Given $P\colon\mathbb{R} \to \mathbb{R}$ , $P$ is injective (one to one) polynomial function. Prove formally that $P$ is onto $\mathbb{R}$ Given $P\colon\mathbb{R}  \to \mathbb{R}$ , $P$ is injective (one to one) polynomial function i need to formally prove that $P$ is onto $\mathbb{R}$
my strategy so far .......
polynomial function is continuous and since it one-to-one function it must be strictly monotonic and now i have no idea what to do .... 
there is a theorem saying all continuous and monotonic functions have an inverse function and another theorem saying function have an inverse function if and only if its one-to-one and onto ... 
but for formal proof of the on-to function property i think i need to show here that every element in the co-domain/target-set have a "source" in the domain .
i don't think its useful for this proof but only a polynomials of odd degrees have the one-to-one property .   
 A: First, we show by induction that if $P$ has degree greater than or equal to 1, $|P(x)|$ can be made arbitrarily large by choosing $x$ arbitrarily large. 
If $P$ is of degree 1, it is of the form $P(x) = a_0 + a_1x$ so for any $y_0$ we wish to obtain we can choose $x = -a_0/a_1$. If $P$ is of degree 2, it is of the form $P(x) = a_0 + a_1x + a_2x^2$ which can be rewritten in the form $P(x) = p(x-h)^2 + k$ ($p,h,k\in\mathbb{R}$ always exist). Then, if $p$ is positive, any $y_0 > k$ can be found using $x = \sqrt{(y_0 - k)/p}+h$. A symmetric argument applies for negative $p$.
Now, if $P$ has degree more that 2, from Gauss we know that any real polynomial can be factored into quadratic and linear factors. Since each of these gets arbitrarily large (in magnitude) as $x$ gets big, it is trivial to show that their product also gets arbitrarily large.
Then, using froggie's comment, we can use the intermediate value theorem to get to the solution.
A: Your reasoning is in the right direction. A polynomial is continuous and given it is one-one it follows that it is strictly monotone. Now let $p(x)$ be that polynomial and let's assume that it is strictly increasing. Clearly we have the following:
1) $\lim_{x \to \infty}p(x) = L$ or $\infty$
2) $\lim_{x \to -\infty}p(x) = M$ or $-\infty$
But given the form of the polynomial the highest power of $x$ dominates other terms when $x \to \pm\infty$ and hence the finite limits $L, M$ are out of question. It follows that $\lim_{x \to \infty}p(x) = \infty$ and $\lim_{x \to -\infty}p(x) = -\infty$ and by intermediate value theorem the range of $p(x)$ is whole of $\mathbb{R}$.
