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The permanent of a matrix adds the products obtained by choosing one entry in each row and one in each column. For an $n\times n$ matrix $A=(a_{ij})$ with entries in the real numbers or complex numbers,Here each permutation $\sigma$ chooses column $\sigma (i)$ in row $i$, and the symmetric group $S_n$ lists every such bijection. The formula resembles the Leibniz formula for determinants, but omits the sign of a permutation. For a two-by-two matrix,The absence of alternating signs makes the permanent of a matrix particularly useful for counting allowed assignments.

For a zero-one matrix, let rows represent people, columns represent jobs, and $a_{ij}=1$ mean that person $i$ can do job $j$. This is the biadjacency matrix of a bipartite graph. A product in the permanent is one exactly when all the selected assignments are allowed; otherwise it is zero. Each surviving permutation assigns every person a different job, which is a perfect matching. Thus the permanent counts perfect matchings. This interpretation and the definition are given in Yufei Zhao, Probabilistic Methods in Combinatorics, §10.2, pp. 178–179.

For a fully enumerated example, takeAmong the six permutations of three columns, exactly three give nonzero products: $(1,2,3)$, $(1,3,2)$ and $(2,1,3)$. These tuples list the chosen column in each successive row. Each product is one, so $\mathrm{perm}A=3$. Their permutation signs are $+1,-1,-1$, giving $\det A=-1$. The diagram displays all three allowed assignments.

A zero determinant need not mean that no assignment exists. The two-by-two all-ones matrix has permanent $2$ and determinant $0$: its two assignments cancel in the determinant. The nonzero determinant criterion for a perfect matching is sufficient for existence, but the converse fails for a fixed zero-one matrix.

For nonnegative entries, the permanent instead sums assignment weights, with each assignment weighted by the product of its selected entries. It is then a total weight rather than necessarily an integer count. With arbitrary entries in the real numbers or complex numbers, even the permanent can have cancellation; the counting interpretation specifically uses zero-one entries. The permanent as probability of distinct independent choices gives another useful nonnegative interpretation.

There are $n!$ terms in the defining sum, linking direct evaluation to the factorial. For the all-ones $n\times n$ matrix, every term is one and the permanent is exactly $n!$; the identity matrix has permanent one. Finding one allowed assignment and counting all allowed assignments are different questions. Their relationship to computational complexity theory was studied in L. G. Valiant, The complexity of computing the permanent, Theoretical Computer Science 8(2), 189–201, 1979. The elementary sum is useful for small examples, while the existing dense permanent approximation describes a more specialized way of estimating large counts.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

For a bijection $f$, composing $f^2=f^{-1}$ with $f$ gives $f^3=\mathrm{id}$, and composing the latter with $f^{-1}$ gives the converse. Every orbit therefore has size one or three: the permutation consists of fixed points and disjoint three-cycles. This applies to an infinite underlying set as well as a finite one, since each individual orbit has at most three elements.

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.

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

The power set $\mathcal{P}(X)$ is the set of all subsets of $X$, including the empty set and $X$ itself. It is useful whenever an object is a selection rather than an individual member: a possible collection of active vertices in a graph, for example, belongs to the power set of the vertex set.

For $X=\{a,b,c\}$, the power set has eight members:Each of the three set elements is either included or excluded. More generally, if $|X|=n<\infty$, these $n$ independent binary choices give $|\mathcal{P}(X)|=2^n$. This is a bijection between subsets and binary strings of length $n$, not a probabilistic independence assumption.

Ordering the subsets by inclusion gives a Boolean lattice. In the picture an upward edge adds one member. Complementation reverses inclusion, and union and intersection combine selections. The empty-input case is worth checking: $\mathcal{P}(\varnothing )=\{\varnothing \}$ has one member, not zero, consistently with $2^0=1$.

For distinct primes $p_1,\dots ,p_r$, let $M$ consist of the positive integers having no other prime factors, including $1$. The map from a number to its exponent tuple is a bijection from $M$ to ${\mathbb{N}}^r$ when $0\in \mathbb{N}$. The fundamental theorem of arithmetic proves that the tuple is uniquely defined and that equality of tuples implies equality of numbers. Every nonnegative tuple gives the allowed number $p_1^{e_1}\cdots p_r^{e_r}$, proving surjectivity. Multiplication of numbers corresponds to coordinatewise addition of tuples, with $1$ corresponding to the zero tuple.

