Function ($f:X\to Y$)

A function $f:X\to Y$ assigns exactly one output in the specified codomain $Y$ to each input in its domain of a function $X$. Its rule may be a formula, a finite table, or another unambiguous assignment. The outputs actually reached form its image of a function $f(X)\subseteq Y$. A codomain states where outputs are allowed; the image of a function records where they actually land. Oscar Levin's introduction makes this distinction explicit. ^[ref].

For example, $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, has the entire real line as its domain of a function and codomain, but its image of a function is $[0,\infty )$. Both $-2$ and $2$ give the output $4$: several inputs sharing an output is permitted. Assigning two distinct outputs to one input is not permitted. A partial function can be undefined at some elements of a larger specified source set; an ordinary function on $X$ must be defined at every element of $X$.

Three properties describe how a function uses its target. An injective function never sends distinct inputs to the same output. A surjective function reaches every member of its specified codomain. A bijection does both, pairing each source element with exactly one target element. These are different requirements, as finite arrow diagrams show.

The chosen codomain matters even when the formula does not change. The function $x\mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is not surjective, whereas the same rule from $\mathbb{R}$ to $[0,\infty )$ is surjective. Both fail injectivity. Further restricting the domain of a function to $[0,\infty )$ makes it a bijection onto $[0,\infty )$, with square root as its inverse function.

Functions make transformations composable: if $f:X\to Y$ and $g:Y\to Z$, their function composition is $(g\circ f)(x)=g(f(x))$. Identifying each intermediate codomain prevents attempting to feed an output into a rule that is not defined there. When a transformation is a bijection, an inverse function recovers the original input without ambiguity. This is useful whenever information must be transported or a calculation reversed.

A function taking values in the complex numbers can be written uniquely as $f=u+iv$ with real-valued functions $u,v$. Its Lebesgue integral is defined by integrating these components when $|f|$ is integrable. Multiplication by a complex exponential of modulus one leaves this integrability unchanged.

The empty function has empty domain and empty graph. For any specified codomain there is exactly one function from the empty set into it. This explains why a cardinal product over an empty index set is one, including the convention $0^0=1$ in cardinal exponentiation.

A variable whose value is specified as a function of other variables. In an ordinary differential equation, the unknown function $y$ is the dependent variable, and its argument $x$ is the independent variable.

For two mathematical functions $f,g:A\to B$ with the same domain and codomain, equality is equivalent to equality of their values at every argument. In the set-theoretic representation, their graphs of functions then contain exactly the same ordered pairs. This permits a function identity to be proved by fixing an arbitrary argument and proving the resulting value equality.

A coordinate of a function on a finite Cartesian product is essential if changing that coordinate, with all the others fixed suitably, can change the output. A coordinate is inessential precisely when such a change never affects the output in any context. If only coordinate $i$ is essential, changing the other coordinates one at a time proves that the output is $g(x_i)$ for a nonconstant unary function $g$. This deterministic dependence definition is not by itself a probabilistic or experimental notion of causal inference.

The function taking the greater value at every common-domain point. For finitely many affine functions it is a convex function, by bounding each affine branch by the convex combination of the endpoint maximum values. This is the finite-affine case of pointwise maximum of convex functions.

The function taking the lesser value at every common-domain point. It forms the lower envelope of two graphs. For finitely many affine functions it is a concave function: each affine branch is at least the corresponding convex combination of the two endpoint minimum values, so taking the minimum preserves that lower bound. A maximizing mixed strategy in a two-player zero-sum game maximizes such a lower envelope of opposing pure-choice expected payoffs.

A scalar-valued function is a function whose codomain is a field, typically the real numbers or complex numbers. Its output is a single scalar, while its input can belong to an arbitrary set. For example, the square of the norm of a vector is real-valued. In a relativistic application, being scalar-valued alone does not establish Lorentz invariance: the transformation of the input and of the complete quantity must also be specified.

An implicit function is a function specified through a relation between its input and output, rather than an explicit output formula. Locally, an equation $F(x,y)=0$ may specify $y=g(x)$ if one output $y$ is selected for each input $x$ in a suitable domain. Different branches can define different functions; the relation itself need not give a unique output globally. For example, $x^2+y^2=1$ gives the two branches $y=\pm \sqrt{1-x^2}$ on $-1<x<1$.

