# Calculus

## Set Theory # Inverse Functions

Suppose $$f : A \to B$$ is a function whose domain is the set $$A$$ and whose codomain is the set $$B.$$ The function $$f$$ is called invertible if there exists a function $$f^{-1} : B \to A$$ with the domain $$B$$ and the codomain $$A$$ such that

${f^{ - 1}}\left( y \right) = x\; \text{ if and only if }\; f\left( x \right) = y,$

where $$x \in A,$$ $$y \in B.$$

The function $$f^{-1}$$ is then called the inverse of $$f.$$

Not all functions have an inverse. If a function $$f$$ is not injective, different elements in its domain may have the same image:

$f\left( {{x_1}} \right) = f\left( {{x_2}} \right) = y_1.$

In this case, the converse relation $${f^{-1}}$$ is not a function because there are two preimages $${x_1}$$ and $${x_2}$$ for the element $${y_1}$$ in the codomain $$B.$$ So, to have an inverse, the function must be injective.

If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. In this case, the converse relation $${f^{-1}}$$ is also not a function.

Recall that a function which is both injective and surjective is called bijective. Hence, to have an inverse, a function $$f$$ must be bijective. The converse is also true. If $$f : A \to B$$ is bijective, then it has an inverse function $${f^{-1}}.$$