Inverse function ($f^{-1}$)

An inverse function reverses a bijection. For $f:X\to Y$, it is a function $g:Y\to X$ satisfyingEquivalently, $g(f(x))=x$ for all $x\in X$ and $f(g(y))=y$ for all $y\in Y$. We write $g=f^{-1}$. The superscript denotes reversal through function composition; it does not mean the pointwise multiplicative inverse $1/f(x)$. The preimage notation $f^{-1}(B)$ for a set $B$ is also distinct: a preimage exists even when no inverse function exists.

For a computed example, let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=2x+1$. Solving $y=2x+1$ for $x$ gives $g(y)=(y-1)/2$. Thus $f(1)=3$ and $g(3)=1$; direct substitution gives $g(f(x))=x$ and $f(g(y))=y$ throughout the real line. In contrast, $1/f(3)=1/7$, which does not undo $f(1)=3$.

Why must $f$ be a bijection? If $f(x_1)=f(x_2)$, applying $g$ gives $x_1=x_2$, proving injectivity. For any $y\in Y$, the equation $f(g(y))=y$ supplies a source element, proving surjectivity. Conversely, a bijection supplies exactly one predecessor for each $y$, defining a unique inverse function. A left inverse or right inverse requires only one of the two composition identities.

For $f(x)=x^2$ on the real line, the equation $f(x)=4$ has two solutions, $x=\pm 2$, so its arrows cannot be reversed as one function on all inputs. The restriction to $[0,\infty )$ removes that ambiguity. As a bijection from $[0,\infty )$ onto $[0,\infty )$, it has inverse function $g(y)=\sqrt{y}$. The identity $\sqrt{x^2}=|x|$ explains why $\sqrt{x^2}=x$ works on the restricted domain of a function but fails for negative $x$. ^[ref].

More generally, $(a,b)$ lies on the graph of a function $f$ precisely when $(b,a)$ lies on the graph of a function of its inverse function. This coordinate swap explains the geometric reflection and the horizontal line test. It also explains why inverse sensitivity is reciprocal locally: the inverse function rule divides by the original derivative at the corresponding input, when that derivative is nonzero.

For a real-valued function drawn in the plane, the horizontal line test asks whether every horizontal line intersects its function graph at most once. This is exactly injectivity, since two intersections at the same height mean two inputs have the same output. A missing height is compatible with injectivity, so an inverse function is then defined on the image of a function; it is an inverse on the entire specified codomain only when surjectivity also holds.

Suppose $f$ is continuously differentiable near $a$ with $f^{\prime }(a)\ne 0$. It has a local inverse function $g$ near $b=f(a)$, andThe chain rule applied to $g(f(x))=x$ gives the formula. The nonzero derivative condition also ensures local invertibility; it does not assert a global inverse function on an arbitrary larger domain of a function. For $f(x)=x^2$ at $a=2$, the corresponding output is $b=4$, and the square root's derivative there is $1/4$. At $a=0$, this division is unavailable: the nonnegative restriction still has a square root inverse, but its finite derivative at zero does not exist. ^[ref].

A right inverse of a function $f:X\to Y$ is a function $g:Y\to X$ satisfying $f\circ g={\mathrm{id}}_Y$. It forces $f$ to be a surjective function. A surjective linear map onto a finite-dimensional vector space has a linear right inverse: lift a basis of the target and extend by linearity.

The assertion that every surjective function has a right inverse is equivalent to the axiom of choice. Choice selects one member of each nonempty fiber of a function, giving a right inverse. Conversely, for a family $(X_i{)}_{i\in I}$ of nonempty sets, the projection from the disjoint union $\{(i,x):i\in I,x\in X_i\}$ onto $I$ is surjective. A right inverse selects one element from each $X_i$, giving a choice function.

A left inverse of a function $f:X\to Y$ is a function $g:Y\to X$ satisfying $g\circ f={\mathrm{id}}_X$. It forces $f$ to be injective, since equal images can be mapped back to equal original elements.