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The image of a function asks which outputs a chosen set of inputs reaches. A preimage asks which inputs reach a chosen set of outputs. Both constructions apply to every function, including one that loses information by sending several inputs to the same output. Their difference is useful whenever a transformation must be followed forward or a condition on its result must be traced back.

Consider the function $f:A\to B$ withEvery input has exactly one output, but zero has two inputs and three has none. The full image of a function is $f(A)=\{0,1,2\}$, a proper subset of the codomain $B$. Selecting inputs $\{a,c\}$ gives image $\{0,1\}$. Selecting outputs $\{0,2\}$ instead gives preimage $\{a,b,d\}$.

The function is not injective, because $f(a)=f(b)$, and is not surjective onto $B$, because three is absent from its image. Restricting only the codomain to $\{0,1,2\}$ makes the same rule surjective; it remains noninjective. This separates the output set actually reached from the target set specified in the definition. The definitions of images of a function and preimages, including the distinction from an inverse function, are given in Richard Hammack, Book of Proof, §12.6: Image and Preimage.

An equation can define a function without giving an explicit formula for it. The implicit function theorem tells us when the constraint $F(x,y)=0$ determines $y$ uniquely from nearby $x$, and makes the resulting function continuously differentiable. It is useful when solving an equation explicitly is cumbersome or impossible: the partial derivatives already reveal whether local recovery is possible.

Let $F$ be a continuously differentiable map from an open subset of ${\mathbb{R}}^m\times {\mathbb{R}}^n$ to ${\mathbb{R}}^n$. Suppose $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map. There are open neighbourhoods $U$ of $x_0$ and $V$ of $y_0$ and a unique continuously differentiable function $g:U\to V$ such that, for $(x,y)\in U\times V$,Thus the nearby level set is a graph of a function. The derivative of this graph isThe matrix inverse acts on the left; the order matters in several dimensions.

For example, take the unit circle, written as the level set $F(x,y)=x^2+y^2-1=0$. At $(x_0,y_0)=(3/5,4/5)$, $F_y=2y_0=8/5\ne 0$, so the local upper branch is $y=g(x)=\sqrt{1-x^2}$. Implicit differentiation givesThis tangent line slope is available directly from the constraint. The negative square root branch is elsewhere on the same circle; it does not destroy uniqueness near the selected upper point.

At $(1,0)$, however, $F_y=0$ and the circle cannot be represented locally as one function $y=g(x)$: just to the left of $x=1$ there are two nearby values of $y$. The circle is still smooth there. Since $F_x=2\ne 0$, use the other coordinate instead: $x=h(y)=\sqrt{1-y^2}$, with $h^{\prime }(0)=0$. A failed coordinate choice is different from a geometric singularity.

The derivative condition is sufficient, not necessary for every possible implicit graph. For $F(x,y)=y^3$, the entire zero set is the smooth graph $y=0$, even though $F_y$ vanishes there. In this case the defining equation itself has a degenerate derivative; the theorem's test gives no conclusion.

The connection to the inverse function theorem is particularly elegant. Augment the constraint to the map $H(x,y)=(x,F(x,y))$, whose Jacobian matrix isIts determinant is $\det D_{y}F$. When this is nonzero, locally invert $H$ and set its second output to zero. The recovered second input is $g(x)$. In this way an implicit constraint becomes an ordinary inverse function problem. ^[ref]

Interval bisection solves a target equation $f(x)=t$ by repeatedly halving a closed interval known to contain a solution. It is the bisection method applied to $g(x)=f(x)-t$, so the target becomes a zero of a function. This reduction from an equation to root finding is described in Demanet's notes.^[ref] Here the emphasis is on a continuous strictly increasing function: the target identifies one input, making interval bisection a way to evaluate its inverse function without an explicit inverse formula.

Let $f:[a_0,b_0]\to \mathbb{R}$ be a continuous function, with $a_0<b_0$ and $f(a_0)\le t\le f(b_0)$. The intermediate value theorem guarantees existence; a strictly increasing function also guarantees uniqueness. Check endpoint equality first. Otherwise, for the current root bracket $[a_k,b_k]$, evaluate its midpoint $m_k=(a_k+b_k)/2$. If $f(m_k)<t$, retain $[m_k,b_k]$; if $f(m_k)>t$, retain $[a_k,m_k]$; if $f(m_k)=t$, return that exact solution. These comparisons preserve the target bracket. General bisection method needs continuity and opposite signs of $g$ at the endpoints, without requiring monotonicity or a unique zero of a function; see limitations of bisection.