A primitive recursive pairing function is a pairing function whose encoding function and both coordinate functions of its inverse function are primitive recursive functions. Thus the bijection can be used in either direction inside a construction by primitive recursion and function composition in recursion theory. A decoder that is merely a computable function would not establish this stronger closure property. Sutner's definition of a primitive recursive pairing system explicitly includes the decoding functions; his more general systems need only be injective, whereas a pairing function here is a bijection.^[ref]

The Cantor pairing function is a primitive recursive pairing function. For its encoding, construct the triangular numbers directly by primitive recursion:Since addition is primitive recursive, so are $T$ and $\pi$ by primitive recursion and function composition in recursion theory. This establishes the claim without treating division by two as an unexplained primitive operation.

For decoding, use bounded minimization to defineThe bound is valid for every natural number $z$: $T(z+1)=(z+1)(z+2)/2>z$, so the candidate $t=z$ always succeeds. The search includes zero; when $z=0$, it immediately returns $s(0)=0$. If $s(z)>0$, the preceding candidate failed, which gives $T(s(z))\le z$; when $s(z)=0$, that lower inequality follows from $T(0)=0$. Thus this search finds exactly the diagonal in the Cantor pairing function inverse, withThe comparison is primitive recursive, and bounded minimization of a primitive recursive predicate is primitive recursive. Therefore $s$ is a primitive recursive function. Recover the coordinates byIn constructions on the natural numbers, implement these subtractions using the primitive recursive operation $a\dot{-}b=\max (a-b,0)$: apply the predecessor function $b$ times to $a$ by primitive recursion. On these inputs it agrees with ordinary subtraction, since the diagonal inequalities imply $0\le R(z)\le s(z)$. Consequently $L$ and $R$ are primitive recursive functions, and $\pi (L(z),R(z))=z$. Zimmermann proves the relevant closure under bounded minimization and the primitive recursive character of the Cantor encoding and coordinate decoding; his convention adds the first coordinate, so its coordinates are exchanged relative to $\pi$ here.^[ref]

For a worked bounded minimization, take $z=18$. Candidates $t=0,1,2,3,4$ give $T(t+1)=1,3,6,10,15$, none exceeding $18$. Candidate $t=5$ gives $T(6)=21>18$, so $s(18)=5$, $R(18)=3$, and $L(18)=2$. The search bound was $18$; it terminates much earlier in this example. The bound proves membership in the primitive recursive functions; it is not a claim that scanning all candidates is the fastest implementation.

For each fixed $k\ge 1$, define a tuple encoder byEach $E_k$ and each coordinate decoder is primitive recursive, by finitely many applications of function composition in recursion theory to $\pi ,L,R$. For the three-entry tuple encoded as $257$, applying $R$ returns $4$, applying $L$ returns $18$, and then $L(18)=2$, $R(18)=3$. The tuple length remains part of the agreed convention, as explained for a general pairing function.

This makes primitive recursive pairing functions useful for encoding several evolving values as one state in primitive recursion. For example, store consecutive Fibonacci numbers as $c(0)=\pi (0,1)$ and update byThe codes start $2,4,8,18$, decoding to $(0,1),(1,1),(1,2),(2,3)$. Every step uses primitive recursive encoding, decoding and addition; a single numerical state therefore carries the two values needed for the next step. This is also the reason a suitable Gödel numbering should permit sufficiently simple decoding: representation should preserve the class of functions being studied.

For $n\ge 2$, a Prüfer sequence is the encoding of a labelled tree on $\{1,\dots ,n\}$ by a sequenceRepeated labels are allowed, and their order matters. To encode the tree, repeatedly remove its smallest-labelled leaf of a graph together with its incident edge, recording the label of its unique neighbour. Stop when two vertices remain; do not record their final edge. Removing a leaf of a graph leaves a tree, so the procedure makes exactly $n-2$ deletions. The choice of the smallest label makes the encoding deterministic. ^[ref]

For a concrete example, take six vertices and the set of edgesInitially $2,4,6$ label the vertices of vertex degree one. Remove $2$ and record its neighbour $1$. Now $1$ has become a leaf of a graph, so remove $1$ and record $3$. Next remove $4$ and record $3$ again; then remove $3$ and record $5$. The final edge joins $5$ and $6$. Thus the Prüfer sequence isThe deletion order is $2,1,4,3$: removing a leaf of a graph can make its neighbour a leaf of a graph, so the eligible labels change as the tree shrinks.

