Let $f(x)=c_1x+c_3x^3+c_5x^5+\cdots+c_{2m+1}x^{2m+1}$, $c_1,c_3,c_5,\ldots,c_{2m+1} \in \mathbb{C}$, $m \in \mathbb{N}$, namely, $f$ is an odd polynomial over $\mathbb{C}$.
When such a polynomial is a one-one map on $\mathbb{C}$?
(What can be said about the coefficients of $f$? Are their absolute values of the same sign? What about the derivative of $f$? is it relevant at all?).
In this question we deal with $\mathbb{R}$ instead of $\mathbb{C}$.
Clearly, in my current question $f$ is not always one-one, for example: $f(x)=x+x^3$. If $f(a)=f(b)$ for $a,b \in \mathbb{C}$, 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 $a=b$, but in the second case, there are two families of solutions with $b \neq a$, so $f$ is not one-one.
Remark: Observe that such $f$ was one-one on $\mathbb{R}$ (= its derivative is positive for all $x \in \mathbb{R}$).
It is one to one if the degree is $1$ otherwise $f(x)=c$ has more than one root for some values of $c$. If $m=deg f>0$, $f$ has $m$ roots with multiplicity counted, if you have two root distincts, done if all the root are equal, $f=(X-a)^m$ and $f(x)=c, c\neq a$ has distinct roots, since if $u$ is a root of $f(x)=c$, $u\neq a$ and $f'(u)\neq 0$ this implies that $u$ is a simple root.