# Inverse Function – Does g(f(x))=x Hold if f(g(x))=x?

formal-power-seriesinverse function

If $$f(x)$$ and $$g(x)$$ are formal power series, i.e. $$f(x)=\sum_{n\ge 0} a_n x^n, g(x)=\sum_{n\ge 0} b_n x^n,$$
and $$f(g(x))=x,$$
can it be proved that always have $$g(f(x))=x?$$
It seems intuitive, then $$f(x)$$ will be the inverse function of $$g(x)$$ and vice versa. But I could not prove it.

It is true if we assume that the ring of coefficients $$R$$ is an integral domain. In this case it holds: let $$h = \sum b_i X^i,g = \sum a_i X^i \in R[[X]]$$ where $$g \neq 0$$ has no constant term, i.e. $$a_0 = 0$$.
If $$h\circ g = 0$$ then $$h = 0$$.
This can be seen by the formula for the coefficients of the composition: $$c_n = \sum_{k \in \mathbb{N}_0, I \in \mathbb{N}^k} b_k a_I$$ where $$a_I = a_{i_1}...a_{i_k}$$ for $$I \in \mathbb{N}^k$$. It follows inductively: if $$c_n = 0$$ for all $$n \in \mathbb{N}_0$$ then $$b_i = 0$$ for all $$i \in \mathbb{N}_0$$ (because $$R$$ has no zero divisors).
Now, let $$f\circ g = X$$. Then $$f$$ has no constant term and $$g \circ f$$ is defined. Therefore $$(g\circ f) \circ g = g$$ thus $$(g\circ f - X)\circ g = 0$$ and by the above $$g\circ f - X = 0$$, i.e. $$g \circ f = X$$.
Probably there is some easy counterexample if $$R$$ has zero divisors, but I don't have one right now.