The implicit function theorem gives sufficient conditions for a unique local differentiable branch, while implicit differentiation computes its derivative by the chain rule. If $F$ is continuously differentiable and $F_y\ne 0$, that derivative is $g^{\prime }(x)=-F_x/F_y$ along the branch.

Function evaluation determines the function output $f(x)$ for a specified function input $x$ in the domain of a function. Numerical integration uses these values at selected quadrature nodes.

A function input to a function $f:X\to Y$ is an element $x$ of its domain of a function $X$, also called its argument. The corresponding function output is $f(x)$.

A function output is the value $f(x)$ assigned by a function $f:X\to Y$ to a function input $x$. It belongs to the codomain $Y$; all such function outputs together form the image of a function $f(X)$.

For functions $f$ and $g$ taking values in the same field, their quotient of functions is defined by $(f/g)(x)=f(x)/g(x)$ on the common domain of a function restricted to points where $g(x)\ne 0$. Thus $f(x)=1$ and $g(x)=x$ on the real line give a quotient of functions defined only for $x\ne 0$.

The codomain of a function is its specified target set. The image of a function is a subset of the codomain, and equals it precisely for a surjective function. Changing a codomain can change whether the same input-output rule is surjective, as the quadratic function examples under function show.

A self-map is a function whose domain of a function and codomain are the same set. It need not be a bijection: a constant function on a two-element set is a self-map that merges its inputs. Self-maps can always be composed with one another, and all self-maps of a specified set form its full transformation monoid.

A self-map $f:X\to X$ is idempotent if its function composition with itself leaves it unchanged: $f(f(x))=f(x)$ for every $x$. Equivalently, every element of its image of a function is a fixed point. For example, the function on $\{a,b,c\}$ sending $a$ to $b$, $b$ to $b$, and $c$ to $c$ is idempotent. A second application makes no further change.

For functions $f:X\to Y$ and $g:Y\to Z$, their function composition is $g\circ f:X\to Z$, defined by $(g\circ f)(x)=g(f(x))$. The rightmost function is applied first. Matching the intermediate codomain and domain of a function makes the expression well-defined.

For self-maps of the real numbers, take $f(x)=x+1$ and $g(x)=2x$. Then $(g\circ f)(x)=2x+2$, whereas $(f\circ g)(x)=2x+1$. Thus function composition need not be commutative. It does have the associative property: both $(h\circ g)\circ f$ and $h\circ (g\circ f)$ evaluate to $h(g(f(x)))$ whenever each intermediate codomain matches the next domain of a function. The identity function leaves composition unchanged, making self-maps particularly useful as a monoid of transformations.

For $f:A\to B$ and $g:B\to C$, the following implications hold. If both are injective, equality of two composite images can be cancelled first through $g$ and then through $f$, proving that $g\circ f$ is injective. Conversely, if $g\circ f$ is injective, equality under $f$ implies equality under the composite, so $f$ is injective. If $g\circ f$ is surjective, every $c\in C$ is $g(f(a))$, so $g$ is surjective.

There are two useful strengthened cancellation rules. If $g\circ f$ is injective and $f$ is surjective, lift any $b,b^{\prime }\in B$ with $g(b)=g(b^{\prime })$ to preimages under $f$; composite injectivity then proves $b=b^{\prime }$. If $g\circ f$ is surjective and $g$ is injective, apply composite surjectivity to $g(b)$ and cancel $g$ to find a preimage of every $b$. Neither strengthened conclusion follows from the composite property alone.

A germ at a point is an equivalence class of functions defined on a neighbourhood of the point: two representatives are equivalent if they agree on some smaller neighbourhood. A germ of a holomorphic function requires its representatives to be holomorphic functions; smooth and continuous germs use their corresponding classes of functions. The germ retains local behavior while forgetting the representative's values farther from the point.

A real-valued function is nonnegative when all its values are at least zero. Products and nonnegative linear combinations of nonnegative functions remain nonnegative. A nonnegative partition of unity gives convex combinations, hence a convex-hull bound for a geometric basis.

