Many of the most natural constructions in geometry and topology begin with an identification: glue two endpoints of an interval to form a circle, collapse the boundary of a disk to a single point to obtain a sphere, or identify opposite edges of a square to build a torus. In each case, we start with a topological space whose structure we understand and produce a new space by declaring that certain distinct points are "the same." The challenge is to equip the resulting set of equivalence classes with a topology that faithfully reflects the geometry of the identification — one that is fine enough to detect the new structure created by gluing, yet coarse enough to ensure that the collapsing map remains [continuous](/page/Continuity).

This problem cannot be solved by the two standard constructions for building topologies from existing ones. The [subspace topology](/page/Subspace%20Topology) restricts to a subset, while the [product topology](/page/Product%20Topology) combines factors — neither captures the act of *collapsing* part of a space to a point or *identifying* separate regions. A new construction is required, one that is governed not by inclusions or projections but by surjections.

[example: Collapsing the Boundary of a Disk] Consider the closed disk $D := \{x \in \mathbb{R}^2 : |x| \le 1\}$ with the subspace topology inherited from $\mathbb{R}^2$. Define an [equivalence relation](/page/Equivalence%20Relation) $\sim$ on $D$ by declaring all points on the boundary circle $S^1 = \{x \in \mathbb{R}^2 : |x| = 1\}$ to be equivalent, while each interior point is equivalent only to itself: \begin{align*} x \sim y \iff x = y \quad \text{or} \quad |x| = |y| = 1. \end{align*} The quotient set $D/{\sim}$ has one class for each interior point and a single additional class $[p]$ for the entire boundary. Geometrically, we expect this space to be a sphere $S^2$: the interior of the disk opens up into the sphere, and the collapsed boundary becomes the "north pole." But what topology should $D/{\sim}$ carry? If we give it the discrete topology, every map out of it is continuous — but the natural projection $\pi: D \to D/{\sim}$ is not, and the space loses all connection to the geometry of $D$. If we give it the indiscrete topology, $\pi$ is continuous but the space is too coarse to distinguish any points. The correct topology must sit between these extremes: it should be the *finest* topology making $\pi$ continuous. [/example]

## Definition

The guiding principle is that the quotient topology should be *as fine as possible* subject to the constraint that the canonical projection is continuous. Since $\pi$ is continuous precisely when preimages of open sets are open, a subset $V$ of the quotient should be declared open if and only if $\pi^{-1}(V)$ is open in the original space. This condition is both necessary (for continuity of $\pi$) and sufficient (one can verify the axioms directly).

[definition: Quotient Topology] Let $(X, \tau)$ be a [topological space](/page/Topology) and let $\sim$ be an equivalence relation on $X$. Let $\pi: X \to X/{\sim}$ denote the canonical projection $\pi(x) = [x]$. The **quotient topology** on $X/{\sim}$ is \begin{align*} \tau_{X/\sim} := \{V \subset X/{\sim} : \pi^{-1}(V) \in \tau\}. \end{align*} The space $(X/{\sim}, \tau_{X/\sim})$ is called the **quotient space** of $X$ by $\sim$. [/definition]

One should verify that $\tau_{X/\sim}$ is indeed a topology. The empty set and $X/{\sim}$ have preimages $\varnothing$ and $X$, both open. For arbitrary unions, $\pi^{-1}(\bigcup_\alpha V_\alpha) = \bigcup_\alpha \pi^{-1}(V_\alpha)$, which is a union of open sets and hence open. For finite intersections, $\pi^{-1}(\bigcap_{i=1}^n V_i) = \bigcap_{i=1}^n \pi^{-1}(V_i)$, which is a finite intersection of open sets and hence open.

The quotient topology has an equivalent characterization in terms of closed sets: a subset $C \subset X/{\sim}$ is closed if and only if $\pi^{-1}(C)$ is closed in $X$. This follows from the identity $\pi^{-1}(X/{\sim} \setminus C) = X \setminus \pi^{-1}(C)$.

