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. The converse fails: quotient maps need be neither open nor closed. The identification map $\pi: [0, 1] \to [0, 1]/\{0, 1\} \cong S^1$ is a quotient map, but it is not open — the image of the open set $[0, 1/4)$ (open in the subspace $[0, 1]$) is not open in $S^1$ because the corresponding arc is missing the other half near the identified point. [example: A Continuous Bijection That Is Not a Quotient Map] Consider $X = [0, 1) \cup [2, 3]$ with the subspace topology from $\mathbb{R}$, and let $Y = [0, 2]$ with the standard topology. Define \begin{align*} q: X &\to Y \\ x &\mapsto \begin{cases} x & \text{if } x \in [0, 1), \\ x - 1 & \text{if } x \in [2, 3]. \end{cases} \end{align*} This map is a continuous bijection: $q$ maps $[0, 1)$ identically onto $[0, 1)$ and shifts $[2, 3]$ onto $[1, 2]$. However, it is *not* a quotient map. To see this, consider the subset $V = [1, 2) \subset Y$. Its preimage is $q^{-1}(V) = [2, 3)$, which is open in $X$ (since $[2, 3) = X \cap (1.5, 3)$). If $q$ were a quotient map, the condition $q^{-1}(V)$ open would force $V$ to be open in $Y$. But $[1, 2)$ is not open in $[0, 2]$: every neighbourhood of $1$ in the subspace $[0, 2]$ contains points less than $1$, which lie outside $[1, 2)$. The source of the failure is topological: $X$ is disconnected (it has a gap between $[0, 1)$ and $[2, 3]$), while $Y = [0, 2]$ is connected. The bijection $q$ maps the "jump" across this gap to the interior point $1 \in Y$, creating open sets in $X$ (such as $[2, 3)$, which is open because it is a component boundary) whose images are not open in $Y$. This example also illustrates why compactness matters: $X$ is not compact ($[0, 1)$ is not closed in $\mathbb{R}$). If $X$ were compact and $Y$ Hausdorff, the closed map lemma would force $q$ to be a homeomorphism, hence a quotient map. [/example] ## The Universal Property of the Quotient Topology The defining feature of the quotient topology — beyond its explicit construction via preimages — is a universal property that governs which maps out of the quotient space are continuous. This property is the topological analogue of the [universal property of the quotient set](/page/Equivalence%20Relation): just as a function $f: X \to Y$ that respects an equivalence relation descends to a unique set-theoretic map $\bar{f}: X/{\sim} \to Y$, a *continuous* function that respects the relation descends to a unique *continuous* map. The question is: given a continuous map $f: X \to Z$ that is constant on each equivalence class (meaning $x \sim y$ implies $f(x) = f(y)$), when is the induced map $\bar{f}: X/{\sim} \to Z$ continuous? The answer, with the quotient topology, is: *always*. [quotetheorem:1031] The proof reduces to unwinding definitions. For the forward direction: if $g$ is continuous, then $g \circ \pi$ is a composition of continuous maps and hence continuous. For the reverse: if $g \circ \pi$ is continuous and $V \subset Z$ is open, then $\pi^{-1}(g^{-1}(V)) = (g \circ \pi)^{-1}(V)$ is open in $X$. By definition of the quotient topology, $g^{-1}(V)$ is open in $X/{\sim}$, so $g$ is continuous. This universal property characterizes the quotient topology uniquely: if $\sigma$ is any topology on $X/{\sim}$ such that $g: (X/{\sim}, \sigma) \to Z$ is continuous if and only if $g \circ \pi: X \to Z$ is continuous for every topological space $Z$, then $\sigma$ must equal $\tau_{X/\sim}$. [explanation: The Universal Property as a Design Principle] The universal property explains *why* the quotient topology is defined as it is. We could have chosen any topology on $X/{\sim}$ making $\pi$ continuous (there are many — the indiscrete topology always works). But the quotient topology is the *only* choice that produces the "correct" notion of continuity for maps out of the quotient. Consider the alternative: if we equipped $X/{\sim}$ with a strictly coarser topology $\sigma \subsetneq \tau_{X/\sim}$, then some non-continuous maps $g: (X/{\sim}, \tau_{X/\sim}) \to Z$ would become continuous when viewed as maps from $(X/{\sim}, \sigma)$. The