Two self-maps $R,S:X\to X$ commute when $R\circ S=S\circ R$. Induction then gives $R^n\circ S=S\circ R^n$ and $R^n\circ S^m=S^m\circ R^n$ for all nonnegative integers $m,n$. Sharing a fixed point or invariant set does not by itself imply commutation. For rational sphere maps of degrees at least two, commutation does imply equality of their Fatou sets and Julia sets, by the commuting rational maps of degree at least two have the same Julia set theorem.

A finite modification agrees with the identity map outside a finite exceptional set. Sorting that set gives a unique finite record of pairs $(x,f(x))$. Real-number pairing by separated digits encodes the record in $(0,1)$, and placing length-$n$ records in disjoint intervals $(n,n+1)$ gives an injection from all finite modifications into the real line. Conversely, changing the value at one fixed point gives a real-parameter family, so the Cantor-Schröder-Bernstein theorem shows this family has continuum cardinality.

A pointwise periodic self-map returns each point to itself after a positive number of iterates which may depend on the point. It is a bijection: each finite cycle supplies predecessors; equality of two images can be undone by an iterate with an exponent one below a common multiple of the two periods. The set consequently decomposes into disjoint finite permutation cycles. Pointwise periodicity does not imply that one iterate is the identity on an infinite set.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.

A vector-valued function is a function with values in a vector space. For finite-dimensional real or complex target spaces, choosing a basis represents it by scalar coordinate functions; continuity and differentiability can then be checked coordinatewise.

A real-valued set function assigns a real number to each subset of a specified ground set.

On nonempty subsets of a nonempty finite set $V$, take the maximum of the contained element weights. Setting $f(\varnothing )={\min }_{v\in V}w(v)$ extends this to a monotone set function and a submodular set function. More generally any value at the empty set at most every element weight works, because ${\Delta }_{x}f(S)=\max \{0,w(x)-f(S)\}$ satisfies diminishing returns.

The marginal increment measures how much a set function changes when one new element is added. Usually $x\notin S$ is required; if $x\in S$, the displayed difference is zero. The marginal increments of a modular set function depend only on $x$. The marginal increments of a submodular set function can only decrease as the input subset grows without acquiring $x$.

On a finite ground set, a modular set function is a set function satisfying the displayed equality for every pair of subsets. Equivalently,To prove this representation, apply the equality to disjoint $S$ and $\{x\}$: $h(S+x)-h(S)=h(\{x\})-h(\varnothing )$. Add these fixed marginal increments along any ordering of $S$. Conversely, the contributions of each element and of the constant cancel in the displayed identity. Thus a modular set function need not be a normalised set function; its normalised instances are precisely the real-valued set functions with finite additivity on a finite power set. Adding or subtracting a modular set function preserves submodularity.

A set function on the power set of a ground set $V$ is symmetric when complementary subsets have equal values. A hypergraph cut function has this property because swapping the two sides changes no crossing hyperedge. If a symmetric set function is also a monotone set function, it is constant: $f(\varnothing )\le f(S)\le f(V)=f(\varnothing )$.

A real-valued set function is monotone, in the nondecreasing convention, when enlarging its input subset cannot decrease its value. Equivalently, each marginal increment ${\Delta }_{x}f(S)$ is nonnegative. This follows by adding elements individually; the converse follows by considering $S\subseteq S\cup \{x\}$. A normalised set function with this property is a nonnegative function.

A set function is normalised when its value at the empty set is zero. Subtracting the constant $f(\varnothing )$ normalises any real-valued set function and preserves submodularity and the monotone set function property.

A real-valued set function on an algebra of sets is finitely additive if it has the displayed property for disjoint pairs. It follows that $F(\varnothing )=0$. On a finite ambient set, induction gives $F(A)={\sum }_{x\in A}F(\{x\})$, so arbitrary signed singleton weights describe all such functions. Summing the pointwise indicator function identity for the inclusion-exclusion principle against these weights proves the same formula for $F$.

