It is well known that the group of diffeomorphisms of the circle contains free non-Abelian subgroups. Is it true (known) that the group of diffeomorphisms of the interval $[0,1]$ contains free subgroups? One approach to get a positive answer can be the following. Consider all functions $f\_a=\frac{\exp(ax)-1}{\exp(a)-1}$, $a>1$, which are smooth diffeomorphisms of $[0,1]$. First prove that the group generated by these functions does not satisfy any non-trivial law. Now for any non-trivial word $w$ in two variables consider the function $w(f\_a,f\_b)$ (for every $a,b$). The set of pairs $(a,b)\in \mathbb{R}^2$ for which this function is the identity function has dimension at most 1. Then by Baire category theorem, since the set of words $w$ is countable, there are $a,b$ such that $f\_a,f\_b$ freely generate a free subgroup. Does this argument actually work?



Free subgroup of Diff([0,1])?
Much more is true. The compactly supported diffeomorphism group of any (positive-dimensional, nonempty) manifold contains free subgroups of uncountable rank. In fact, there are such subgroups that are generated by sets which are arcwise connected! See the paper


MR0974661 (90b:58031)
Grabowski, Janusz(PL-WASW)
Free subgroups of diffeomorphism groups. 
Fund. Math. 131 (1988), no. 2, 103–121. 


which is available online [here](http://matwbn.icm.edu.pl/ksiazki/fm/fm131/fm131110.pdf).


