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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?

MEG
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2 Answers2

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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.

zhw.
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  • I learned from here that one can also use the Liouville's theorem to show the denseness of $g(\mathbb C)$ (+1). – cqfd Jul 02 '20 at 07:44
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    @ShiveringSoldier Right, that step is used in proving CW. Actually the density of $g(\mathbb C)$ is not needed here. I edited my answer. – zhw. Jul 02 '20 at 16:49
  • Once you conclude $f$ is injective, the remainder of the proof is not overly difficult. If $f$ is not a polynomial, then its power series expansion at 0 must have an infinite number of terms. So $f$ will have an essential singularity at $\infty$. Then $f$ will not be injective by Picard's big theorem. If $f$ is a polynomial of degree 2 or higher, then it would not be injective by the fundamental theorem of algebra. So $f$ must be linear. – MEG Jul 03 '20 at 02:37
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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 $f(z)=a^{-1}z+b$.

egreg
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  • You can use Casorati-Weierstrass to see an injective entire function cannot have an essential singularity at infinity – zhw. Jul 01 '20 at 19:35
  • @zhw. “The theorem is considerably strengthened by Picard's great theorem,...” – egreg Jul 01 '20 at 19:38
  • just pointing out the problem at hand has an elementary solution. – zhw. Jul 01 '20 at 19:45