Quotient Map Characterization

Quotient maps are the central notion for constructing new spaces from old ones by identification. This theorem provides multiple equivalent conditions for recognizing quotient maps, which is essential since verifying the definition directly can be difficult.

Main Characterization

Theorem3.8Characterization of Quotient Maps

Let $p: X \to Y$ be a surjective continuous map. The following are equivalent:

$p$ is a quotient map (i.e., $U \subseteq Y$ is open $\iff$ $p^{-1}(U)$ is open in $X$ ).
$C \subseteq Y$ is closed $\iff$ $p^{-1}(C)$ is closed in $X$ .
$p$ maps saturated open sets to open sets.
$p$ maps saturated closed sets to closed sets.

Recall that $S \subseteq X$ is saturated with respect to $p$ if $S = p^{-1}(p(S))$ .

Proof

(1 $\Leftrightarrow$ 2): Taking complements: $U$ is open in $Y$ iff $Y \setminus U$ is closed in $Y$ , and $p^{-1}(U)$ is open in $X$ iff $p^{-1}(Y \setminus U) = X \setminus p^{-1}(U)$ is closed in $X$ . So condition (1) for open sets is equivalent to condition (2) for closed sets.

(1 $\Leftrightarrow$ 3): Suppose (1) holds. If $U \subseteq X$ is saturated and open, then $U = p^{-1}(p(U))$ , so $p^{-1}(p(U))$ is open in $X$ . By (1), $p(U)$ is open in $Y$ .

Conversely, suppose (3) holds. Since $p$ is continuous, the forward direction of (1) is clear. For the converse, suppose $p^{-1}(V)$ is open in $X$ for some $V \subseteq Y$ . The set $p^{-1}(V)$ is saturated (preimages are always saturated), so by (3), $p(p^{-1}(V)) = V$ (using surjectivity) is open in $Y$ .

(2 $\Leftrightarrow$ 4): Identical argument with "open" replaced by "closed."

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Sufficient Conditions

Theorem3.9Sufficient Conditions for Quotient Maps

Each of the following is sufficient for a continuous surjection $p: X \to Y$ to be a quotient map:

$p$ is an open map.
$p$ is a closed map.
$p$ admits a continuous right inverse (i.e., a continuous section $s: Y \to X$ with $p \circ s = \operatorname{id}_Y$ ).

Proof

(1): If $p$ is open, then for any saturated open $U \subseteq X$ , $p(U)$ is open. This is condition (3) above.

(2): If $p$ is closed, then for any saturated closed $C \subseteq X$ , $p(C)$ is closed. This is condition (4) above.

(3): Suppose $s: Y \to X$ is a continuous section. For any $V \subseteq Y$ : if $p^{-1}(V)$ is open in $X$ , then $V = p(s(V)) \subseteq p(p^{-1}(V))$ . But also $s^{-1}(p^{-1}(V)) = (p \circ s)^{-1}(V) = V$ , and since $s$ is continuous, $V$ is open if $p^{-1}(V)$ is open. (More precisely: $V = s^{-1}(p^{-1}(V))$ is open since $s$ is continuous.)

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Quotients and Composition

Theorem3.10Composition of Quotient Maps

If $p: X \to Y$ and $q: Y \to Z$ are quotient maps, then $q \circ p: X \to Z$ is a quotient map.

Proof

First, $q \circ p$ is surjective (composition of surjections) and continuous (composition of continuous maps). Let $W \subseteq Z$ and suppose $(q \circ p)^{-1}(W) = p^{-1}(q^{-1}(W))$ is open in $X$ . Since $p$ is a quotient map, $q^{-1}(W)$ is open in $Y$ . Since $q$ is a quotient map, $W$ is open in $Z$ .

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Quotient Maps and the Universal Property

Theorem3.11Passing to the Quotient

Let $p: X \to Y$ be a quotient map and $f: X \to Z$ a continuous map that is constant on the fibers of $p$ (i.e., $p(x_1) = p(x_2) \implies f(x_1) = f(x_2)$ ). Then there exists a unique continuous map $\bar{f}: Y \to Z$ with $f = \bar{f} \circ p$ .

$\begin{array}{ccc} X & \xrightarrow{f} & Z \\ \downarrow^{p} & \nearrow_{\bar{f}} & \\ Y & & \end{array}$

Moreover:

$\bar{f}$ is a quotient map if $f$ is a quotient map.
$\bar{f}$ is injective if and only if $f$ is constant on fibers and distinct fibers have distinct images.
$\bar{f}$ is a homeomorphism if and only if $f$ is a quotient map and $\bar{f}$ is a bijection.

Proof

Existence: Define $\bar{f}(y) = f(x)$ for any $x \in p^{-1}(y)$ . This is well-defined since $f$ is constant on fibers.

Uniqueness: If $\bar{f} \circ p = f = \bar{g} \circ p$ , then $\bar{f}(y) = \bar{f}(p(x)) = f(x) = \bar{g}(p(x)) = \bar{g}(y)$ for any $x \in p^{-1}(y)$ .

Continuity: For $W$ open in $Z$ , $p^{-1}(\bar{f}^{-1}(W)) = (\bar{f} \circ p)^{-1}(W) = f^{-1}(W)$ , which is open since $f$ is continuous. Since $p$ is a quotient map, $\bar{f}^{-1}(W)$ is open.

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ExampleInduced Map on Projective Space

The map $f: S^n \to \mathbb{R}P^n$ given by $f(x) = [x]$ (equivalence class under the antipodal relation $x \sim -x$ ) is a quotient map. If $g: S^n \to Y$ is continuous with $g(x) = g(-x)$ for all $x$ , then $g$ factors uniquely through $\mathbb{R}P^n$ : there exists a unique continuous $\bar{g}: \mathbb{R}P^n \to Y$ with $g = \bar{g} \circ f$ .