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Bijection and Inverse Function

Definition

A function ff is invertible (or bijective) if it is both injective and surjective. If f ⁣:X→Yf\colon X\to Y is a bijection, then for every y∈Y,y\in Y\text{,} there is exactly one xx such that f(x)=yf(x) = y (there is at least one because of surjectivity, and at most one because of injectivity). This value of xx is denoted f−1(y);f^{-1}(y)\text{;} thus f−1f^{-1} is a function from YY to X.X\text{.} We call f−1f^{-1} the inverse of f.f\text{.}

Interpretation

A bijection establishes a perfect correspondence between two sets: every element of the codomain is paired with exactly one element of the domain. Invertibility formalizes the idea that such a function can be run backwards without ambiguity: from each output one recovers the unique input that produced it.

Two sets X and Y with four points each and arrows from X to Y that pair every point of X with exactly one point of Y, so the arrows can be reversed to give the inverse function.

In the figure every point of YY receives exactly one arrow; for instance f(x1)=y2f(x_1)=y_2, so f−1(y2)=x1f^{-1}(y_2)=x_1. Reversing every arrow gives the inverse f−1 ⁣:Y→Xf^{-1}\colon Y\to X. It satisfies f−1(f(x))=xf^{-1}(f(x))=x for x∈Xx\in X and f(f−1(y))=yf(f^{-1}(y))=y for y∈Yy\in Y.

Each half of the definition is needed for the reversed arrows to form a function. If ff is not injective, some yy receives two arrows, and f−1(y)f^{-1}(y) would have two values. If ff is not surjective, some yy receives no arrow, and f−1(y)f^{-1}(y) is undefined. Whether ff is a bijection therefore depends on the chosen domain and codomain, not only on the formula: x2x^2 is not a bijection from R\mathbb{R} to R\mathbb{R}, but it is one from [0,∞)[0,\infty) to [0,∞)[0,\infty), with inverse y\sqrt{y}, the nn-th root for n=2n=2. In the same way the logarithm with base bb is the inverse of the exponential function bxb^x for b>0b>0, b≠1b\neq1.

A common slip is to read f−1(y)f^{-1}(y) as the reciprocal 1/f(y)1/f(y). For f(x)=x3f(x)=x^3 the inverse gives f−1(8)=2f^{-1}(8)=2, while 1/f(8)=15121/f(8)=\tfrac{1}{512}.

Key Results

  • A function admits an inverse if and only if it is bijective.
  • The inverse of a bijection is itself a bijection.
  • Composition with an inverse recovers the identity function on both domain and codomain.

Examples

  • The function f ⁣:R→Rf\colon\mathbb{R}\to\mathbb{R}, f(x)=x3f(x)=x^3, is bijective and has inverse f−1(y)=y3.f^{-1}(y)=\sqrt[3]{y}\text{.}
  • The function f ⁣:R→Rf\colon\mathbb{R}\to\mathbb{R}, f(x)=x2f(x)=x^2, is not bijective and therefore not invertible.

Prerequisites

Sources

Aliases: bijective function, bijection, one-to-one correspondence, invertible function, invertible, bijective, inverse mapping, bijective map, invertibility, bijectivity, inverse function, one to one correspondence Keywords: inverse function, one-to-one and onto, finding an inverse, conditions for invertibility, when is a function invertible, proving a function is bijective, prove injective and surjective Symbols: element of, infinity, real numbers, square root

Last updated: 27 October 2026 11:52