A real-valued set function is supermodular if it satisfies the displayed inequality. Its negative is a submodular set function. Equivalently, the gain from adding an element cannot decrease as the set grows. This is the defining property of a convex cooperative game, and makes coalition marginal allocations lie in the core of a cooperative game.

A set function is submodular if $f(A)+f(B)\ge f(A\cup B)+f(A\cap B)$ for every pair of subsets. Entropy submodularity is a fundamental example.

Every real-valued submodular set function on a finite ground set $V$ is the sum of a normalised set function $g$ that is nonnegative, monotone and submodular, and a modular set function $h$. A useful explicit choice isProof. The formula makes $h$ a modular set function, so $g$ is a submodular set function. It is a normalised set function because $h(\varnothing )=f(\varnothing )$. For $x\notin S$, diminishing returns gives ${\Delta }_{x}f(S)\ge {\Delta }_{x}f(V\setminus \{x\})=d_x$, hence ${\Delta }_{x}g(S)\ge 0$. Thus $g$ is a monotone set function, and starting at zero makes it a nonnegative function. The decomposition separates all potentially negative terminal marginal increments into the simple additive term. A constant term in $h$ is necessary unless $f$ is already a normalised set function.

Suppose $f$ is a normalised set function and a submodular set function, and $T_1,\dots ,T_t\subseteq T$ cover every element of $T$ exactly $k$ times, counting repetitions in the indexed collection. Then the displayed inequality holds. Neither the monotone set function property nor the nonnegative function property is required.

Proof. Fix an ordering of $T$ and write $a_j=f(P_j)-f(P_{j-1})$ for its successive marginal increments. The chain lower bound for a submodular set function gives $f(T_i)\ge {\sum }_{x_j\in T_i}{a}_j$. Summing counts each $a_j$ exactly $k$ times, and ${\sum }_j{a}_j=f(T)$. Exact coverage matters because the $a_j$ may be negative. For a monotone set function, the same argument also allows each element to occur at least $k$ times.

Let $f$ be a normalised set function that is a submodular set function, and order a finite subset $T=\{x_1,\dots ,x_n\}$. Set $P_j=\{x_1,\dots ,x_j\}$ and $P_0=\varnothing$. Then the displayed inequality holds for every $U\subseteq T$, with equality for $U=T$. The right side is a modular set function lying below $f$ on the power set of $T$ and touching it at the top.

Proof. Reveal the elements of $U$ in the chosen order. Immediately before adding $x_j\in U$, the conditioning set is $U\cap P_{j-1}\subseteq P_{j-1}$. By diminishing returns, its marginal increment is at least $f(P_j)-f(P_{j-1})$. Sum and use $f(\varnothing )=0$. The same telescoping sum gives equality at $T$. The individual increments need not be nonnegative.

For a real-valued set function on a finite ground set, the displayed marginal increment comparison is equivalent to submodularity. The interpretation is that an additional item contributes no more after more items have already been chosen. This concerns relative increments, which may be negative; it does not assert the monotone set function property.

Proof. Applying the defining inequality for a submodular set function to $A+x$ and $B$ gives the comparison immediately. Conversely, list $A\setminus B=\{x_1,\dots ,x_m\}$ and compare adding these elements, in the same order, first to $A\cap B$ and then to $B$. At every step the first conditioning set is contained in the second, so the first marginal increment is at least the second. The telescoping sums givewhich rearranges to submodularity.

A function collision is a pair of distinct inputs $x\ne x^{\prime }$ with $f(x)=f(x^{\prime })$. An injective function has no collisions. For a two-to-one function every nonempty fiber has two elements and hence one unordered collision pair. Finding such a pair through an oracle is the task of quantum collision finding.

In the finite example under image and preimage of a function, the pair $\{a,b\}$ is a function collision: both inputs give zero. The function retains their shared output but loses the distinction between them. If a finite set has more inputs than the number of available outputs, the pigeonhole principle forces a function collision. Equal source and target cardinalities alone do not prevent one.

For a finite function, let $m_y$ be the cardinality of its fiber of a function over $y$. The number of unordered function collisions is ${\sum }_y\left(\genfrac{}{}{0ex}{}{m_y}{2}\right)$, because each pair must lie in one common fiber. The example's nonempty fiber sizes are $2,1,1$, giving exactly one collision pair.

