Composition of entire functions is identity, then functions are linear

Question

A question from a previous qualifying exam at my university reads:

"Suppose that f and g are entire functions such that $f \circ g(x) = x$ when $x \in \mathbb{R}$. Show that $f$ and $g$ are linear functions."

One can conclude that the composition of $f$ and $g$ is the identity on all of $\mathbb{C}$, by the uniqueness principle. I know how to solve the problem if one assumes that $f$ is injective. However, there are examples of functions that have a right inverse but are not injective. However, entire functions have many properties, so is there a way of showing $f$ must be injective from the information above, or should I approach the problem differently?

Answer 1

Injectivity of $f:$ Note that $$(g\circ f) (g(z)) = g[(f\circ g)(z)] = g(z).$$ Thus $g\circ f$ is the identity on $g(\mathbb C).$ Since $g(\mathbb C)$ is a set with limit point, the identity principle shows $g\circ f$ is the identity on $\mathbb C.$ This proves $f$ is injective.

Answer 2

Since $f\circ g$ is the identity, $g$ is injective. An injective entire function is a linear polynomial.

Indeed, by the great Picard theorem, if it has an essential singularity at infinity, it is not injective. So it must have a pole at infinity (not a removable singularity, or it would be constant). An entire function that has a pole at infinity is a polynomial.

If it is not a linear polynomial it is not injective: indeed, it cannot have distinct roots, so it is of the form $a(z-b)^n$, with $a\ne0$. The equation $a(z-b)^n=a$ has $n$ distinct solutions, so we conclude $n=1$.

Thus $g(z)=a(z-b)$, so $f(z)=a^{-1}z+b$.