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A semiconjugacy from $f:X\to X$ to $g:Y\to Y$ is a map $h:X\to Y$ satisfying $h\circ f=g\circ h$. It transports forward orbits but need not be invertible, A bijection instead gives conjugate functions.

Set theory studies sets, their set elements, and the constructions and sizes possible for them. It supplies a shared vocabulary for mathematics: a function connects two sets, a binary relation specifies which pairs of objects are related, and a probability space starts with a set of possible outcomes. Giving these objects a common language lets a construction from one subject be used in another without repeating its underlying bookkeeping.

A particularly beautiful question is how to compare sizes without first assigning a number. Two sets have the same cardinality when a bijection pairs every member of one with exactly one member of the other. For finite sets, this agrees with ordinary counting. For infinite sets, it produces surprises: $n\mapsto 2n$ bijects the natural numbers with the nonnegative even integers, and the Cantor pairing function bijects the Cartesian product ${\mathbb{N}}^2$ with $\mathbb{N}$. A proper subset can have the same cardinality as its containing set.

This does not make all infinite sets equally large. Cantor's diagonal argument shows that all infinite binary sequences cannot be listed by the natural numbers. Cantor's theorem gives the wider principle: the power set of any set has strictly greater cardinality. The contrast between an explicit enumeration and an impossibility argument makes the meaning of infinity much sharper than a picture of something merely continuing forever.

Set formation itself requires care. Russell's paradox shows why not every condition defines a set of all objects satisfying it. Zermelo–Fraenkel set theory instead uses axioms specifying permitted constructions, including set comprehension within a previously given set. The restrictions are useful: they let us form the collections ordinary arguments need while blocking the unrestricted construction that produces the paradox.

The odd integers form an abelian group under the displayed operation. The identity element is $-1$ and the inverse element of $a$ is $-a-2$. The bijection $a\mapsto (a+1)/2$ is a group isomorphism to $(\mathbb{Z},+)$, since it converts the shifted operation into addition. This illustrates how an unfamiliar identity and inverse can arise from ordinary addition by a change of coordinates.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

A simplicial homeomorphism is the homeomorphism of geometric realizations induced by an isomorphism of simplicial complexes: a bijection of their vertex sets such that a finite vertex set is a simplex exactly when its image is a simplex. The map acts linearly in each barycentric coordinate system. It and its inverse are continuous by the realization topology. Between the boundaries of two triangles, any bijection of the three vertices extends linearly on edges to a simplicial homeomorphism. This ensures that the edge identifications agree at common endpoints.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

A countable model of set theory can internally assert that an uncountable set exists. There is no contradiction because “there is a bijection” ranges only over functions in the model. An external enumeration of its domain need not itself be an element of that domain. Löwenheim-Skolem theorem controls external cardinality; Cantor's theorem remains valid for the sets and functions that the model contains.

Take $X=\{0,1,2,\dots \}$, the nonnegative integers, and $Y=\{1,2,3,\dots \}$, the positive integers. Choose the injectionsHere $g$ is an inclusion map, while $f$ reaches only the positive even integers. In particular, $f$ is not a bijection: it misses every positive odd integer. Simply using the inverse function of $g$ on its image misses input zero.

The remedy is to follow the alternating injections starting at the missing input zero:The subscripts distinguish an input in $X$ from an output in $Y$, even when their numerical values agree. The resulting subset of $X$ is $A=\{0,2,6,14,30,\dots \}$. Its sequence of members satisfies $a_0=0$ and $a_{k+1}=2a_k+2$, giving $a_k=2^{k+1}-2$ by mathematical induction. Define the function

The image $h(A)=\{2,6,14,30,\dots \}=A\setminus \{0\}$ covers all the positive members of $A$ exactly once. The other inputs cover $Y\setminus A$ exactly once by staying fixed. These output sets are disjoint and their union is $Y$, proving that $h$ is a bijection. In particular, $h(0)=2$, $h(2)=6$, $h(6)=14$, while $h(1)=1$ and $h(4)=4$.

For completeness, its inverse function isThe familiar bijection $n\mapsto n+1$ also works, but this example shows something more specific: the two chosen injections can repair the missing zero by moving only a sparse sequence of inputs. The repair never has to shift the entire set.

