Inverse functions

Let \(A\) and \(B\) be sets and let \(f\subset A\times B\) be a function (\(f:A\to B\) in the alternative notation). Since \(f\) is a relation, one can consider the inverse relation \(f^{-1}\). Sometimes the inverse relation \(f^{-1}\) is a function, and sometimes it is not a function.

Examples

Let \(R\) be the relation \(\{(x,x^2):x\in\mathbb{R}\}\subset\mathbb{R}\times\mathbb{R}\). - \(R\) is a function because for every \(x\in\mathbb{R}\) there is a unique \(y=x^2\) in \(\mathbb{R}\) so that \((x,y)\in R\).

\(R^{-1}\) is not a function because both \((1,-1)\) and \((1,1)\) are in \(R^{-1}\).

Example

Let \(R\) be the relation \(\{(x,\frac{1}{1+x^2}):x\in\mathbb{R}\}\subset\mathbb{R}\times\mathbb{R}\).

\(R\) is a function.
\(R^{-1}\) is not a function because \(0<\frac{1}{1+x^2}\le 1\) for all \(x\), and therefore there is no pair \((x,y)\in R^{-1}\) with \(x=2\).

Examples

Let \(R\) be the relation \(\{(x,x^3):x\in\mathbb{R}\}\subset\mathbb{R}\times\mathbb{R}\).

\(R\) is a function because for every \(x\in\mathbb{R}\) there is a unique \(y=x^3\) in \(\mathbb{R}\) so that \((x,y)\in R\).

\(R^{-1}\) is also a function because for every \(x\in\mathbb{R}\) there is a unique \(y=x^{1/3}\) for every \(x\in\mathbb{R}\) so that \((x,y)\in R^{-1}\).

Examples (p. 239)

The Inverse Function Theorem

Theorem: Let \(F\subset A\times B\) be a function. The inverse relation \(F^{-1}\subset B\times A\) is also a function if and only if \(F\) is bijective.

Inverse functions (definition)

Definition: If \(f:A\to B\) is bijective, then its inverse is the function \[f^{-1}:B\to A.\] We have \[ f^{-1}\circ f:A\to A = i_A. \] and \[ f\circ f^{-1}:B\to B = i_B \]