For a function $f:X\to Y$ and $y\in Y$, its fiber of a function over $y$ is the set of all inputs producing that one output:Here $f^{-1}(y)$ is shorthand for a preimage of a singleton set, rather than a claim that an inverse function exists. A fiber can contain several inputs or be empty. It describes exactly which distinctions the function forgets.

For the finite function under image and preimage of a function, the fibers over $0,1,2,3$ are $\{a,b\}$, $\{c\}$, $\{d\}$ and the empty set, respectively. The nonempty fibers form a partition of a set: they are disjoint sets whose union is the domain of a function. Every input belongs to the fiber over its unique output, and two fibers with distinct output labels cannot share an input. A function is injective exactly when every fiber has at most one member; it is surjective exactly when every member of its specified codomain has a nonempty fiber.

More generally, $f^{-1}(U)$ is the union of the fibers over all $y\in U$. The relation $x\sim x^{\prime }$ defined by $f(x)=f(x^{\prime })$ is an equivalence relation, and its equivalence classes are precisely the nonempty fibers. The resulting quotient set is naturally in bijection with the image of a function. This links an input partition to the output values actually reached, without choosing one representative from each class.

Fibers are useful when an output is known but the input is not uniquely determined. A solution set $g^{-1}(\{0\})$ is one fiber, and factorization through a surjection explains when another function depends only on the retained output. Quantum hidden subgroup problem algorithms use functions whose fibers are cosets, giving the same idea a structured form in group theory.

For a surjective function $f:B\to A$ and a function $g:B\to C$, there is a unique function $h:A\to C$ with $g=h\circ f$ if and only if $g$ is constant on each fiber of a function of $f$. Define $h(a)$ to be the common value on $f^{-1}(a)$; it is well-defined and unique because that fiber is nonempty. This uniquely specified value requires no simultaneous arbitrary choices.

For a concrete example, restrict the codomain of the finite function above to its image of a function $I=\{0,1,2\}$, making $f:A\to I$ surjective. Define $g(a)=g(b)=5$, $g(c)=7$ and $g(d)=9$. Then $h(y)=2y+5$ on $I$ gives $g=h\circ f$: knowing the output of $f$ is enough to compute $g$. If instead $g(a)\ne g(b)$, no such $h$ can exist, because $h(f(a))=h(f(b))$ must be the same value.

The surjectivity assumption supplies uniqueness on the full target. If the original unused target value three were retained, the factorization would impose no condition on $h(3)$, and different extensions could agree on every actual input. Thus factorization through a surjection formalizes when a compressed description retains all information needed for a later calculation.

// Target: combinatorics.bigb

Translation shifts the argument without changing the shape of a function. On the real line, the displayed convention shifts a graph to the right by $h$. Lebesgue measure invariance preserves every finite Lp norm; its Fourier transform is multiplied by $e^{-ih\xi }$.

For a function $f:A\to B$, its graph is the subset $\{(x,f(x)):x\in A\}$ of $A\times B$. It differs from a graph in graph theory. A solution curve of a scalar differential equation can be a graph over its independent coordinate.

The map $x\mapsto (x,f(x))$ is continuous into the product topology. It maps a compact space $X$ onto the graph of a function of a continuous $f$. Hence that graph is compact, whether or not the codomain is a Hausdorff space. Projection to $X$ is its continuous inverse.

For a continuous map $f:X\to Y$ with Hausdorff space $Y$, its graph of a function is closed in the product topology. If $y\ne f(x)$, choose disjoint neighborhoods of those two points in $Y$. Continuity supplies a neighborhood of $x$ whose image lies in the neighborhood of $f(x)$; its product with the other neighborhood misses the graph.

A partial function from $X$ to $Y$ assigns at most one value in $Y$ to each element of $X$; it may be undefined outside a subset of $X$.

The union of a chain of partial-function graphs is a partial function: two values at the same input occur together in a larger chain member and therefore agree. Any closure or algebraic identity involving finitely many inputs passes to the union if it holds in every chain member, because one member contains all those inputs.

