# There cannot exist a rational function $f: \mathbb{R} \to \mathbb{R}$ injective, not surjective

I was looking for a rational function $f: \mathbb{R} \to \mathbb{R}$ that looks like $\arctan$, in that it is

• injective
• not surjective
• well-defined on all $x\in \mathbb{R}$ (no vertical asymptotes)

I believe I have proven that one can't exist.

My thinking is this: it can't be a polynomial, because an odd-degree polynomial is surjective and an even degree polynomial is not injective. Therefore it must have a nontrivial denominator. The denominator can't be an odd degree polynomial, or else it has a zero somewhere. If the numerator is a polynomial of degree greater than the denominator $N>D$, then we get surjectivity (if $N-D$ is odd), or we lose injectivity (if $N-D$ is even). If $N\leq D$, then we lose injectivity, since the function will approach the same limit as $x \to \pm \infty$ (I claim that together with continuity, this implies that we lose injectivity).

Do you agree? Is there a general result about such functions?

• $1/(x^2+1)$ is not surjective. So, "The numerator must be a polynomial of the same degree, or else we get surjectivity" is not true. – lhf Jun 16 '14 at 15:05
• Consider the function $$f(x)=\int_{0}^{x}\frac{dt}{1+t^{4}}$$ – TheOscillator Jun 16 '14 at 15:07
• @lhf You are correct. I attempted to fix this – Eric Auld Jun 16 '14 at 15:07
• @TheOscillator That is not a rational function. – Eric Auld Jun 16 '14 at 15:07
• Looks good to me. – TonyK Jun 16 '14 at 15:15

This looks good to me as well, though the various cases could be cleaned up a bit: the key point is that the range of an injective continuous function is the interval between its two limits, and that for a rational function $f(x)/g(x)$ the two limits can only differ when $\deg f > \deg g$ in which case the interval is $(- \infty, \infty)$.
It's hard to say how to generalize this result: in $\mathbb{C}$ for instance all continuous rational functions are polynomials, and all injective polynomials are surjective. Over any proper subinterval of $\mathbb{R}$, you can find an injective rational functions with whatever values you like on the two boundaries.