To decode, keep a set $L$ of remaining labels, initially $\{1,\dots ,n\}$, and the remaining sequence, initially $p$. While the sequence is nonempty, let $v$ be the smallest label in $L$ absent from the entire remaining sequence. Add an edge between $v$ and its first entry, then remove $v$ from $L$ and delete that first entry. Finally join the two labels left in $L$. For the example:

The decoder always has a choice: if the remaining sequence has length $m$, then $L$ has $m+2$ labels, so at least two of them are absent. Its first entry is still in $L$, and differs from the chosen absent label. Every removed vertex attaches to a vertex that survives longer; following these attachments eventually reaches the final pair. Consequently the decoded graph is connected, and its $n-1$ edges make it a tree. In this construction, each occurrence of $v$ contributes an incident edge, and $v$ gets one more edge at its final attachment. Thus a remaining label is absent from the remaining sequence exactly when it labels a leaf of a graph. Encoding and decoding choose the same smallest label at each step and undo each other. This bijection proves Cayley's formula and explains why every freely chosen sequence is valid. ^[ref]

For $n=2$ there is exactly one empty sequence, and decoding adds the sole edge. For $n=1$ there is one tree, with vertex degree zero, but no length-$n-2$ Prüfer sequence. This is the exception both to the encoding convention and to the Prüfer degree formula.

Heinz Prüfer's 1918 paper "Neuer Beweis eines Satzes über Permutationen" appeared in Archiv der Mathematik und Physik, third series, volume 27, pages 142–144. It gives an encoding proof of the labelled tree count. The historical title means "New proof of a theorem on permutations"; the modern Prüfer sequence permits repeated labels, so it is not a permutation. ^[ref]

A nonzero congruence class modulo an odd prime is a quadratic nonresidue when it is not a square. In the convention of quadratic residue, zero belongs to neither class. Thus $3,5,6$ are the quadratic nonresidues modulo seven, while $2,6,7,8,10$ are the quadratic nonresidues modulo eleven. No integer $x$ satisfies $x^2\equiv 3(\mathrm{mod}7)$, as the complete square map under quadratic residue shows.

A quadratic nonresidue is not an arithmetic dead end: multiplying by any fixed quadratic nonresidue $n$ exchanges the quadratic residues and the quadratic nonresidues. If $nq$ were a square for a square $q$, division by $q$ would make $n$ a square. Hence $n{Q}_p$ consists entirely of quadratic nonresidues; it has $(p-1)/2$ elements, because multiplication by nonzero $n$ is a bijection. It therefore equals the entire other coset of the quadratic residue subgroup. Modulo seven, $3\{1,2,4\}=\{3,6,5\}$.

The product of two quadratic nonresidues is consequently a quadratic residue: write them as $n{q}_1,n{q}_2$ in that same coset, and their product is $n^2{q}_1{q}_2$, a square. For example, $3\cdot 5\equiv 1(\mathrm{mod}7)$, whereas $2\cdot 3\equiv 6(\mathrm{mod}7)$ remains a quadratic nonresidue. The multiplicativity of the Legendre symbol packages these rules as ordinary sign multiplication.

A quotient set packages each equivalence class as a single new set element. It remembers which class an original set element belongs to and forgets which member of that class it was. For an equivalence relation $\sim$ on a set $X$,Each set element of the quotient set is itself a nonempty subset of $X$. The canonical projection of a quotient set sends an original set element to this entire equivalence class. ^[ref]

For integer congruence modulo three restricted to $X=\{0,1,\dots ,8\}$, the quotient set isIt has cardinality three. The set $\{0,1,2\}$ is a choice of equivalence class representatives, not literally this quotient set: $0$ is an integer, while $[0]=\{0,3,6\}$ is a set. The function sending each chosen equivalence class representative to its equivalence class is a bijection between these two different sets.

A quotient set alone specifies no arithmetic. To obtain the familiar modular arithmetic example with addition, use integer congruence on the whole set of integers. Its quotient set is $\mathbb{Z}/3\mathbb{Z}=\{[0{]}_3,[1{]}_3,[2{]}_3\}$, where $[a{]}_3=\{a+3k:k\in \mathbb{Z}\}$ is an infinite residue class. Define addition byThe answer must be independent of the equivalence class representatives. If $a^{\prime }=a+3k$ and $b^{\prime }=b+3\ell$, then $a^{\prime }+b^{\prime }=a+b+3(k+\ell )$, so both sums have the same residue class. This is what it means for the group operation to be well-defined. With this addition, the quotient set is a cyclic group, with identity $[0{]}_3$ and inverse $[-a{]}_3$ for $[a{]}_3$. ^[ref]