A weakly linear partial function on the real numbers has a domain closed under rational scalar multiplication and addition, and it preserves addition on that domain. A nonempty domain is a vector space over the rational numbers, and additivity implies rational homogeneity of an additive function by integer multiplication and division. The empty function also satisfies the defining conditions vacuously. This terminology does not assert homogeneity under arbitrary real scalars.

A partial unary operation on a set $A$ is a partial function with specified domain $D\subseteq A$ and values in $A$. A map preserving such operations must send each point of the source domain into the target domain and preserve its assigned value. It need not reflect whether the operation is defined.

Composition of partial functions is strict: $g(h_1(x),\dots ,h_r(x))$ is defined only if every $h_i(x)$ is defined and $g$ is defined on the resulting tuple. An argument discarded by the outer function still has to be defined. A representation in a non-strict programming language therefore needs an explicit sequencing guard.

The domain of a function is the set of inputs on which it is defined.

An algebraic simplification proves equality only where both original and transformed expressions are defined. If a cancelled factor vanishes at a point, the simplified expression may instead describe an extension to that point. For example,holds only when $\cos x\ne 0$ and $\cos x+\sin x\ne 0$. The right side is zero at $x=\pi /2+k\pi$, but the left side is undefined there. A graph of the original function must retain those holes. The extension may be useful, but it is a different function with a larger domain of a function.

For a partial function $f:X\rightharpoonup Y$, the domain is the subset $\{x\in X:f(x)\text{ is defined}\}$. Outside this subset, evaluation of $f$ is undefined.

For a function $f:X\to Y$ and a subset $A\subseteq X$, its restriction $f{|}_A$ keeps the same values but allows only inputs in $A$. Restricting the domain of a function can remove collisions and permit an inverse function, as the nonnegative quadratic function example shows. This changes the allowed inputs; it is different from keeping the inputs and changing the codomain.

A function class is a set whose elements are functions, usually sharing a common domain of a function and codomain.

The support of a function is the closure of the set on which it is nonzero.

If two functions are zero outside finite supports and a binary operation satisfies $F(0,0)=0$, the pointwise result is zero outside the union of those supports. Enumerating their union is therefore sufficient for addition, subtraction, maximum and minimum of signed count functions. It is also sufficient to test their pointwise order, since $0\le 0$ elsewhere. Neither scanning only the left support nor intersecting supports is generally valid when negative values are allowed.

A function has compact support when its support is a compact set.

The identity function on a set $X$ maps each $x\in X$ to itself: ${\mathrm{id}}_X(x)=x$. It changes no information. For every function $f:X\to Y$, function composition satisfies $f\circ {\mathrm{id}}_X=f$ and ${\mathrm{id}}_Y\circ f=f$. The subscripts specify the appropriate source or target set. An inverse function reverses a bijection exactly when the two possible compositions are these identity functions.

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.

An inverse branch of a function $f$ on a set $J$ is an inverse function $g:J\to I$ of a bijective restriction $f{|}_I:I\to J$. Different restrictions can give different branches on the same $J$. Their function compositions describe successive preimages and itinerary cylinders in an interval map.

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.

A projection is a function that fixes the points in its image: applying it twice has the same effect as applying it once. A coordinate projection can be represented this way when its target is identified with a retract; a linear projection additionally respects the vector-space operations.

A projection map from a Cartesian product selects specified coordinates, for example $p_j:{\prod }_i{X}_i\to X_j$ with $p_j((x_i{)}_i)=x_j$.

For a coordinate projection map onto $X_i$, write $X_j$ for the other factor. The map is surjective exactly when $X_i$ is empty or $X_j$ is nonempty. It is injective exactly when $X_i$ is empty or $X_j$ has at most one element. In particular, a surjection onto an empty Cartesian product need not induce a surjection onto a nonempty factor.

A fixed point of a function $f:X\to X$ is an element $x\in X$ satisfying $f(x)=x$.

