hence $f$ is 1-to-1.

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    Oh! Ok thank you, that made it much clearer. Thank you very much!2012-03-21
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If $f(a)=f(b)$ for $a \ne b$, choose $c$ such that $f(c)=a$ and $d$ such that $f(d)=b$. You can do this because $f$ is onto. Then $f(f(f(c)))=f(f(f(d)))$ but $f(c) \ne f(d)$