universal property would fail — there would exist functions $g$ for which $g \circ \pi$ is continuous but $g$ is not. In category-theoretic language, the quotient space $X/{\sim}$ with its quotient topology is the **coequalizer** of the two maps $R \rightrightarrows X$ (where $R = \{(x, y) \in X \times X : x \sim y\}$ and the two maps are the coordinate projections), or equivalently the **pushout** of the diagram $\{*\} \leftarrow A \rightarrow X$ when collapsing a subspace $A$ to a point. This abstract perspective unifies the quotient construction with the product and subspace constructions, placing them into the framework of limits and colimits. [/explanation] ## Constructing Spaces by Identification The real power of the quotient topology lies in its ability to build new topological spaces from explicit geometric instructions: "glue this edge to that edge," "collapse this subspace to a point," or "identify points related by a group action." Each of these prescriptions defines an equivalence relation, and the quotient topology translates the geometric instruction into a rigorous topological space. The difficulty in each case is verifying that the resulting quotient has the properties one expects from geometric intuition — and recognizing when it does not. ### Collapsing a Subspace Given a topological space $X$ and a nonempty subspace $A \subset X$, we can form a new space by collapsing $A$ to a single point. The equivalence relation is: $x \sim y$ if and only if $x = y$ or both $x, y \in A$. The quotient $X/A := X/{\sim}$ has one equivalence class for each point of $X \setminus A$, plus one class $[A]$ for the entire subspace $A$. [example: The Sphere as a Quotient of the Disk] Returning to the opening example, we verify that $D/S^1$ is homeomorphic to $S^2$. Define \begin{align*} f: D &\to S^2 \\ (r, \theta) &\mapsto (\sin(\pi r) \cos \theta,\, \sin(\pi r) \sin \theta,\, \cos(\pi r)), \end{align*} where $(r, \theta)$ are polar coordinates on $D$ (with $0 \le r \le 1$). This map sends the origin to the south pole $(0, 0, -1)$ and the entire boundary circle $r = 1$ to the north pole $(0, 0, 1)$. It is continuous (as a composition of continuous functions), surjective, and constant on equivalence classes: if $|x| = |y| = 1$, then $f(x) = f(y) = (0, 0, 1)$. By the universal property, $f$ descends to a continuous bijection $\bar{f}: D/S^1 \to S^2$. Since $D$ is [compact](/page/Compact%20Space) and $S^2$ is [Hausdorff](/page/Topology), and quotient maps preserve compactness (the continuous image of a compact space is compact), $D/S^1$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore $\bar{f}$ is a homeomorphism: $D/S^1 \cong S^2$. [/example] ### Edge Identifications on Polygons A rich source of surfaces comes from identifying edges of a polygon. Consider the unit square $I^2 = [0, 1]^2$ with various identification patterns on its boundary. [example: The Torus from a Square] Define an equivalence relation on $I^2$ by identifying opposite edges with the same orientation: \begin{align*} (0, t) \sim (1, t) \quad &\text{for all } t \in [0, 1], \\ (s, 0) \sim (s, 1) \quad &\text{for all } s \in [0, 1]. \end{align*} The resulting quotient $I^2/{\sim}$ is the **torus** $T^2$. To see this, we construct an explicit homeomorphism. Define \begin{align*} f: I^2 &\to T^2 \subset \mathbb{R}^3 \\ (s, t) &\mapsto \big((2 + \cos 2\pi t)\cos 2\pi s,\; (2 + \cos 2\pi t)\sin 2\pi s,\; \sin 2\pi t\big), \end{align*} where $T^2$ denotes the standard torus of revolution with major radius $2$ and minor radius $1$. This map is continuous, surjective, and respects the equivalence relation: $f(0, t) = f(1, t)$ and $f(s, 0) = f(s, 1)$ because $\cos$ and $\sin$ have period $1$ in the argument $2\pi s$ (resp. $2\pi t$). By the universal property, $f$ descends to a continuous bijection $\bar{f}: I^2/{\sim} \to T^2$. Since $I^2$ is compact and $T^2 \subset \mathbb{R}^3$ is Hausdorff, $\bar{f}$ is a homeomorphism. [/example] Changing the identification pattern produces topologically distinct surfaces. The key variable is the *relative orientation* of