# If $f(g(x))=x$, does $g(f(x))=x$ hold?

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.

• In general, it's not true that $fg=\mathrm{id}$ (the identity function) implies that $gf=\mathrm{id}$. For example, $f:\{0\}\to\{0,1\}$ and $g:\{0,1\}\to\{0\}$; it follows that $gf=\mathrm{id}$, but $fg\neq\mathrm{id}$. In this specific context, I don't know. May 15 at 15:52
• In "formal power series": in general, if $b_0 \ne 0$, then substitution $f(g(x))$ need not even be defined. Example $f(x) = \sum n x^n$ and $g(x) = 2+x$. May 15 at 15:59
• If $a_0$ and $b_0$ are supposed to be $0$, then $n \ge 0$ should be replaced with either $n \ge 1$ or $n > 0$. May 15 at 16:46

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.

• You mean $a_0 = 0$, not $a_0 \ne 0$. May 15 at 16:44
• @RobertIsrael yes. May 15 at 16:45
• @psl2Z sorry I don’t understand the part about the $c_n$ and $a_I$, with the notations like $N_0$, $N^k$. Do you happen to know some links that I can read that up pls? May 17 at 21:53
• @athos $\mathbb{N}$ natural numbers (without $0$) and $\mathbb{N}_0$ natural numbers with $0$. $\mathbb{N}^k$ are $k$-tuples of natural numbers. Basically, you get the formula if you formally plug $g$ in $h$ and multiply all out, like for polynomials. May 21 at 14:24