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An equivalence class representative is any set element belonging to that equivalence class. In the finite modulo-three example, either $0$, $3$ or $6$ represents $\{0,3,6\}$. Changing the equivalence class representative changes its name but preserves the equivalence class itself. Choosing $0,1,2$ gives one representative of an equivalence class from each class; the quotient set consists of the three whole sets.

Choose the first unclassified element as the next equivalence class representative:Assume the Eq instance is an equivalence relation. The stable list partition collects every later occurrence equivalent to x, and rest contains none of that equivalence class. Induction therefore gives exactly one group per class, preserving order within each group and ordering groups by their first occurrence. For $d$ distinct classes, each input element is tested against at most $d$ representatives: the cost is $O(nd)$ and $\Theta (n^2)$ in the all-distinct worst case; a single class takes $\Theta (n)$.

The equality-only grouping lower bound shows that this worst case is unavoidable in the unrestricted equality-oracle model. On an all-distinct input, an untested pair could instead be equal without changing any observed comparison. Certifying that all $n$ elements belong to different classes therefore requires all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise equality tests.

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.

The quotient topology turns an instruction to glue points together into a precise topological space. Start with a topological space $(X,{\tau }_X)$ and an equivalence relation $\sim$. The quotient set $X/\sim$ consists of equivalence classes, and its canonical projection of a quotient set isThe quotient topology is defined by the rule using preimages of open sets:Thus a subset of the quotient space is open if and only if its entire preimage is open in $X$. Each fiber $q^{-1}(\{[x]\})$ is the whole equivalence class $[x]$, so the preimage includes every point glued to any of its members.^[ref]

This is the finest topology making $q$ a continuous map: every topology with that property can contain only subsets whose preimages are open. It really is a topology, because preimages preserve the empty set, the whole set, arbitrary unions, and finite intersections. The gluing example below shows why checking a complete preimage matters, especially at the seam.

The quotient topology also makes continuous maps out of a glued space easy to construct. A continuous map $f:X\to Z$ descends to a unique continuous map $\bar{f}:X/\sim \to Z$ exactly when it is constant on every fiber of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar{f}([x])=f(x)$; the equivalence class representative cannot affect the answer. For every open set $V\subseteq Z$,The right side is open, so the quotient topology makes ${\bar{f}}^{-1}(V)$ open. Conversely, if $f=\bar{f}\circ q$ for a continuous map $\bar{f}$, then composition of continuous functions makes $f$ continuous, and equal equivalence classes force equal values. This is the universal property of the quotient topology. ^[ref]

A surjective continuous map is not automatically a quotient map. On the same two-point set $D=\{a,b\}$, take the identity map from the discrete topology to the indiscrete topology. It is continuous and surjective, but $\{a\}$ is not open in the target even though its preimage is open in the domain. The quotient topology for this identity map would be the discrete topology. Thus continuity supplies only one direction of the rule using preimages of open sets; a quotient map requires both.