Calculus

Set Theory

Set Theory Logo

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:

\[f\left( {{x_1}} \right) = f\left( {{x_2}} \right) = y_1.\]
Not injective function does not have an inverse.
Figure 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.

Not injective function does not have an inverse.
Figure 2.

Thus, to have an inverse, the function must be surjective.

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 \({f^{-1}}.\)

Any bijective function has an inverse function.
Figure 3.

See solved problems on Page 2.

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