the identified edges. [example: The Mobius Band and Klein Bottle from a Square] Modify the identification to reverse the orientation of one pair of edges: \begin{align*} (0, t) \sim (1, 1 - t) \quad \text{for all } t \in [0, 1]. \end{align*} The top and bottom edges remain free (unidentified). The resulting quotient is the **Mobius band**. An explicit embedding into $\mathbb{R}^3$ is given by \begin{align*} f: I^2 &\to \mathbb{R}^3 \\ (s, t) &\mapsto \left(\left(1 + \left(t - \tfrac{1}{2}\right)\cos \pi s\right)\cos 2\pi s,\; \left(1 + \left(t - \tfrac{1}{2}\right)\cos \pi s\right)\sin 2\pi s,\; \left(t - \tfrac{1}{2}\right)\sin \pi s\right). \end{align*} One can verify that $f(0, t) = f(1, 1 - t)$ by direct substitution: at $s = 0$, $f(0, t) = (1 + (t - 1/2), 0, 0)$, and at $s = 1$, $f(1, 1 - t) = (1 + ((1 - t) - 1/2)\cos\pi, 0, ((1-t) - 1/2)\sin\pi) = (1 + (t - 1/2), 0, 0)$. The map $f$ respects the equivalence relation, and the compact-Hausdorff argument yields a homeomorphism from $I^2/{\sim}$ onto the image. The Mobius band is non-orientable and has only *one* boundary component — unlike a cylinder (same orientation identification), which has two. If we additionally identify the top and bottom edges with the same orientation, $(s, 0) \sim (s, 1)$, the resulting quotient is the **Klein bottle** — a closed non-orientable surface that cannot be embedded in $\mathbb{R}^3$ (though it can be immersed with a single self-intersection circle). [/example] ### Orbit Spaces of Group Actions When a [group](/page/Group) $G$ acts on a topological space $X$ by homeomorphisms, the orbits of the action define an equivalence relation: $x \sim y$ if and only if there exists $g \in G$ with $g \cdot x = y$. The quotient $X/G := X/{\sim}$ is the **orbit space**, and its quotient topology captures the geometry of the action. [example: Real Projective Space] Let $S^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$ be the unit sphere, and let the group $G = \mathbb{Z}/2\mathbb{Z}$ act by the antipodal map: $x \mapsto -x$. The orbits are pairs of antipodal points $\{x, -x\}$, and the orbit space is **real projective space**: \begin{align*} \mathbb{R}P^n := S^n / (x \sim -x). \end{align*} Each point of $\mathbb{R}P^n$ represents a line through the origin in $\mathbb{R}^{n+1}$ (since a line is determined by either of its two unit direction vectors). The quotient map $\pi: S^n \to \mathbb{R}P^n$ is a two-sheeted covering map. For $n = 1$, $\mathbb{R}P^1 \cong S^1$: the circle with antipodal identification is again a circle (the map $e^{i\theta} \mapsto e^{2i\theta}$ wraps around twice). For $n = 2$, $\mathbb{R}P^2$ is a genuinely new space — a non-orientable closed surface that contains a Mobius band as a subspace. Since $S^n$ is compact and the quotient map is continuous, $\mathbb{R}P^n$ is compact. Since the antipodal action is free (no non-identity element fixes any point) and $S^n$ is Hausdorff, $\mathbb{R}P^n$ is also Hausdorff (we verify this in the next section). [/example] The same orbit-space construction applies to actions of infinite groups. The simplest and most important case is the action of $\mathbb{Z}$ on $\mathbb{R}$ by translation, which recovers the circle. [example: The Circle as a Quotient of the Real Line] The additive group $G = \mathbb{Z}$ acts on $\mathbb{R}$ by translation: $n \cdot t = t + n$. The orbits are the cosets $t + \mathbb{Z}$, and the orbit space is \begin{align*} \mathbb{R}/\mathbb{Z} \cong S^1. \end{align*} The quotient map $\pi: \mathbb{R} \to \mathbb{R}/\mathbb{Z}$ coincides with the exponential map $t \mapsto e^{2\pi i t}$, which we verified is a quotient map in the previous section. More generally, $\mathbb{R}^n / \mathbb{Z}^n \cong T^n$ (the $n$-torus), where $\mathbb{Z}^n$ acts on $\mathbb{R}^n$ by coordinate-wise translation. [/example] ## Interaction with Separation Axioms One of the most striking features of the quotient construction is its tendency to *destroy* separation properties. A quotient of a metrizable space need not be metrizable; a quotient of a Hausdorff space need not be Hausdorff. This is