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.

Best Answer

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.

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