The composition of two involution functions Is the composition of two involution functions always an involution? I think this is probably not the case but would like if someone could provide me with some counter-examples. 
 A: Consider
$$f(z)=\frac{1-z}{1+z}\quad\text{and}\quad\mu(z)=\sqrt{1-z^2}$$
Both are involutions but their composition is not:
$$h(z):=(\mu\circ f)(z)=\sqrt{1-\Big(\frac{1-z}{1+z}\Big)^2}=\frac{2\sqrt{z}}{1+z}$$
In fact, if you start with any value on the positive real line, repeatedly applying $h$ to it gives you a nonrepeating sequence that converges to 1. So for example,
$$2\rightarrow h(2)\rightarrow h^2(2)\rightarrow h^3(2)\rightarrow \ ...\ $$
approaches 1 quadratically and does not repeat. If you are curious, these functions come from the arithmetic-geometric-mean iteration and are related to elliptic integral substitutions.
A: Taking $f(x)=-x$ and $g(x)=1-x$ then neither $g(f(x)) = 1+x$ nor $f(g(x)) = x - 1 $ are involutions. 
A: This is not true in general because our two involutions $h$ and $g$ do not commute in general.
Let $h$ and $g$ be our involutions. They are both invertible, so their composition $h \circ g$ is invertible. However, its inverse is $g \circ h$, not $h \circ g$ .
$$ (h \circ g) ^ {-1} = g^{-1} \circ h^{-1} = g \circ h $$
So $ h \circ g $ is in involution if and only if it is equal to $g \circ h$ .
However, if you have two arbitrary involutions with the same domain/codomain, then you can produce a new involution $k$ through conjugation.
$$ k \stackrel{df}{=} h \circ g \circ h^{-1} = h \circ g \circ h $$
To verify this, let's check $k^{-1}$ .
$$ k^{-1} = (h \circ g \circ h)^{-1} = h^{-1} \circ g^{-1} \circ h^{-1} = h \circ g \circ h $$
To work through an example, let's take reed_de_la_mer's $f$ and $g$ .
$$ f(x) = -x $$
$$ g(x) = 1 - x$$
$$ (f \circ g \circ f)(x) = -(1 - (-x)) = -1 - x $$
$$ -1 - (-1 - x) = -1 + 1 + x = x $$
and the other direction
$$ (g \circ f \circ g)(x) = 1-(-(1-x)) = 2-x $$
$$ 2 - (2 - x) = x $$