in sharp contrast to the subspace and product topologies, which both preserve the Hausdorff property. Understanding when separation survives a quotient — and diagnosing why it fails — is essential for working with quotient spaces in practice. ### When Quotients Fail to Be Hausdorff The failure mechanism is geometric: if two equivalence classes $[x]$ and $[y]$ cannot be separated by saturated open sets in $X$, then $[x]$ and $[y]$ cannot be separated by open sets in $X/{\sim}$. The issue is that open sets in the quotient correspond to *saturated* open sets in the original space, and the saturation requirement can force any open set containing one fibre to intersect another fibre. [example: The Line with Two Origins] Let $X = \mathbb{R} \sqcup \mathbb{R}$ be the disjoint union of two copies of the real line (with the disjoint union topology), and write points as $(x, 1)$ and $(x, 2)$ for $x \in \mathbb{R}$. Define an equivalence relation by \begin{align*} (x, 1) \sim (x, 2) \quad \text{for all } x \neq 0. \end{align*} This identifies the two copies everywhere except at the origin. The quotient $Y = X/{\sim}$ is called the **line with two origins**: it looks like $\mathbb{R}$ except that the origin has been "doubled" into two points $o_1 = [(0, 1)]$ and $o_2 = [(0, 2)]$. The space $Y$ is *not* Hausdorff. To see this, suppose $U$ and $V$ are open sets in $Y$ with $o_1 \in U$ and $o_2 \in V$. Their preimages $\pi^{-1}(U)$ and $\pi^{-1}(V)$ are open in $X$ and contain $(0, 1)$ and $(0, 2)$, respectively. By openness, $\pi^{-1}(U)$ contains an interval $(-\varepsilon, \varepsilon) \times \{1\}$ in the first copy of $\mathbb{R}$, and $\pi^{-1}(V)$ contains $(-\delta, \delta) \times \{2\}$ in the second copy. For any $t$ with $0 < |t| < \min(\varepsilon, \delta)$, the point $(t, 1)$ lies in $\pi^{-1}(U)$ and $(t, 2)$ lies in $\pi^{-1}(V)$. But $(t, 1) \sim (t, 2)$, so $[t] \in U \cap V$. Therefore $U \cap V \neq \varnothing$: the two origins cannot be separated. [/example] The line with two origins is not a pathological curiosity — it arises naturally when gluing charts in differential geometry. If two coordinate patches overlap on an open set and the transition function is a homeomorphism, the resulting quotient is Hausdorff. But if the gluing is done carelessly (e.g., the overlap set is not specified correctly), the result can be a non-Hausdorff manifold. ### Criteria for Hausdorff Quotients The following theorem gives a precise characterization. [quotetheorem:1032] The set $R$ is the **graph of the equivalence relation** — it consists of all pairs of points that are identified. When $R$ is closed, the diagonal in $Y \times Y$ (which equals $(q \times q)(R)$) is "well-behaved" enough that points outside it can be separated. For the line with two origins, $R \subset (\mathbb{R} \sqcup \mathbb{R})^2$ is not closed: the sequence $((1/n, 1), (1/n, 2))$ lies in $R$ (since $(1/n, 1) \sim (1/n, 2)$ for $n \ge 1$), but its limit $((0, 1), (0, 2))$ does not (since $(0, 1) \not\sim (0, 2)$). This confirms the failure of the Hausdorff property. For the projective space $\mathbb{R}P^n$, the relation set is $R = \{(x, -x) : x \in S^n\}$, which is the image of the continuous map $x \mapsto (x, -x)$ from the compact space $S^n$ into the Hausdorff space $S^n \times S^n$. A continuous image of a compact space in a Hausdorff space is closed, so $R$ is closed, confirming that $\mathbb{R}P^n$ is Hausdorff. ### Quotients and Compactness While quotients can destroy the Hausdorff property, they *always* preserve compactness and [connectedness](/page/Connectedness). This is an immediate consequence of the fact that the quotient map $\pi$ is a continuous surjection: the continuous image of a compact space is compact, and the continuous image of a connected space is connected. [quotetheorem:1033] This theorem, combined with the Hausdorff criterion, gives a practical strategy for identifying quotient spaces with known spaces: if $q: X \to Y$ is a continuous bijection, $X$ is compact, and $Y$ is Hausdorff, then $q$ is automatically a homeomorphism. We used this