Thread: Commutativity of Inverse Functions.

Let, $\displaystyle f(x)$ be a real valued function definited for all reals. Proof (or disproof) that is there exists a function such as $\displaystyle g(f(x))=x$ is true for all reals. Then, $\displaystyle f(g(x))=x$ is always true.

Originally Posted by ThePerfectHacker

Let, $\displaystyle f(x)$ be a real valued function definited for all reals. Proof (or disproof) that is there exists a function such as $\displaystyle g(f(x))=x$ is true for all reals. Then, $\displaystyle f(g(x))=x$ is always true.

Let $\displaystyle f(x)$ be a sigmoid function something like the cumulative
normal functiom, then $\displaystyle \mathbb{R}$ is mapped one one onto
$\displaystyle (0,1)$, and so there is an inverse function $\displaystyle g(y)$ from
$\displaystyle (0,1)$ onto $\displaystyle \mathbb{R}$

Then $\displaystyle f(g(x))=x$ holds for all $\displaystyle x \in (0,1)$ but not for all $\displaystyle x \in \mathbb{R}$

That is the domains of $\displaystyle f\circ g$ and $\displaystyle g \circ f$ are not equal.

RonL