When an odd polynomial is a one-one map on $\mathbb{R}$ Let $f(x)=c_1x+c_3x^3+c_5x^5+\cdots+c_{2m+1}x^{2m+1}$, $c_1,c_3,c_5,\ldots,c_{2m+1} \in \mathbb{R}$, $m \in \mathbb{N}$,
namely, $f$ is an odd polynomial over $\mathbb{R}$.

When such a polynomial is a one-one map on $\mathbb{R}$?
  Is it always one-one?

Example:
$f(x)=x+x^3$. If $f(a)=f(b)$ for $a,b \in \mathbb{R}$, then
$a+a^3=b+b^3$, so $(a-b)+(a^3-b^3)=0$,
and then, $(a-b)(1+(a^2+ab+b^2))=0$.
Therefore, $a-b=0$ or $1+(a^2+ab+b^2)=0$.
In the first case $a=b$ and we are done, while in the second case,
the discriminant is $-4-3a^2 < 0$, so there are no $a,b \in \mathbb{R}$ satisfying $1+(a^2+ab+b^2)=0$.
I guess that a general solution should be similar to the special case of the example: If 
$c_1a+c_3a^3+c_5a^5+\cdots+c_{2m+1}a^{2m+1}=c_1b+c_3b^3+c_5b^5+\cdots+c_{2m+1}b^{2m+1}$, then $(a-b)(c_1+c_3(a^4+a^3b+a^2b^2+ab^3+b^4)+\cdots)=0$,
so (first case) $a-b=0$ and we are done, or (second case) $c_1+c_3(a^4+a^3b+a^2b^2+ab^3+b^4)+\cdots=0$, which I am not sure I know how to show that it is necessarily non-zero (it is an even polynomial).

The answer is no, ti is not always one-to-one. 
Consider the following  odd polynomial
$$P(x)=x(x-5)(x-10)= x^3-15x^2+50x$$
$$ P(0)=P(5)=P(10)=0$$
It is  not one-to-one.
However, a function will be one-to-one if it's derivative is always positive or always negative. 
