A self inverse function is a function $f$, such that $y=f(x)$, with the special property that $ff(x)=x$, or written another way, $f(x) = f^-1(x)$


Imagine there's a function $f$, such that $y = 1/x$

$f^-1(x) = 1/x$, which means that $f(x) = f^-1(x)$, therefore this specific function is said to be a self-inverse function.

Another example:

Let the function $f$ be such that $y = 3-x$

$ff(x) = 3-x$, { so, $ff(x) = x$ } therefore this specific function is a self-inverse function.

What is/are the domain(s) of these types of functions?

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    $\begingroup$ The domain must equal the range. Other than that it's up to the definition of the involution. $\endgroup$ – dxiv Oct 9 '16 at 0:36
  • $\begingroup$ @dxiv, No idea how I missed that, wow. Not everything that's simple is also obvious, I guess. Consider posting this as an answer and I'll accept it. $\endgroup$ – Buffer Over Read Oct 9 '16 at 0:44

Such self-inverse functions are called involutions. Since the function $f : X \to Y$ and its inverse $f^{-1} : Y \to X$ coincide for an involution, the domain and codomain must be the same $X = Y$. Moreover, since $f$ must be a bijection (in order to have an inverse), the range must equal the codomain. So in the end, the domain of an involution $f$ must equal its range $f(X) = X$.

Other than that, the domain of an involution depends on its definition. For example, each of the following is a valid definition of an involution on the respective domain:

  • $f : \mathbb{R} \to \mathbb{R} \quad f(x) = x$
  • $f : [-1,1] \cap \mathbb{Q} \to [-1,1] \cap \mathbb{Q} \quad f(x) = -x$
  • $f : \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\} \quad f(x) = \frac{1}{x}$
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