Quotient topology

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.

If an equivalence relation is closed in a space's square and the quotient map is open, the quotient is Hausdorff. An inequivalent pair has an open rectangle avoiding the relation. The two factor images are open quotient neighborhoods and are disjoint, since an intersection would give equivalent representatives in that rectangle. Conversely, a Hausdorff quotient always has a closed equivalence relation by pulling back its closed diagonal.

For a topological space $X$, the displayed relation is an equivalence relation. Give its quotient the quotient topology. Every inverse image under a continuous map from $X$ to a Hausdorff space is a saturation of a subset. Distinct quotient classes can be distinguished by some such map; disjoint open neighborhoods of their images pull back to disjoint saturated open sets, whose quotient images are open. Thus the quotient is Hausdorff. Each continuous map from $X$ to a Hausdorff space factors uniquely through this quotient, and the factor is continuous by the quotient topology. This is its universal reflection property.

Suppose a group acts by isometries on a metric space $X$, and for points $a,b$ in different orbits one has ${\inf }_{g}d(a,gb)>0$. Then the orbit space with its quotient topology is Hausdorff. Indeed, choose $\epsilon$ less than one third of this positive infimum and saturate the open radius-$\epsilon$ balls about $a$ and $b$ by the action. These saturated sets are open because every acting isometry is a homeomorphism. Their translates cannot meet: their centres stay at least the given positive distance apart. Their disjoint open images in the quotient separate the two orbit points, since the preimage of each image is its saturated set. This applies to the glide reflection $(x,y)\mapsto (x+1,-y)$ on a closed strip: the distance from one point to the translates of another tends to infinity as the integer translation grows, so two distinct orbits have a positive minimum distance.

A quotient space of a space carrying the cofinite topology has a topology contained in the cofinite topology on its underlying quotient set. If $q:X\to Y$ is the quotient map and $U$ is a nonempty open subset of $Y$, its preimage has finite complement. Since $q$ is surjective, the complement of $U$ is the image of that finite complement, hence is finite. Equality need not hold: partition $\mathbb{Z}$ into its even and odd integers. Both classes are infinite, so their two-point quotient space has the indiscrete topology, whereas the cofinite topology on a finite set is the discrete topology.

Identify points of the circle having the same real coordinate, equivalently points related by complex conjugation. The continuous surjection $h(z)=(1+\mathrm{Re}z)/2$ has exactly these equivalence classes as its fibers. It induces a continuous bijection from the quotient space to the closed interval. The quotient space is compact as a continuous image of a compact space, while the closed interval is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes the induced map a homeomorphism. The two fixed points of the reflection become the two interval endpoints.

For a nonempty subset $A$ of a topological space $X$, identify all points of $A$ and leave each point of $X\setminus A$ in its own equivalence class. The resulting quotient space is denoted $X/A$. For every subset $K\subseteq X$, its saturation of a subset isThe quotient topology therefore gives a useful test: $q_A(K)$ is a closed set if and only if the displayed saturation of a subset is a closed set of $X$. In particular, every closed set disjoint from $A$ has closed image, even if $A$ is not closed. When $A$ is closed, $q_A$ is a closed map, because both possibilities in the formula are then closed sets for closed $K$.

Choose $b_n\in (n,n+1/2)\setminus A$, possible because each interval is uncountable. Every tail $K_n=\{b_k:k\ge n\}$ is closed and disjoint from $A$. Therefore $U_n=\mathbb{R}\setminus K_n$ is an open saturated set for collapsing a subset to a point, so its image in the quotient topology is open. These increasing images cover the quotient, but no finite selection covers it, because the largest selected image still misses an infinite tail of distinct singleton classes. This is an open cover with no finite subcover, proving noncompactness. The argument does not require the countable subset to be closed or nondense. For the empty subset, read the quotient as the original line.

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.

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.

Collapse the equator of a sphere $S^2$ to a single equivalence class and leave every other point in its own class. The two closed hemispheres are disks sharing that equator. A disk with its boundary collapsed is a sphere, so the two hemisphere quotients are spheres sharing just one point: their wedge sum. An explicit map sends height $t\in [0,1]$ and longitude $\theta$ on the upper hemisphere to $(1-\cos \pi t,\sin \pi t\cos \theta ,\sin \pi t\sin \theta )$, and sends the lower hemisphere to the sphere centred at $(-1,0,0)$ by reversing the first coordinate. The equator goes to the common origin and the poles go to $(\pm 2,0,0)$. This map is continuous and has precisely the collapsed equator as its only nonsingleton fibre. Its induced map is a homeomorphism by the compact-to-Hausdorff continuous bijection theorem.

Let $X$ be a compact Hausdorff space and $R\subseteq X\times X$ a closed equivalence relation. The quotient space $X/R$ is compact and Hausdorff. Indeed, the saturation $R[C]$ of any closed $C\subseteq X$ is the projection of the compact set $R\cap (C\times X)$, so is closed. Thus the quotient map is a closed map.

Two different equivalence classes are disjoint compact subsets of $X$ and have disjoint open neighbourhoods $U,V$. To see the latter assertion, separate each pair of points by disjoint neighbourhoods and use compactness twice to extract finite covers. Now $U^{\prime }=X\setminus R[X\setminus U]$ and $V^{\prime }=X\setminus R[X\setminus V]$ are open saturated subsets containing the respective classes, and lie inside $U,V$. Their quotient images are disjoint open neighbourhoods, proving the Hausdorff condition. Compactness follows from the continuous quotient map. This applies to finite gluing along closed boundary pieces when the generated equivalence relation is closed; that last condition must be checked.

A quotient map is a surjective continuous map $q:X\to Y$ for which $U\subseteq Y$ is open exactly when $q^{-1}(U)$ is open.

If $q:E\to B$ is a quotient map, then $q\times {\mathrm{id}}_{[0,1]}$ is a quotient map. More generally the second factor may be any locally compact space that is Hausdorff. To prove the assertion, let $W\subseteq B\times [0,1]$ have open preimage $O$, and take $(b,t)\in W$. Choose $e\in q^{-1}(b)$ and a compact neighbourhood $K$ of $t$ with $\{e\}\times K\subseteq O$. The set $E_K=\{u:\{u\}\times K\subseteq O\}$ is open by compactness of $K$, and is saturated under $q$ because membership in $O$ depends only on $q(u)$ and the second coordinate. Hence $q(E_K)$ is open by the quotient topology. The rectangle $q(E_K)\times \mathrm{int}K$ is a neighbourhood of $(b,t)$ contained in $W$, proving that $W$ is open. This justifies parameterized homotopies defined on quotient spaces.

If $q:X\to X/R$ is the quotient map and a continuous map $f:X\to Y$ is constant on each equivalence class, there is a unique continuous map $F:X/R\to Y$ satisfying $Fq=f$. It is given by $F([x])=f(x)$; continuity follows becausefor every open $U\subseteq Y$.

Identify $x,y\in \mathbb{R}$ when $x-y\in \mathbb{Q}$. Every nonempty saturated open subset of $\mathbb{R}$ meets every other one because rational translates of any open interval meet any prescribed nonempty open interval. The quotient $\mathbb{R}/\mathbb{Q}$ has multiple points but no two nonempty disjoint open subsets, so it is not Hausdorff.

Identify the vertical sides of $[0,1{]}^2$, then collapse each horizontal side separately to a point. The resulting quotient is homeomorphic to $S^2$ throughEquivalently, the quotient is the suspension of the circle.