A standard Young tableau of shape $\lambda ⊢n$ is a Young tableau that uses $1,\dots ,n$ exactly once and strictly increases left to right along every row and top to bottom down every column. Formally, a bijection $T:Y(\lambda )\to \{1,\dots ,n\}$ is standard whenwhenever the indicated cells belong to the Young diagram. We write $\mathrm{SYT}(\lambda )$ for the set of such fillings and $f^{\lambda }=|\mathrm{SYT}(\lambda )|$ for its cardinality.

For $(3,2)$, the five standard Young tableaux are listed below, with each pair giving the top row followed by the bottom row:There are no others: the upper-left entry must be $1$, and choosing the two bottom-row entries fixes both increasing row orders. Of the six choices of two entries from $\{2,3,4,5\}$, only bottom row $(2,3)$ fails, because its $3$ would lie below $4$. Hence $f^{(3,2)}=5$ by direct enumeration, independently of the hook-length formula.

The cells occupied by $1,\dots ,r$ form a Young diagram: every cell above or to the left of an occupied cell has a smaller label. Consequently a standard Young tableau is equivalently a path from the empty shape to $\lambda$ in the Young branching graph, adjoining one addable node of a Young diagram at each step. For $T_2$, the successive shapes areConversely, number the cell added at step $r$ by $r$ to recover the standard Young tableau. This path interpretation also makes a standard Young tableau a linear extension of a partially ordered set defined by the row and column requirements; see Stanley's Algebraic Combinatorics, Chapter 8, on tableaux and paths in Young's lattice.

The usefulness goes beyond counting: the standard polytabloid basis of a Specht module $S^{\lambda }$ over the complex numbers is indexed by $\mathrm{SYT}(\lambda )$, so ${\dim }_{\mathbb{C}}{S}^{\lambda }=f^{\lambda }$. Each of the five pictured standard Young tableaux therefore labels one basis element of the irreducible representation $S^{(3,2)}$ of the symmetric group $S_5$, whose dimension over the complex numbers is five. The filled diagrams index the basis; the actual vectors are polytabloids. The hook-length formula computes this dimension without listing all the standard Young tableaux.

Stars and bars counts tuples of nonnegative integers with a prescribed sum. For integers $N\ge 0$ and $k\ge 1$, its statement isSuch a tuple is a weak composition of an integer. Think of $k$ labelled bins: $x_1$ counts identical objects in the first bin, $x_2$ in the second, and so on. Bin names distinguish positions even when counts agree. Stars and bars counts these count tuples; assigning individually distinct objects requires additional information. ^[ref]

Computed example. With $N=5$ and $k=3$, the number of weak compositions of an integer is $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$. The string ••|•|•• encodes $(2,1,2)$: two dots before the first bar, one between the bars, and two after the second. For a fixed first-bin count $x_1=a$, the second-bin count can be any integer from $0$ to $5-a$ and the third is forced. Thus the direct count is $6+5+4+3+2+1=21$.

Adjacent bars encode an empty middle bin: ••||••• gives $(2,0,3)$. A bar at the left end encodes an empty first bin, as in |••|••• for $(0,2,3)$; a bar at the right end encodes an empty third bin, as in ••|•••| for $(2,3,0)$. These are valid weak compositions of an integer. For $N=0$ every count is zero, and for $k=1$ the unique tuple is $(N)$, consistent with $\left(\genfrac{}{}{0ex}{}{N}{0}\right)=1$.

Proof outline. Replace each count $x_i$ by that many dots and separate successive groups with $k-1$ bars. Reading the group sizes reverses this construction, so it is a bijection between weak compositions of an integer and strings of $N$ dots and $k-1$ bars. Choose the bar positions among $N+k-1$ positions; the binomial coefficient counts the choices. The beauty is that empty bins need no separate cases: end bars and adjacent bars encode them automatically.

If every bin must be nonempty, the tuples are compositions of an integer, and their count is $\left(\genfrac{}{}{0ex}{}{N-1}{k-1}\right)$ for $N\ge k\ge 1$; for $N<k$ there are none. Subtract one from every count, setting $y_i=x_i-1$. This is a bijection to weak compositions of an integer of $N-k$, so stars and bars gives $\left(\genfrac{}{}{0ex}{}{(N-k)+k-1}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{N-1}{k-1}\right)$. For five objects and three bins, this is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, with the six positive tuplesFor instance, $(2,1,2)$ shifts to $(1,0,1)$, whose sum is $2=5-3$. Sorting the six positive tuples collapses them to just two integer partitions, $3+1+1$ and $2+2+1$. Thus named bins and unordered summands answer different questions.

