Characterize the polynomial bijections from $(0,1)$ to$ (0,1)$ That is, show which polynomials, taken as functions, truncated to $(0,1)$, biject with $(0,1)$. I know for the increasing case, it is equivalent to (the derivative is greater than or equal to $0$ on $[0,1]$, $p$ is non-constant, and $p(0)=0$). The problem is with the first part. Bonus: characterize the continuous bijections, or differentiable bijections.  
 A: Firstly, your characterization of polynomials is incorrect. Notice that the polynomial $1-x$ is a bijection, yet does not satisfy your condition.
Second, for continuous function, the relevant condition is monotonicity combined with the statement that $f(\{0, 1\}) = \{0, 1\}.$
A: My thoughts on the polynomial case are the following.
Let $f:I\to I$ be a polynomial function on $I=[0,1]$ such that $f(0)=0$, $f(1)=1$ and $f'(x)\ge 0$ for all $x\in I$. Since $f$ is polynomial, it has the form
$$ f(x) = \sum_{i=0}^n a_i x^i $$
for some $n\in\mathbb N$ and $a_i\in\mathbb R$. From $f(0)=0$ we get $a_0=0$ and from $f(1)=1$ we get
$$ \sum_{i=1}^n a_i = 0. $$
The monotonicity becomes
$$ f'(x) = \sum_{i=1}^n i a_i x^{i-1} \ge 0 $$
for all $x\in I$. As $\mathbb Q$ is dense in $\mathbb R$, it is enough if this inequality holds at rational points, i.e. for all $p,q\in\mathbb N$ with $p\le q$ we have
$$ \sum_{i=1}^n i a_i \left(\frac p q\right)^{i-1} \ge 0 $$
or equivalently
$$ \sum_{i=1}^n i a_i p^{i-1} q^{n-i} \ge 0.$$
At this point I'm not sure how to weaken the condition further.
Does this help you in any way?
A: nd^(n-1)  (n-1)d^(n-2) ...  1
nd^(2n-2) (n-1)d^(2n-4) ... 1
nd^(3n-3) (n-1)d^(3n-6) ... 1
.
.
.                                a=*0*
.
.                           1
where a=(a1, a2, a3, ..., an)T and d is arbitrarily close to 1. If we can solve this, and interpret it correctly in terms of the inequality we should be done.
