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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.

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.

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.

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.

Identify the initial endpoints of countably many closed intervals to a vertex $v$, using the quotient topology. This is the geometric realization of the complex whose only edges are $\{v,u_n\}$. A neighbourhood of $v$ may contain a separately chosen initial interval $[0,{\epsilon }_n)$ on each arm, with no positive lower bound on the ${\epsilon }_n$.

The space is not first countable. Given any proposed countable neighbourhood basis $(U_n)$ at $v$, choose a point $x_n$ at coordinate $t_n>0$ on arm $n$ with $x_n\in U_n$. This is possible by the interval topology on that arm. The set whose intersection with arm $n$ is $[0,t_n/2)$ is an open neighbourhood $V$ of $v$, by the simplexwise definition. But $x_n\in U_n\setminus V$ for every $n$, so no $U_n$ lies inside $V$. Consequently this weak star cannot embed in the plane, or in any metric space.

The terminal vertices form an infinite closed discrete subset: on each closed arm they meet only its terminal endpoint. Hence the star is not compact, by closed discrete subsets of compact spaces are finite.

Give the geometric realization of an abstract simplicial complex the quotient topology obtained by gluing the disjoint union of its standard simplexes along their labelled faces. A subset $U$ is open exactly when $U\cap |\sigma |$ is open in every closed geometric simplex $|\sigma |$. This is the weak realization topology. It is the topology of its CW complex structure; the word weak here refers to the simplexwise test, rather than to the weak topology of a topological vector space. A map out of the realization is continuous exactly when its restriction to every simplex is continuous.