Checking whether a polynomial of high degree is bijective or not. Let $P(x)$ be a polynomial of degree $101$. Then $x\mapsto P(x)$ cannot be a one-one onto mapping, i.e., bijective function from $\Bbb{R}$ to $\Bbb{R}$. True or false?
I think is when we take $P(x)=P(y)$, we get :
$$
ax^{101}+bx^{100}+\ldots+cx+d = ay^{101}+by^{100}+\ldots+cy+d
$$
or
$$
a\left(x^{101} - y^{101}\right)+b\left(x^{100} - y^{100}\right)+\ldots+c(x-y) = 0
$$
or
$$
(x-y)\cdot(\textrm{some equation in }x,y) = 0.
$$
Thus we get either $x=y$ or some other expression other than $x=y$.
As a result, $P(x)$ can't be one-one as had it been so, we should have got only $x=y$ as the solution to $P(x)=P(y)$.
Can you please let me know if I am doing it right. If not, what's the mistake I am making?
To check if $P(x)$ is onto, we need to write $y=P(x)$ and then find the expression in terms of $x$ and see if the range is equal to the co-domain. But I am not able to visualize how to check the surjectivity of the function. Can someone help me check if $P(x)$ is onto or not? I am a bit confused here. Thanks for all the help.

Although (as Javier demonstrated) not all polynomials of odd degree are injective, they are all surjective. Say $p(x) = a_n x^n + \ldots + a_0$, where $n$ is odd. Then if $a_n > 0$, we will have $\lim_{x\to+\infty} p(x) = +\infty$ and $\lim_{x\to-\infty} p(x) = -\infty$. Since polynomials are continuous, we have to have hit everything in between somewhere along the way, so $p(x)$ is surjective (if $a_n < 0$, then $\lim_{x\to+\infty} p(x) = -\infty$ and $\lim_{x\to-\infty} p(x) = +\infty$). This happens because as $\left|x\right|$ becomes large, the highest degree term starts to contribute more and more compared to the other terms, so eventually it overwhelms them and the polynomial will simply take the sign of $a_n x^n$.