Stars and bars is useful for allocating identical resources or enumerating integer solutions without listing them individually. Prescribed lower bounds can be subtracted just like the nonempty-bin requirement. Upper bounds require removing forbidden tuples; the existing bounded weak compositions article applies the inclusion-exclusion principle to those violations.

Stereographic projection identifies a sphere minus one pole with the plane and supplies the standard charts of the Riemann sphere.

In the equatorial-plane convention of the Riemann sphere, draw the straight line through $N=(0,0,1)$ and a point $P=(X,Y,Z)\ne N$ on the unit sphere. Its intersection with the plane $Z=0$ has complex coordinateThe inverse function, for $z=x+iy$ and $r^2=x^2+y^2$, isHere positive $x$ gives positive $X$, and positive $y$ gives positive $Y$. Thus $1$ and $i$ lie on the positive $X$ and $Y$ ends of the equator. The omitted north pole is assigned $z=\infty$.^[ref]

For original worked points, the real numbers $1/2$ and $2$ lift to points with the same $X$ coordinate but opposite $Z$ coordinates:Projecting back gives $(4/5)/(1+3/5)=1/2$ and $(4/5)/(1-3/5)=2$. A point off the real axis checks the other sign:The coordinates satisfy $4/9+1/9+4/9=1$, as a point on the unit sphere must. Also $P(0)=S$, $P(-1)=(-1,0,0)$ and $P(i)=(0,1,0)$.

A short derivation starts with the line $N+t(P-N)=(tX,tY,1+t(Z-1))$. Setting its last coordinate to zero gives $t=1/(1-Z)$, hence the projection formula. Conversely the line from $N$ to $(x,y,0)$ is $(tx,ty,1-t)$. Substitution in the unit sphere equation givesThe root $t=0$ is $N$; the other root $t=2/(1+r^2)$ gives $P(z)$. These line intersections establish both the formula and its bijection property without guessing the inverse.

The same algebra explains how large planar distances become small spherical ones:Consequently $|z|\to \infty$ brings $P(z)$ to $N$ independently of direction. The last expression is twice the chordal metric distance $\chi (z,\infty )$. A whole unbounded exterior in the complex plane therefore becomes a small cap on the unit sphere.

For nonnegative integers $n,k$, the Stirling number of the second kindcounts the set partitions of an $n$-element finite set into exactly $k$ unlabelled blocks of a set partition. Set $S(0,0)=1$, $S(n,0)=0$ for $n>0$, and $S(n,k)=0$ for $k>n$. Also $S(n,1)=S(n,n)=1$ for $n\ge 1$. Grassl and Levin develop these counts through explicit set partitions and a recurrence relation. ^[ref]

Insert one distinguished element. For $n\ge 1$ and $1\le k\le n$, the Stirling numbers of the second kind obeyProof outline. Inspect the block of a set partition containing $n$. If it contains another element, removing $n$ leaves a set partition into $k$ blocks of a set partition; conversely, insert $n$ into any of those $k$ blocks of a set partition. If $n$ is alone, remove its singleton set and obtain a set partition into $k-1$ blocks of a set partition. The cases cannot overlap, and deletion reverses each construction. This explains the factor $k$ even though the blocks of a set partition have no names. ^[ref]

For $n=4,k=2$, start with the three two-block set partitions of $\{1,2,3\}$. Inserting $4$ into either block of a set partition gives six possibilities. The remaining possibility is $\{\{1,2,3\},\{4\}\}$, obtained by making $4$ a singleton set. Hence $S(4,2)=2\cdot 3+1=7$. The seven set partitions in the diagram comprise four with sizes $3+1$ and three with sizes $2+2$; this grouping by sizes independently gives $4+3=7$.

Three different counting questions. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ selects one $k$-element subset. The Stirling number of the second kind $S(n,k)$ groups all $n$ distinct elements into $k$ nonempty unnamed blocks of a set partition. The power $k^n$ counts all functions to a specified $k$-element finite set, allowing some destination labels to go unused. At $n=4,k=2$, these counts are respectively $6$, $7$, and $16$. Choosing two people, making two unnamed nonempty teams, and assigning each person a red or blue label are different tasks.