A fixed point of a function on a partially ordered set that is greater than or equal to every other fixed point. For the order-preserving function $H(X)=A\cap \{s:\mathrm{Post}(s)\subseteq X\}$ on the power set of a finite transition system with no terminal states, its greatest fixed point is the set of states from which every infinite path in a transition system remains in $A$. Iteration from the whole state space decreases to this set; every strict decrease removes at least one state.

Functions $f:X\to X$ and $g:Y\to Y$ are conjugate through a bijection $\sigma :X\to Y$ when $g=\sigma f{\sigma }^{-1}$. The bijection transports orbits and fixed points of every iterate of $f$ to the corresponding objects for $g$.

A piecewise-linear function is linear on each member of a finite or locally finite partition of its domain into intervals or polyhedral pieces.

Between two prescribed values, linear interpolation assigns $(1-t)f(a)+tf(b)$ at $(1-t)a+tb$, for $0\le t\le 1$. Joining successive samples produces a piecewise linear function. For a continuous function on a compact interval, interpolants along grids with mesh tending to zero converge in the uniform norm; their error is bounded by the modulus of continuity at the mesh size.

A linear interpolant is the unique polynomial of polynomial degree at most one taking prescribed values $y_a,y_b$ at distinct interpolation nodes $a,b$. It is an affine function, including the constant function case. For $a\le x\le b$, its two displayed coefficients are nonnegative and sum to one, so changing each sampled value by at most $\delta$ changes every interpolated value by at most $\delta$. Differentiation instead uses the difference of the two samples divided by $b-a$, causing noise amplification by differentiation of a linear interpolant.

A real- or complex-valued function is bounded when the moduli of all its values have one finite upper bound.

An unbounded function has no finite bound on the moduli of its values.

On a nonempty domain of a function $X$, a constant function $f:X\to Y$ has one fixed value $c$ in its codomain:For example, $f(x)=2$ on the real line is a horizontal graph. Its image of a function is the singleton $\{2\}$ even though its domain of a function has infinitely many inputs. A constant function from the real line to the real numbers has derivative zero and every nonzero real period of a function; there is no smallest positive period. It supplies a reference level from which variation can be measured.

A constant map between vector spaces is a linear map only when its value is the zero vector, since a linear map must preserve zero. A nonzero constant map is still an affine map. In a Fourier series, the constant term records the period average, while the other terms describe oscillation around that level.

A function is nonconstant when two inputs in its domain of a function have different outputs. For example, $f(x)=x^2$ on the real line is nonconstant, whereas the same rule restricted to $\{-1,1\}$ is a constant function.

A real-valued function has codomain contained in the real numbers.

A global maximum of a real-valued function is a value $f(x_0)$ satisfying $f(x)\le f(x_0)$ throughout its domain.

A global maximum is one type of extremum of a real-valued function; global minima reverse the inequality.

The positive part of a real-valued function $f$ is $f_{+}(x)=\max (0,f(x))$. Together with its negative part, it gives $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$.

The negative part of a real-valued function $f$ is the nonnegative function $f_{-}(x)=\max (0,-f(x))$. It records the magnitude of the negative values, rather than retaining their sign: if $f(x)=-3$, then $f_{-}(x)=3$. Together with the positive part, it gives $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. This decomposition defines the Lebesgue integral of a signed function and makes the integrability condition explicit.

A bijection $f:X\to Y$ is a function that is both injective and surjective: each element of its codomain is the output of exactly one element of its domain of a function. The two requirements exclude unused targets and merged inputs respectively. The finite example $f(A)=1$, $f(B)=2$, $f(C)=3$ is a bijection from $\{A,B,C\}$ to $\{1,2,3\}$, whereas adding an unused target $4$ makes that same rule fail surjectivity.

A bijection gives an inverse function by reversing every pairing. Conversely, a two-sided inverse function implies a bijection, as the proof under inverse function shows. This is useful for transporting structures without losing or identifying elements. It also defines equality of cardinality: two sets have equal cardinality exactly when a bijection pairs them. An infinite example is $n\mapsto 2n$ from the integers to the even integers, with inverse $m\mapsto m/2$ on that specified codomain. A proper subset can therefore have the same cardinality as an infinite set. ^[ref].