An original dimensionless polynomial example isHere $f(0)=0<10<30=f(3)$ and the derivative $f^{\prime }(x)=3x^2+1>0$, so there is exactly one solution. In fact $x_{\ast }=2$; knowing this lets us check the computed absolute error independently of the guarantee. The first four halvings give the following exact rational numbers:

The beauty of interval bisection is that the enclosure itself supplies the certificate: one comparison removes half the possible inputs, regardless of the steepness of the function graph. In exact arithmetic, after $k$ completed halvings without exact termination, the width and the bisection error bound for the new midpoint areThe reported midpoint after update four is $m_4=63/32$, rather than the last tested point $m_3=33/16$. Its certified absolute error is at most $3/32$, and its actual absolute error is $1/32$. For a requested position tolerance $10^{-3}$, the bisection stopping criterion requires eleven halvings: ten give $3/2048>10^{-3}$, while eleven giveThe actual absolute error is $1/4096$. The new midpoint need not be evaluated to obtain this certificate.

The signed output residual is $r=f(m)-t$; the position approximation error is $m-x_{\ast }$. They answer different questions: how close is the output to the target, and how close is the input to the solution? Driscoll and Braun explain why a small residual needs information about the derivative to imply small position error.^[ref] For this particular polynomial, direct factorization givesThus at $m_4=63/32$ the residual is $-13121/32768\approx -0.400421$, despite position absolute error $1/32=0.03125$. If a differentiable function satisfies $0<\ell \le f^{\prime }(x)\le L$ throughout the bracket, the mean value theorem gives the rigorous conversionOur example permits $\ell =1$ and $L=28$ on $[0,3]$. Strict increase alone supplies no positive lower derivative bound: $f(x)=x^3$ is strictly increasing but $f^{\prime }(0)=0$. A small residual alone is therefore not a universal position certificate; even the target equation $10^{-6}(x-2)=0$ has residual $10^{-6}$ at $x=3$, a full unit from its solution.

The guarantees require continuity and correct evaluations, not just narrowing endpoints. For the strictly increasing function $h(x)=x$ when $x<0$ and $h(x)=x+1$ when $x\ge 0$, the target $1/2$ lies between $h(-1)$ and $h(1)$ but is never attained: a jump discontinuity defeats the existence argument. With floating-point arithmetic or an approximate evaluation, an incorrect comparison can discard the true solution. A certified interval enclosure of $f(m)$ wholly below or above $t$ certifies the corresponding decision; an enclosure containing $t$ is inconclusive, rather than proof of equality. If rounding makes the midpoint equal an endpoint, report the precision limit; see bisection stopping criterion for the rounded position bound. Both figures are exact mathematical examples, not measured data.

An inverse element of $a$ in a monoid is an element $b$ satisfying $a\ast b=b\ast a=e$, where $e$ is the identity element. Its notation is $a^{-1}$. Some elements of a monoid can lack an inverse element; a group requires one for every element. Associativity makes a two-sided inverse element unique: if $b,c$ both invert $a$, then $b=b\ast (a\ast c)=(b\ast a)\ast c=c$.

For the full transformation monoid, a self-map has an inverse element exactly when it is a bijection; that element is its inverse function. On $\{0,1\}$, the swap $s(0)=1,s(1)=0$ is its own inverse element, while a constant function has none because it merges the two inputs.

Because the real error function is strictly increasing from $-1$ to $1$, its real inverse function maps $(-1,1)$ onto the real line. For $0<q<1$, the standard normal quantile is thereforeMore generally, the $q$-quantile of a normal distribution with expected value $\mu$ and positive standard deviation $\sigma$ is $\mu +\sigma \sqrt{2}{\mathrm{erf}}^{-1}(2q-1)$. This reverses the area calculation: instead of asking how much probability lies below an endpoint, a quantile asks which endpoint encloses a specified probability. The domain restriction matters; $q=0$ and $q=1$ correspond to infinite endpoints, rather than finite real values of the inverse error function.

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.

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].

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]

A local diffeomorphism is a differentiable map that restricts near every point to a diffeomorphism onto an open subset of its target. The local inverse functions need not combine into a single global inverse function: different parts of the domain of a function can map onto the same output region.

For continuously differentiable maps between open subsets of ${\mathbb{R}}^n$, being a local diffeomorphism is equivalent to having an invertible derivative at every point. The inverse function theorem proves one direction. For the other, the chain rule for each local inverse function gives two inverse linear maps. The periodic angle example under the inverse function theorem has this property everywhere, even though it has infinitely many inputs for every point of the punctured plane. The noninjective map with constant Jacobian determinant gives another useful distinction: even a constant nonzero Jacobian determinant need not give global injectivity.

Every local diffeomorphism is an open map. To see this, take an open subset of the source and intersect it with the neighbourhoods where the map is a diffeomorphism. Each local image is open, and their union is the full image. Thus nearby outputs remain available locally, despite the possible global ambiguity.