When the destinations have labels. For fixed finite sets $X,Y$ of sizes $n,k$, respectively, the number of surjective functions $X\to Y$ isHere every element of $Y$ must be used. Each surjective function gives a set partition of $X$ into its nonempty preimages $f^{-1}(\{y\})$. Conversely, assign the $k$ distinct destination labels to the $k$ blocks of a set partition in $k!$ ways, where $k!$ is a factorial. For two labelled teams of four people this gives $2!\cdot 7=14$: of the $2^4=16$ assignments, exactly two use only one label. Dividing arbitrary function counts by $k!$ fails in general because unused labels can be permuted without changing a function. For example, $3^4/3!=13.5$ cannot count set partitions. The empty-to-empty function is a surjection, consistent with $0!S(0,0)=1$. Stanley gives the bijection underlying this qualification.^[ref]

For example, the inclusion-exclusion principle counts the surjections from four elements to three labelled destinations as $3^4-3\cdot 2^4+3\cdot 1^4=36$. Thus $S(4,3)=36/3!=6$. An independent check chooses the only two-element block of a set partition in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways; the other two elements form singleton sets.

Why these numbers also occur in algebra. Classifying all functions to an $m$-element finite set by the size $k$ of their image givesFor a fixed set partition, the falling factorial $(m{)}_k$ counts assignments of distinct labels to its $k$ blocks of a set partition, and is zero when $k>m$. At $n=3,m=4$, this reads $64=1\cdot 4+3\cdot 12+1\cdot 24$. Replacing $m$ by $x$ yields a polynomial identity: both sides agree at every nonnegative integer. Thus the Stirling numbers of the second kind translate powers into falling factorials. Grouping by equal values turns a counting rule into an algebraic change of coordinates. Summing $S(n,k)$ without destination labels instead gives the Bell number.

The map $j\mapsto n-j$ is a bijection of the integers between one and $n-1$ that are coprime to $n$. If their sum is $S$, this bijection gives $S=n\phi (n)-S$, so $S=n\phi (n)/2$. This works even at $n=2$, where the single residue is fixed. The same identity follows from the weighted inclusion-exclusion principle by excluding multiples of the prime factors of $n$.

A symmetry group collects bijections of an object to itself that preserve a specified structure in both directions, with function composition as the operation. For a geometric object one may require Euclidean distances to be preserved, giving an isometry group; for a graph, one preserves adjacency, giving a graph automorphism group. For an unstructured finite set, every permutation is allowed. The structure to be preserved is part of the definition.

For a square with its usual geometric structure, the symmetry group consists of four rotations and four reflections. These form the dihedral group of group order eight. The existing group action example shows how its eight elements move the vertices of a polygon; changing the object to its edges or a marked configuration changes the group action being studied.^[ref]

Specifying the preserved data makes it possible to test a proposed symmetry exactly. After the preserved data are declared, function composition, the identity map, and inverse maps automatically preserve them too. Symmetry transformations therefore form a group, and questions about equivalent configurations become questions about group orbits, stabilizer subgroups, and group homomorphisms. This supplies reusable mathematics for comparing many different kinds of objects.

A symmetry transformation is a bijection preserving the data that define the object's symmetry group. Its individual group elements are symmetry transformations; the group operation combines them by function composition.

The first-order theory of a bijection $f$ with named inverse $g$ and no finite cycles has first-order models consisting of disjoint integer orbits. Atomic logical formulas reduce to $x_j=f^k(x_i)$ for integer $k$. An existential logical variable is either fixed by such a logical equality or must avoid finitely many points. This proves quantifier elimination and completeness. There is one complete one-type; the complete two-types are the integer offsets and the nonprincipal different-orbit partial type. There are countably many finite-arity partial types but countable first-order models with different numbers of orbits are not isomorphic.

A finite simple graph that is a cubic graph and has a perfect matching admits a partition of a set of its edges into graph paths of length three.

