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For the canonical projection of a quotient set $p$, defineThe inverse images of the empty set and whole target are $\varnothing$ and $X$. Inverse images commute with arbitrary unions and finite intersections, so the topology axioms follow from those of $\mathcal{T}$. This is the quotient topology, and it makes $p$ continuous by definition.

The crucial criterion is an equivalence: $U$ is open precisely when $p^{-1}(U)$ is open. This is stronger than continuity alone and is what makes the next part work.

The quotient set $X/\sim$ consists of the equivalence classes $[x]=\{y\in X:y\sim x\}$. Its collapsing map is the canonical projection of a quotient setEquip this set with the quotient topology:This is a topology because preimages preserve arbitrary unions, finite intersections, the empty set, and the whole set. The resulting topological space is the quotient space. The displayed rule immediately implies that the preimage of every open set is open, so $p$ is continuous; it is also surjective.

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

True. The canonical projection of a quotient set $p:X\to X/\sim$ is a continuous surjection. Its image is compact by the continuous image of a compact space theorem.

For connectedness, if $U,V$ formed a separation of a topological space of $X/\sim$, then the preimages $p^{-1}(U),p^{-1}(V)$ would be disjoint nonempty open sets covering $X$, contradicting its connectedness. This proves the continuous image of a connected space theorem in this case. Hence the quotient space is both compact and connected.

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.

For the canonical projection of a quotient set $p:X\to X/\sim$, the saturation of $A\subseteq X$ is the union of all equivalence classes meeting $A$, namely $p^{-1}(p(A))$. A subset is called saturated when it equals its saturation. The quotient topology says that $p(A)$ is closed exactly when the saturation of $A$ is closed, and similarly for openness. A closed set need not have closed saturation: collapsing $(0,1)\subset \mathbb{R}$ to one point makes the saturation of $\{1/2\}$ equal to $(0,1)$. Hence a quotient map need not be a closed map.