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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.

A linear map $T:V\to W$ between finite-dimensional vector spaces admits a linear left inverse $S:W\to V$ if and only if its kernel of a linear map is $\{0\}$, equivalently if and only if it is injective.

Necessity follows from $Tv=0⟹v=STv=0$. For sufficiency, choose a basis $v_1,\dots ,v_n$ of $V$. Triviality of the kernel of a linear map implies that $T{v}_1,\dots ,T{v}_n$ are linearly independent. By basis extension, they extend to a basis $T{v}_1,\dots ,T{v}_n,w_{n+1},\dots ,w_m$ of $W$. Define $S(T{v}_i)=v_i$ and $S(w_j)=0$, and extend by linearity. Then $ST{v}_i=v_i$ on a basis, proving $ST=I_V$.

The freedom in $S$ lies in its values on a direct-sum complement of the image of a linear map. For example, for $T(x,y)=(x,y,0)$, every linear map $S(u,v,w)=(u+aw,v+bw)$ is a left inverse, with arbitrary scalars $a,b$. A right inverse instead requires $T$ to be a surjective linear map.

True. The function composition $j=g\circ f$ is an injection: $j(a)=j(a^{\prime })$ first implies $f(a)=f(a^{\prime })$ by the injectivity of $g$, then $a=a^{\prime }$ by the injectivity of $f$. Fix one $a_0\in A$, using the given nonemptiness, and defineThe first case is well defined because $j$ is an injection. For every $a\in A$, $h(j(a))=a$. Thus this left inverse satisfies $\boxed{h\circ g\circ f={\mathrm{id}}_A}$, the identity function on $A$.

A left inverse of $f$ is a function $\ell :B\to A$ satisfying $\ell \circ f={\mathrm{id}}_A$. If such $\ell$ exists and $f(a)=f(a^{\prime })$, applying $\ell$ gives $a=a^{\prime }$, so $f$ is an injection. Conversely, suppose $f$ is an injection and fix $a_0\in A$, possible because $A$ is nonempty. DefineThe first value is unique by injectivity, and $\ell (f(a))=a$ for every $a$. ThusNo choice from a family of nonempty sets is needed here: each preimage in the first case is unique, and the second case uses one fixed element.

A right inverse is a function $r:B\to A$ with $f\circ r={\mathrm{id}}_B$. Its existence makes every $b$ the image of $r(b)$, proving that $f$ is a surjection. Conversely, if $f$ is a surjection, every fiber of a function $f^{-1}(\{b\})$ is nonempty. Using the usual axiom of choice for arbitrary sets, choose $r(b)\in f^{-1}(\{b\})$ for each $b$. Then $f(r(b))=b$, givingThe converse in this second equivalence is the right-inverse characterization of the axiom of choice; it is not an unrestricted choice-free theorem about arbitrary sets.

Take $fg=f\circ g$. The assertion can fail on an infinite set. On $X={\mathbb{N}}_0$, set $g(n)=n+1$, $f(0)=0$, and $f(n)=n-1$ for $n\ge 1$. Then $f(g(n))=n$ for every $n$, but $g(f(0))=1\ne 0$. Thus a right inverse need not be a left inverse.

On a finite set it is always true. From $fg={\mathrm{id}}_X$, $g$ is an injection: equality of two $g$-values gives equality after applying $f$. An injective self-map of a finite set is a bijection. Composing $fg={\mathrm{id}}_X$ with $g^{-1}$ shows $f=g^{-1}$, and hence $gf={\mathrm{id}}_X$. This includes the empty set.

False. A left inverse would force $\alpha$ to be injective: if $\alpha u=0$, then $u=\beta \alpha u=0$. Take the surjective linear map $\alpha :{\mathbb{R}}^2\to \mathbb{R}$ given by $\alpha (x,y)=x$. Since $(0,1)$ belongs to its kernel, every composite $\beta \alpha$ annihilates $(0,1)$ and cannot be the identity map on ${\mathbb{R}}^2$.

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.