argument to identify $D/S^1 \cong S^2$ and $I^2/{\sim} \cong T^2$ in the examples above. ## When Quotient Maps Are Open or Closed A quotient map is, by definition, a surjection that satisfies the "if and only if" condition for open preimages. But knowing whether a quotient map is additionally *open* (maps open sets to open sets) or *closed* (maps closed sets to closed sets) is important for two reasons: it simplifies the verification of topological properties of the quotient, and it interacts with the Hausdorff criterion. The canonical projection $\pi: X \to X/{\sim}$ is open if and only if for every open set $U \subset X$, the *saturation* $\pi^{-1}(\pi(U)) = \bigcup_{x \in U} [x]$ is open in $X$. This is because $\pi(U)$ is open in $X/{\sim}$ precisely when its preimage — which is the saturation of $U$ — is open. [quotetheorem:1034] The verification is direct: for any open set $U \subset X$, the saturation is \begin{align*} \pi^{-1}(\pi(U)) = \bigcup_{g \in G} g \cdot U. \end{align*} Each $g \cdot U$ is the image of the open set $U$ under the homeomorphism $x \mapsto g \cdot x$, hence open. The union of open sets is open, so the saturation is open, and therefore $\pi(U)$ is open in $X/G$. This result applies to all the orbit space examples above: the projections $\mathbb{R} \to \mathbb{R}/\mathbb{Z}$, $S^n \to \mathbb{R}P^n$, and $\mathbb{R}^n \to \mathbb{R}^n/\mathbb{Z}^n$ are all open maps. Openness of the quotient map is a powerful structural property — it implies, for instance, that the quotient of a locally compact space by a group action is locally compact, and that the quotient of a first-countable space by a group action is first-countable. In contrast, quotient maps arising from collapsing a subspace $A$ to a point are typically *closed* (when $A$ is closed in $X$ and $X$ is a normal space) but not open. The projection $\pi: D \to D/S^1$ is closed but not open: the open set $B(p, \varepsilon) \cap D$ for a boundary point $p$ saturates to $B(p, \varepsilon) \cap D$ union the entire boundary $S^1$, which need not be open in $D$. ## Quotients in Algebra and Analysis When a [group](/page/Group) or vector space carries both algebraic structure and a topology, forming an algebraic quotient (by a normal subgroup, by an ideal, by a closed subspace) raises a fundamental compatibility question: does the quotient topology interact correctly with the algebraic operations? Specifically, if the group operation or vector space addition is continuous on the original space, is it still continuous on the quotient? The answer is yes — provided the equivalence relation is compatible with the algebra — and the key is that the quotient map inherits enough structure (typically openness) to transfer continuity. ### Quotient Groups If $G$ is a topological group and $N \trianglelefteq G$ is a normal subgroup, the quotient group $G/N$ carries the quotient topology induced by the projection $\pi: G \to G/N$. Since the group operation on $G$ is continuous and the left (or right) translation maps $g \mapsto ng$ are homeomorphisms for each $n \in N$, the projection $\pi$ is an open map. Therefore $G/N$ is again a topological group under the quotient topology, provided $N$ is closed. The requirement that $N$ be closed is necessary for $G/N$ to be Hausdorff: the quotient $G/N$ is Hausdorff if and only if $N$ is a closed subgroup of $G$. If $N$ is not closed, the equivalence relation graph $R = \{(g, h) : g^{-1}h \in N\}$ fails to be closed in $G \times G$, and the Hausdorff criterion fails. [example: Quotient of the Reals by the Integers] The group $(\mathbb{R}, +)$ with the standard topology is a topological group. The subgroup $\mathbb{Z}$ is closed and normal (since $\mathbb{R}$ is abelian). The quotient group $\mathbb{R}/\mathbb{Z}$ carries the quotient topology, and we have already seen that $\mathbb{R}/\mathbb{Z} \cong S^1$. The group operation on $S^1$ inherited from this quotient is multiplication of complex numbers: $e^{2\pi i s} \cdot e^{2\pi i t} = e^{2\pi i(s+t)}$. [/example] ### Quotient Banach Spaces In functional analysis, if $X$ is a [Banach