Proof. Delete a perfect matching $M$. The remaining spanning subgraph has vertex degree two at every vertex, so it consists of disjoint graph cycles. Orient each cyclically, giving a successor bijection $s$. Replace each $uv\in M$ by the graph path $s(u),u,v,s(v)$. The four vertices are distinct: the successor edges are not in $M$, neither is a self-loop, and injectivity of $s$ excludes $s(u)=s(v)$. Every edge of $M$ appears once in the middle, and every other edge appears once at its unique tail. Thus the graph paths cover every edge exactly once. In particular, the perfect matching theorem for a cubic graph with at most two bridges guarantees this decomposition under its hypotheses.

A topological order of a finite directed graph $G=(V,E)$ is a sequence $(v_1,\dots ,v_n)$ containing every vertex exactly once, withThus every directed edge points forward in the list. Equivalently, a bijection $p:V\to \{1,\dots ,n\}$ assigning positions satisfies $p(u)<p(v)$ for every directed edge $u\to v$. A topological ordering names the same kind of list. ^[ref]

The toy Directed acyclic graph has exactly five topological orders: $ABCDE$, $ABDCE$, $ACBDE$, $BACDE$, and $BADCE$. These lists were checked by enumerating all $5!=120$ permutations and retaining precisely those satisfying all five directed edge inequalities. Every list ends with $E$, since all other tasks have a directed path to publication. A topological order need not put adjacent entries next to one another in the directed graph: $A$ and $B$ are consecutive in $ABCDE$ without a directed edge between them.

A directed cycle would force the impossible chain of position inequalities $p(u_1)<p(u_2)<\cdots <p(u_k)<p(u_1)$. In the toy example, adding $E\to C$ requires review before publication and publication before review, so there are no topological orders. Together with the construction in topological sorting, this proves that a finite directed graph admits a topological order if and only if it is a Directed acyclic graph. The empty directed graph has the empty sequence as its topological order.

There is also a useful connection to partial orders: define $u⪯v$ when $u=v$ or there is a directed path from $u$ to $v$. For a Directed acyclic graph, this reachability relation is a partial order, and its linear extensions are exactly the topological orders. The Directed acyclic graph specifies which pairs must be ordered; a topological order chooses positions for all the remaining pairs.

A topological space is a set $X$ together with a topology $\tau \subseteq \mathcal{P}(X)$, a family of subsets satisfyingHere $\mathcal{P}(X)$ is the power set; the union may have arbitrarily many members, while the intersection must be finite. The members of $\tau$ are the open sets. These are axioms about subsets, with no assumed metric or surrounding Euclidean space. ^[ref]

For $X=\{a,b\}$, the discrete topology is ${\tau }_d=\{\varnothing ,\{a\},\{b\},X\}$, while the indiscrete topology is ${\tau }_i=\{\varnothing ,X\}$. Both satisfy the axioms. They describe different topological spaces on the same set: the first is Hausdorff and disconnected, while the second is connected and not Hausdorff. Each is compact because $X$ is a finite set.

The identity function $i:(X,{\tau }_d)\to (X,{\tau }_i)$ is a continuous map: the only target open sets have inverse images $\varnothing$ and $X$. Its inverse is not a continuous map, since the target open set $\{a\}$ has inverse image $\{a\}$, which is not an open set in ${\tau }_i$. Thus even a continuous bijection need not be a homeomorphism.

Topology studies topological spaces through open sets and continuous maps. Its central equivalence is a homeomorphism: a bijection that preserves continuity in both directions. This lets us distinguish spaces by connectedness or compactness even when numerical metric values are irrelevant. The open interval $(0,1)$ and the real line have the same topology in this sense, although only the former is a bounded set with its usual metric.

The upper panels show an explicit homeomorphism and the inverse image test for a continuous map. The lower panels exhibit an obstruction to a homeomorphism: removing any point from a circle leaves a connected space, whereas removing an interior point from a closed interval gives two connected components.

Homotopy is a different notion: it varies a continuous map continuously with a parameter and may collapse points. For example, $H(x,t)=(1-t)x$ contracts $[0,1]$ to zero. Thus this closed interval and a singleton have the same homotopy type, but there is no homeomorphism between them because there is no bijection. Homotopy equivalence need not preserve compactness. ^[ref]

The abstract viewpoint already appears explicitly in Tietze's 1923 paper: he describes topology as the study of properties invariant under invertible continuous maps with continuous inverses, and discusses axioms for neighbourhoods, citing Hausdorff's 1914 book, page 213. ^[ref]