[remark: Maximality of the Quotient Topology] The quotient topology is the **finest** (largest) topology on $X/{\sim}$ making $\pi$ continuous. Any topology $\sigma$ on $X/{\sim}$ for which $\pi: X \to (X/{\sim}, \sigma)$ is continuous must satisfy $\sigma \subset \tau_{X/\sim}$, because every $\sigma$-open set has an open preimage and therefore belongs to $\tau_{X/\sim}$. This maximality stands in contrast to the subspace and product topologies, which are *minimal* (coarsest) topologies making the inclusion or projections continuous. The quotient topology is a **final** topology: it is determined by a map *from* a known space, whereas the subspace and product topologies are **initial** topologies, determined by maps *into* known spaces. [/remark]

## Quotient Maps and Saturated Sets

The definition of the quotient topology depends on an equivalence relation, but in practice, quotient spaces often arise from a surjective map $q: X \to Y$ that may not come from an explicit equivalence relation. For instance, the map $q: \mathbb{R} \to S^1$ defined by $q(t) = (\cos 2\pi t, \sin 2\pi t)$ collapses each integer translate to the same point, and we want to recognize that $S^1$ carries the quotient topology induced by $q$ — even though we did not start by writing down an equivalence relation on $\mathbb{R}$.

This motivates an intrinsic characterization of maps that induce quotient topologies.

[definition: Quotient Map] A surjective map $q: X \to Y$ between topological spaces is a **quotient map** if $Y$ carries the quotient topology induced by $q$: a subset $V \subset Y$ is open in $Y$ if and only if $q^{-1}(V)$ is open in $X$. \begin{align*} V \in \tau_Y \iff q^{-1}(V) \in \tau_X. \end{align*} [/definition]

The key subtlety is the "only if" direction: every continuous surjection satisfies the forward implication ($V$ open $\Rightarrow$ $q^{-1}(V)$ open), but a quotient map additionally requires the reverse ($q^{-1}(V)$ open $\Rightarrow$ $V$ open). A continuous surjection that is not a quotient map fails this reverse condition — it carries *strictly fewer* open sets in $Y$ than the quotient topology would prescribe.

[definition: Saturated Set] Let $q: X \to Y$ be a surjection. A subset $U \subset X$ is **saturated** (with respect to $q$) if $U = q^{-1}(q(U))$. Equivalently, $U$ is saturated if whenever $x \in U$ and $q(x) = q(x')$, then $x' \in U$. [/definition]

Saturated sets are the preimages of subsets of $Y$ — they are the unions of entire fibres $q^{-1}(\{y\})$. The quotient topology can be rephrased using this language: $V \subset Y$ is open if and only if $q^{-1}(V)$ is a saturated open set in $X$. This reformulation clarifies the role of saturation: a quotient map $q$ is a continuous surjection with the property that it maps saturated open sets to open sets.

[example: The Exponential Map as a Quotient Map] Define the map \begin{align*} q: \mathbb{R} &\to S^1 \\ t &\mapsto (\cos 2\pi t, \sin 2\pi t), \end{align*} where $S^1 \subset \mathbb{R}^2$ carries the subspace topology. This map is surjective: every point of $S^1$ is the image of some $t \in [0, 1)$. It is continuous because $\cos$ and $\sin$ are continuous. To verify that $q$ is a quotient map, we must show that if $q^{-1}(V)$ is open in $\mathbb{R}$, then $V$ is open in $S^1$. Fix a point $p = (\cos 2\pi t_0, \sin 2\pi t_0) \in V$. Since $q^{-1}(V)$ is open and contains $t_0$, there exists $\varepsilon > 0$ such that $(t_0 - \varepsilon, t_0 + \varepsilon) \subset q^{-1}(V)$. Since $q$ restricted to any interval of length less than $1$ is a homeomorphism onto its image, $q((t_0 - \varepsilon, t_0 + \varepsilon))$ is an open arc in $S^1$ containing $p$ and contained in $V$. Since $p$ was arbitrary, $V$ is open. The equivalence relation induced by $q$ is $t_1 \sim t_2$ if and only if $t_1 - t_2 \in \mathbb{Z}$. The saturated sets are precisely the sets $U \subset \mathbb{R}$ satisfying $U + 1 = U$ (invariant under integer translation). For instance, the interval $(0, 1/2)$ is *not* saturated because it contains $1/4$ but not $1/4 + 1 = 5/4$. The set $\bigcup_{n \in \mathbb{Z}} (n, n + 1/2)$ is saturated. [/example]

Not every continuous surjection is a quotient map. Two important classes of maps, however, are always quotient maps.

[quotetheorem:1030]

For an open surjection: if $q^{-1}(V)$ is open, then $V = q(q^{-1}(V))$ is the image of an open set, hence open. For a closed surjection: if $q^{-1}(V)$ is open, then $q^{-1}(Y \setminus V) = X \setminus q^{-1}(V)$ is closed, so $Y \setminus V = q(q^{-1}(Y \setminus V))$ is closed (using surjectivity), hence $V$ is open.