Proposition 7

Let $f$ be a mapping of A into B. Then $\overset{-1}{f}$ is a function if and only if $f$ is bijective.

If $\overset{-1}{f}$ is a function, its source B is equal to its domain, i.e., to $f(\mathrm{A})$; hence $f$ is surjective. To show that $f$ is injective, let $x$ and $y$ be two elements of A such that $f(x) = f(y)$. If F denotes the graph of $f$, we have

hence

so that $x = y$, which proves the assertion.

¶ Conversely, if $f$ is bijective, it is immediate that $\overset{-1}{\mathrm{F}}$ is functional and that the domain of $\overset{-1}{f}$ is equal to B.

When $f$ is bijective, $\overset{-1}{f}$ is called the inverse mapping of $f$; $\overset{-1}{f}$ is bijective, $\overset{-1}{f} \circ f$ is the identity mapping of A, and $f \circ \overset{-1}{f}$ is the identity mapping of B.

¶ If a permutation is the same as its inverse, it is said to be involutory.