If $f$ is a homeomorphism If $f$ is a homeomorphism  on $[0,1]\to[0,1]$,then I need to show existance of a sequence of polynomials $p_n(x)$ such that $p_n\to f$ uniformly on $[0,1]$ and each $p_n$ is also a homeomorphism.
Thanks for help
 A: WLOG $f$ is strictly increasing; then $f(0)=0,f(1)=1.$
Suppose first $f\in C^1([0,1]).$ Then $f'\ge 0.$ By Weierstrass there is a sequence of polynomials $p_n\to f'$ uniformly on $[0,1].$ Since $f'\ge 0,$ we can assume these polynomials are positive (add on some small positive constants if necessary). Define
$$q_n(x) = \int_0^x p_n(t)\, dt.$$
Then each $q_n$ is a polynomial. We have $q_n \to \int_0^x f'(t)\, dt = f(x)$ uniformly on $[0,1].$ Because $q_n' = p_n >0,$ each $q_n$ is strictly increasing, hence is a homeomorphism of $[0,1]$ ont0 $[0,q_n(1)].$ The polynomials $q_n/q_n(1)$ then are homeomorphisms of $[0,1]$ to $[0,1]$ that converge to $f$ uniformly on $[0,1].$
So we're done if $f\in C^1([0,1]).$ To finish, try to prove that if $f$ is only continuous, it can be approximated uniformly by strictly increasing $C^1$ functions. One approach here would be through mollifiers. Another one would be to approximate $f$ with a piecewise linear function $g$ and then approximate $g$ with a $C^1$ function.