A periodic function repeats exactly after a nonzero shift of its input. For $f:\mathbb{R}\to Y$, a positive period of a function $T$ satisfiesOne interval of length $T$ therefore determines the whole function. For another domain of a function $D$, a period must also preserve the domain, $D+T=D$; equality only where two domains happen to overlap is weaker. Periodicity is an exact translation of a function symmetry, rather than merely two similar-looking pieces of its graph.

The sine $\sin (2\pi x)$ has fundamental period one. The function ${\sin }^2(2\pi x)$ repeats after half a unit: the sine changes sign under that shift, but squaring removes the sign. The mixed function $g(x)=2+\sin (2\pi x)+\frac{1}{2}\cos (4\pi x)$ again has fundamental period one. All positive integer multiples of a period are periods, so specifying one period need not identify the shortest repetition.

If the input is time and $T_0$ is the fundamental period, the frequency is $1/T_0$ and the angular frequency is $2\pi /T_0$. Repetition lets a Fourier series describe a cycle using sines and cosines; the period average separates the reference level from the oscillating terms. Strang develops this representation and its Fourier coefficients in §4.1 of Computational Science and Engineering by Gilbert Strang.^[ref]

This connection grew out of a physical problem. In his 1822 book Théorie analytique de la chaleur by Joseph Fourier, Joseph Fourier used periodic functions to extend profiles beyond their initial interval while treating heat conduction. Chapter III, §§233–234 combines even functions and odd functions into sine–cosine expansions and rescales their period.^[ref] A modern Fourier series interpretation still requires the appropriate convergence conditions.

A constant function has every real shift as a period and no fundamental period. Without continuity, even a nonconstant periodic function can lack a smallest positive period; periodic function without a least positive period gives a concrete example. The period group of a function organizes all these possibilities.

For an integrable periodic function of period $P>0$, its integral over an interval of length $P$ is independent of the starting point. Write $a=mP+r$ with $m$ an integer and $0\le r<P$. By a change of variables and the defining identity of a periodic function,No differentiability is required. This also allows the Fourier coefficients of a periodic function to be computed on any interval of one full period.

Periodicity is invariance under translation by a nonzero period $P$: a periodic function satisfies $f(x+P)=f(x)$ at every point in its domain, with translation preserving the domain. For an angular coordinate on a full circle, single valuedness requires $2\pi$-periodicity. The sine and cosine factors $\sin (k\theta )$ and $\cos (k\theta )$ are $2\pi$-periodic for nonzero real $k$ precisely when $k$ is an integer. Their Fourier series supply the angular modes for the harmonic expansion on a disk.

For $T>0$, values of a function $f$ on a half-open interval $[a,a+T)$ determine its periodic extension to the real line: choose the unique integer $k$ with $x-kT\in [a,a+T)$ and set $F(x)=f(x-kT)$. The resulting periodic function has period $T$. If the endpoint limits of $f$ differ, the extension has a jump discontinuity at every period boundary. Under the usual piecewise-smooth convergence hypotheses, its Fourier series takes the average of the two one-sided limits there; an assigned value at an isolated endpoint does not change the Fourier coefficients.

For $h\in L^1(0,1)$ with integral zero, $Ph$ is the unique absolutely continuous function with $(Ph{)}^{\prime }=-h$, equal endpoint values, and integral zero. The first formula gives the derivative; zero integral of $h$ gives equal endpoint values, while the added constant makes the mean zero. Uniqueness follows because the difference of two such primitives has zero weak derivative and therefore is constant, then its mean is zero. Its periodic extension has distributional derivative $-\overline{h}$, with no endpoint Dirac delta distribution.

Writing $Rh(x)=x{\int }_0^{1}h-{\int }_0^{x}h$, the same operator on zero-mean inputs is $Ph=Rh-{\int }_0^{1}Rh$. For arbitrary $h\in L^1(0,1)$,The first inequality follows from the integral kernel $x-1_{\{t<x\}}$, whose absolute value has integral $(t^2+(1-t{)}^2)/2\le 1/2$ in $x$. The second follows by the triangle inequality. This bound permits arbitrarily many iterations without increasing the L1 space norm.