Quotient Topology and Identification

The quotient topology provides a rigorous framework for "gluing" and "identifying" points in a topological space. It is the categorical dual of the subspace topology and is essential for constructing surfaces, manifolds, and CW complexes.

Quotient Maps and Quotient Topology

Definition3.3Quotient Map

A surjective map $p: X \to Y$ between topological spaces is a quotient map if $U \subseteq Y$ is open if and only if $p^{-1}(U)$ is open in $X$ . Equivalently, $Y$ carries the finest topology making $p$ continuous.

Definition3.4Quotient Topology

Let $X$ be a topological space and let $\sim$ be an equivalence relation on $X$ . The quotient space $X/{\sim}$ is the set of equivalence classes $\{[x] : x \in X\}$ equipped with the quotient topology: $\tau_{X/\sim} = \{U \subseteq X/{\sim} : q^{-1}(U) \text{ is open in } X\},$ where $q: X \to X/{\sim}$ is the canonical projection $q(x) = [x]$ .

The map $q$ is automatically a quotient map.

RemarkQuotient Topology is the Finest

The quotient topology is the finest topology on $X/{\sim}$ making the projection $q$ continuous. Any topology on $X/{\sim}$ making $q$ continuous is contained in the quotient topology.

Examples of Quotient Spaces

ExampleClassical Quotient Constructions

Circle from interval: $[0,1]/(0 \sim 1) \cong S^1$ . Identifying the endpoints of a closed interval yields the circle.
Torus from square: $[0,1]^2/(x,0) \sim (x,1), (0,y) \sim (1,y)$ gives the torus $T^2$ .
Klein bottle: $[0,1]^2/(x,0) \sim (x,1), (0,y) \sim (1, 1-y)$ gives the Klein bottle.
Real projective space: $\mathbb{R}P^n = S^n/(x \sim -x)$ , identifying antipodal points.
Collapsing a subspace: If $A \subseteq X$ , the quotient $X/A$ identifies all points of $A$ to a single point. For example, $D^2/S^1 \cong S^2$ (collapsing the boundary of the disk to a point yields the sphere).

Universal Property

Theorem3.3Universal Property of Quotient Spaces

Let $q: X \to X/{\sim}$ be a quotient map. A function $f: X/{\sim} \to Y$ is continuous if and only if $f \circ q: X \to Y$ is continuous.

Proof

( $\Rightarrow$ ): If $f$ is continuous, then $f \circ q$ is a composition of continuous maps.

( $\Leftarrow$ ): Suppose $f \circ q$ is continuous. Let $V \subseteq Y$ be open. Then $q^{-1}(f^{-1}(V)) = (f \circ q)^{-1}(V)$ is open in $X$ . By definition of the quotient topology, $f^{-1}(V)$ is open in $X/{\sim}$ . So $f$ is continuous.

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This universal property states that $X/{\sim}$ solves the problem: it is the "finest" space through which continuous maps that are constant on equivalence classes can factor.

Conditions for Quotient Maps

Theorem3.4Sufficient Conditions for Quotient Maps

A continuous surjection $p: X \to Y$ is a quotient map if any of the following hold:

$p$ is an open map.
$p$ is a closed map.
$X$ is compact and $Y$ is Hausdorff.

Proof

For (1): Let $p^{-1}(U)$ be open in $X$ . We need $U$ open in $Y$ . Since $p$ is surjective, $p(p^{-1}(U)) = U$ . Since $p$ is open, $U$ is open.

For (2): Let $p^{-1}(C)$ be closed in $X$ . Then $p(p^{-1}(C)) = C$ . Since $p$ is closed, $C$ is closed in $Y$ . The characterization with open sets follows by taking complements.

For (3): By Theorem 2.11, a continuous bijection from compact to Hausdorff is a homeomorphism. More generally, a continuous surjection from compact to Hausdorff is a closed map (image of a compact subset is compact, hence closed in Hausdorff), so it is a quotient map by (2).

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ExampleQuotient Maps Need Not Be Open

Consider $q: [0, 2] \to [0, 2]/(1 \sim 2)$ collapsing $\{1, 2\}$ to a point. The set $(1/2, 3/2)$ is open in $[0, 2]$ with subspace topology. Its saturation is $(1/2, 3/2) \cup \{2\}$ , which is not open. Hence $q((1/2, 3/2)) = q((1/2, 3/2) \cup \{2\})$ is not necessarily open. In fact, $q$ is a closed map (continuous from compact to Hausdorff-like space) but not open.

Identification Spaces

Definition3.5Identification (Attaching) Map

Let $X$ and $Y$ be topological spaces, $A \subseteq X$ a subspace, and $f: A \to Y$ a continuous map. The adjunction space $Y \cup_f X$ is the quotient: $Y \cup_f X = (X \sqcup Y)/{\sim}$ where $a \sim f(a)$ for all $a \in A$ , and $\sqcup$ denotes disjoint union.

This construction "attaches" $X$ to $Y$ along $A$ via the map $f$ .

ExampleAttaching a Cell

Let $Y$ be a space and $f: S^{n-1} \to Y$ a continuous map. The adjunction space $Y \cup_f D^n$ is obtained by attaching an $n$ -cell to $Y$ via $f$ . The interior of $D^n$ maps homeomorphically into the quotient, and $S^{n-1}$ is identified with its image in $Y$ .

CW complexes are built inductively by such cell attachments: $X^{(n)} = X^{(n-1)} \cup_f \bigsqcup_\alpha D^n_\alpha$ .