space](/page/Banach%20Space) and $M \subset X$ is a closed subspace, the quotient vector space $X/M = \{x + M : x \in X\}$ carries the quotient norm \begin{align*} \|x + M\|_{X/M} := \inf_{m \in M} \|x - m\|_X. \end{align*} This norm induces the quotient topology (the finest topology making $\pi: X \to X/M$ continuous), and $X/M$ is complete — hence a Banach space. The relationship between the norm topology and the quotient topology is not a coincidence: for any topological vector space, the quotient topology by a closed subspace coincides with the topology induced by the quotient seminorm, which is a true norm precisely when $M$ is closed. The [dual space](/page/Dual%20Space) of the quotient is related to the dual of the original space by $(X/M)^* \cong M^\perp$, where $M^\perp = \{f \in X^* : f|_M = 0\}$ is the annihilator of $M$. This duality is a cornerstone of the theory of [reflexive spaces](/page/Reflexive%20Space). ## Working with Quotient Spaces ### Verifying Homeomorphism via the Compact-Hausdorff Argument The most common technique for proving that a quotient space is homeomorphic to a known space $Y$ proceeds in three steps: 1. **Construct a continuous surjection** $f: X \to Y$ that respects the equivalence relation (i.e., $x \sim x'$ implies $f(x) = f(x')$). 2. **Verify bijectivity** of the induced map $\bar{f}: X/{\sim} \to Y$. 3. **Apply the compact-Hausdorff lemma**: if $X$ is compact and $Y$ is Hausdorff, then $\bar{f}$ is automatically a homeomorphism. Step 3 works because $X/{\sim}$ is compact (as the continuous image of $X$), and a continuous bijection from a compact space to a Hausdorff space is a closed map (closed subsets of a compact space are compact, and compact subsets of a Hausdorff space are closed), hence a homeomorphism. ### Testing Whether a Given Map Is a Quotient Map To determine whether a continuous surjection $q: X \to Y$ is a quotient map, try the following approaches in order: 1. **Is $q$ open or closed?** If yes, it is a quotient map by the theorem above. 2. **Is $X$ compact and $Y$ Hausdorff?** If yes, $q$ is a closed map (hence a quotient map) by the closed map lemma. 3. **Does $q$ arise from a group action** by homeomorphisms? If yes, it is an open map. 4. **Direct verification**: find a set $V \subset Y$ with $q^{-1}(V)$ open in $X$ and show that $V$ is open in $Y$. This is typically the hardest approach and is used as a last resort. ### Proving a Quotient Space Has a Desired Property To show that $X/{\sim}$ has a topological property $P$: - **If $P$ is preserved by continuous surjections** (compactness, connectedness, path-connectedness, separability, second countability), then $P$ for $X$ implies $P$ for $X/{\sim}$ automatically. - **If $P$ requires separation** (Hausdorff, regularity, normality, metrizability), verify the Hausdorff criterion (closedness of the equivalence relation graph $R \subset X \times X$) and any additional conditions. These properties are *not* automatically preserved. - **If $P$ is local** (local compactness, local connectedness, being a manifold), check that the quotient map is open and that $P$ holds locally in $X$. Open quotient maps (such as those from group actions) preserve many local properties. ### The Danger of Non-Hausdorff Quotients When working with quotients in geometry (e.g., constructing manifolds by gluing charts), always verify the Hausdorff condition explicitly. The two most reliable strategies are: 1. **Closed graph**: show the equivalence relation $R \subset X \times X$ is closed. This works well for finite group actions on Hausdorff spaces (the relation is a finite union of graphs of homeomorphisms, each closed). 2. **Proper action**: if a group $G$ acts properly on a Hausdorff space $X$ (meaning the map $G \times X \to X \times X$, $(g, x) \mapsto (g \cdot x, x)$, is proper), then the orbit space $X/G$ is Hausdorff. ## References 1. Munkres, J. R., *Topology* (2nd ed., 2000). 2. Lee, J. M., *Introduction to Topological Manifolds* (2nd ed., 2011). 3. Willard, S., *General Topology* (1970). 4. Bredon, G. E., *Topology and Geometry* (1993).