Function composition - solving for functions

Question

I have a simple, continuous, real-valued function $\sigma$, whose functional form I know, and I know that two other invertible functions, $f$ and $g$, satisfy the following relationships:

• $f\circ\sigma\circ\sigma\circ f^{-1}=1$,
• $g\circ\sigma\circ\sigma\circ g^{-1}=1$,
• $g^{-1}\circ\sigma\circ f\circ f\circ \sigma \circ g^{-1}=1$,
• $f^{-1}\circ\sigma\circ g\circ g\circ \sigma \circ f^{-1}=1$.

where $(a\circ b)(x)=a(b(x))$ is function composition, $a^{-1}$ is the inverse of function $a$, and $1$ is the identity function.

How do I go about finding $f$ and $g$, or any further relationships between them? If $\sigma=1$ then $f=g=h$ for any invertible function $h$ seems to work, but how about more generally? Any suggestions of topics to investigate very warmly received.

Answer 1

We have that

$$f\circ\sigma\circ\sigma\circ f^{-1}=1\iff \sigma\circ\sigma=1$$

$$g\circ\sigma\circ\sigma\circ g^{-1}=1\iff \sigma\circ\sigma=1$$

$$g^{-1}\circ\sigma\circ f\circ f\circ \sigma \circ g^{-1}=1\iff \sigma\circ f\circ f\circ \sigma=g\circ g$$

$$f^{-1}\circ\sigma\circ g\circ g\circ \sigma \circ f^{-1}=1\iff \sigma\circ g\circ g\circ \sigma=f\circ f$$

If we assume $\sigma =1$ then we have

$$f\circ f=g\circ g.$$ But we can get $f,g$ from the relation above.

Some solutions:

• If $f=g$ we are done.
• Assuming the domain is $\mathbb{R}$ and that $f\circ f=g\circ g=1$ we can have $f(x)=x, g(x)=k-x$ or $f(x)=a-x, g(x)=b-x.$
• Assuming the domain is $\mathbb{R}\setminus\{0\}$ and that $f\circ f=g\circ g=1$ we can have $f(x)=\dfrac 1x, g(x)=k-x$ or $f(x)=a-x, g(x)=b-x.$