It is well known that the group of diffeomorphisms of the circle contains free nonAbelian 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 nontrivial law. Now for any nontrivial 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?

Much more is true. The compactly supported diffeomorphism group of any (positivedimensional, 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(PLWASW) Free subgroups of diffeomorphism groups. Fund. Math. 131 (1988), no. 2, 103–121. which is available online here. 

