Ïnverse functions and composition

[mnb96]

Hi,
given two functions f and g, is there any known condition under which the following is valid:

$$(f \circ g^{-1}) = (f^{-1} \circ g)$$

Basically I have to find out the requirements for $$f$$ and $$g$$ for which composition is commutative in respect to inversion.

 

[mnb96]

attempt to a solution

I probably did a step forward but I'm still stuck.
We know that: $$(f \circ g)^{-1}=(g^{-1} \circ f^{-1})$$.

This obviously implies that the following must be true:
$$(f \circ g^{-1})=(f \circ g^{-1})^{-1}=(g \circ f^{-1})$$

If consider only the body in the parentheses $$h = f \circ g$$, we have the condition:

$$h = h^{-1}$$

OK! Now, what is the family of functions that fulfills the property $$f = f^{-1}$$?
I already found $$f(x)=a-x$$ and $$f(x)=a/x$$
Are there others?

 

[mnb96]

I just found out that a function such that $$f=f^{-1}$$ is called involution.
I also sketched out a proof which proves that the composition of two involutions is not an involution.

This would partly answer my question, implying that in order to satisfy $$(f \circ g^{-1}) = (f^{-1} \circ g)$$ only one of the two function can be an involution; but still nothing useful is known about the nature of f and g

And still, what remains unanswered is: are $$f(x)=a-x$$ and $$f(x)=\frac{a}{x}$$ the only existing involutions for real functions?

 

[mnb96]

...actually I found out also $$f(x) = \frac{1}{x-a}+a$$
The task of finding out the involutions of real functions is probably less trivial than I thought and I created a new thread for this:
https://www.physicsforums.com/showthread.php?p=2094676#post2094676