For example, $[2{]}_3+[2{]}_3=[4{]}_3=[1{]}_3$. Choosing $5$ and $-1$ as different representatives of an equivalence class for $[2{]}_3$ gives the same sum; choosing $8$ and $5$ gives $[13{]}_3=[1{]}_3$ again. These calculations use residue classes of all integers. Raw addition on the finite $X=\{0,\dots ,8\}$ does not stay in $X$: $8+8=16\notin X$. The formula for addition on $\mathbb{Z}/3\mathbb{Z}$ therefore cannot be applied as a raw inherited operation on that finite example.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

Map each real into $(0,1)$ by a fixed bijection, choose its canonical binary expansion, and alternate the two bit sequences as ternary digits $0$ and $2$. If two such outputs first differ at position $k$, the leading difference has size $2\cdot 3^{-k}$ and the remaining tail has size at most $3^{-k}$, so they are unequal. This gives an explicit injection pairing two reals. Iteration encodes every fixed finite real tuple without invoking ambiguous decimal expansions.

There are $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ paths with $n$ up steps and $n$ down steps, ignoring the height constraint. For $n\ge 1$, reflect the prefix of a bad path through its first visit to height $-1$, swapping its up and down steps. The resulting path has $n+1$ up steps and $n-1$ down steps, hence final height two. Its first visit to height one identifies the inverse reflection, so this is a bijection. There are $\left(\genfrac{}{}{0ex}{}{2n}{n+1}\right)$ bad paths. Subtracting givesThis count connects a global nonnegative-height requirement to an unrestricted binomial coefficient. The case $n=0$ is separately the unique empty path.

For a subgroup $H$ of a group $G$ and $g\in G$, its right coset is $Hg=\{hg:h\in H\}$. The map $h\mapsto hg$ is a bijection from $H$ onto $Hg$, with inverse $x\mapsto x{g}^{-1}$; in particular all right cosets have the same cardinality as $H$. Multiplication by subgroup elements is on the left in this expression because $g$ is the factor on the right.

Up to an operation-preserving bijection, there are two rngs with two elements. Write the elements as $0,u$. The additive abelian group must satisfy $u+u=0$. The distributive property forces all products involving zero to vanish, leaving only the choice $u^2=0$ or $u^2=u$. The first gives a zero-multiplication rng; the second gives the prime field ${\mathbb{F}}_2$, with identity $u$. Both satisfy the axioms, and they are not isomorphic because an isomorphism must send the unique nonzero element to the unique nonzero element and preserve its square. In particular, there is just one unital ring with two elements.

Insert the successive letters of a permutation by row insertion to obtain an insertion tableau $P$, recording the insertion times in the new boxes of $Q$. This is a bijection between permutations and pairs of standard Young tableaux of a common shape. Reverse insertion removes the box marked by the latest recording time. The first row of $P$ has length equal to the longest increasing subsequence.

List each pair $(i,j)$ with multiplicity $a_{ij}$, sorted first by $i$ and then by $j$. Apply row insertion to the lower letters $j$, and record the upper letters $i$ in the new cells. The row insertion comparison is strictly greater than the carried entry. The result is a pair of Semistandard Young tableaux of the same shape. Reverse insertion removes the rightmost cell containing the largest recording entry, then replaces the rightmost strictly smaller entry in each preceding row. It reconstructs the ordered list, proving the bijection. The insertion and recording contents are respectively the column sums and row sums of $A$.

A compatible transfinite family of bijections can enumerate the square of an ordinal by adjoining a new row and then the remaining new column. Starting with $g_0=\varnothing$, extend at a successor by $g_{\alpha +1}(\alpha ,\beta )={\gamma }_{\alpha }+\beta$ for $\beta \le \alpha$ and $g_{\alpha +1}(\beta ,\alpha )={\gamma }_{\alpha }+(\alpha +1)+\beta$ for $\beta <\alpha$. Thus ${\gamma }_{\alpha +1}={\gamma }_{\alpha }+(\alpha +1)+\alpha$. At limits take compatible unions. The two new domain pieces are disjoint; the column is empty at the initial step $\alpha =0$. Each step enumerates consecutive ordinal intervals, proving bijectivity by transfinite induction. Finite stages have ${\gamma }_n=n^2$. Successor and limit cardinality arguments then give the square equality for well-orderable infinite cardinals without using that equality to justify the construction.

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.