Sequence of polynomials converging to homeomorphism Let $f:[0,1]\to [0,1]$ be a homeomorphism. Show there is a sequence $(p_n)$ of polynomials such that $p_n$ converges uniformly to $f$ on $[0,1]$ and each $p_n$ is a homeomorphism from $[0,1]$ onto $[0,1]$.
My attempt:First, we note that in order to show $p_n:[0,1]\to [0,1]$ is a homeomorphism, we only need to show that each $p_n$ is bijective (since continuous bijections  from compact spaces onto hausdorff spaces are homeomorphisms). Since $f$ is continuous and injective, and $[0,1]$ is connected, we have that $f$ is monotone. There are then two possiblities since $f$ maps onto $[0,1]$: $f(0)=0$ and $f(1)=1$, or $f(0)=1$ and $f(1)=0$. WLOG assume the former case. Since $f$ is continuous on a compact space, there exists a sequence $(\tilde{p}_n)$ that converges uniformly to $f$. In particular, $\tilde{p}_n(0)\to f(0)=0$. If we define $p_n=\tilde{p}_n-\tilde{p}_n(0)$, then $p_n(0)=0$ for all $n$ and $p_n$ to $f$ uniformly. Now if we can show that $1$ is in the range of $p_n$, we can use the intermediate value theorem to conclude $p_n$ maps onto $[0,1]$ for each $n$.
I wasn't sure how to proceed from here. I think I essentially have the idea of how to show each $p_n$ is surjective, but I'm having particular trouble showing each $p_n$ is one-to-one.
 A: Let $\|\cdot\|_\infty$ deonte the supremum norm on $[0,1]$.
Wlog. $f(0)=0$ and $f(1)=1$. Define $\tilde f\colon[-1,2]\to[-1,2]$ by letting
$$ \tilde f(x)=\begin{cases}f(x)&\text{if }0\le x\le 1,\\
-f(-x)&\text{if }-1\le x\le 0,\\
2-f(2-x)&\text{if }1\le x\le 2.
\end{cases}$$
That is, we extend $f$ by copies of itself rotated by $180^\circ$ around $(0,0)$ and $(1,1)$, respectively.
For $0<h<1$, define $g_h\colon[0,1]\to[0,1]$ as
$$g_h(x)=\frac1{2h}\int_{x-h}^ {x+h}\tilde f(t)\,\mathrm dt.$$
Then $g_h$ is $C^1$ with derivative $g_h'(x)=\frac1{2h}(\tilde f(x+h)-\tilde f(x-h))>0$. Also $g_h(0)=0$ and $g_h(1)=1$, so $g_h$ is a $C^1$ homeomrphism $[0,1]\to[0,1]$.
One verifies that $\|g_h-f\|_\infty\to 0$ as $h\to 0$.
Hence it suffices to approximate $g_h$, i.e., we may assume wlog. that $f$ is a $C^1$ function with strictly positive derivative. The latter means that $\min f'>0$. 
We can approximate $f'$ by polynomials, i.e., given $\epsilon>0$, we find a polynomial $p$ with $\|f'-p\|_\infty < \frac\epsilon2$.
In particular, if we have $\epsilon<2\min f'$, then $p(x)>0$ for all $x\in[0,1]$.
Let $q(x)$ be the polynomial $\int_0^xp(t)\,\mathrm dt$.
Then for all $x\in [0,1]$, 
$$|f(x)-q(x)|\le\int_0^x|f'(t)-p(t)|\,\mathrm dt\le \frac\epsilon2 x\le \frac\epsilon2.$$
In particular, $|q(1)-1|<\frac\epsilon2<1$.
Consider the polynomial $r(x)=\frac1{1-q(1)}q(x)$.
Then $r(0)=0$, $r(1)=1$, and $\|r-q\|_\infty<\frac\epsilon2$ so that $\|r-f\|_\infty<\epsilon$.
As $r'(x)=\frac1{1-q(1)}p(x)>0$, the polynomial $r$ is the desired approximation.
