2.10Quotient Topology

Let $X, Y$ be topological spaces, and let $p: X \to Y$ be surjective. Then $p$ is a quotient map if every subset $U \subseteq Y$ is open if and only if $p^{-1}(U)$ is open.

A quotient map must be surjective, because then $p^{-1}(\emptyset) = \emptyset$. If $p^{-1}(\emptyset)$ weren't empty, then we can force a non-open set to be the preimage of the open set, contradicting the definition.

Note this is stronger than continuity. $p$ would be continuous if $p^{-1}(U) \subseteq X$ was open given that $U \subseteq Y$ is open, which is only the reverse direction for the definition of a quotient map. For $p$ to be a quotient map, it must be continuous, surjective, and it must be true that if $p^{-1}(U) \subseteq X$ is open, then $U \subseteq Y$ is too.

Note that a quotient map does not necessarily take an open $O \subseteq X$ to an open $p(O) \subseteq Y$. By definition, we'd need $p^{-1}(p(O))$ to be open, but $p^{-1}(p(O)) \subseteq O$, and subsets of $O$ need not be open. $p^{-1}(p(O)) = O$ is forced when $p$ is injective, but this would make $p$ a bijective map that is continuous both ways, making $p$ a homeomorphism. Indeed, a quotient map is simply a less restrictive homeomorphism, one that gets rid of injectivity in the hypothesis.

A set $C \subseteq X$ is saturated with respect to $f$ if it has the property that if it intersects some fiber $f^{-1}(\{y\})$, then $C$ contains the entire fiber. That is, $C \subseteq X$ is saturated if it is the union of fibers of $f$.

Let $f: X \to Y$ be a map between two topological spaces. $f$ is an open map if, for every open $U \subseteq X$, $f(U)$ is open. Analogously, $f$ is a closed map if, for every closed $U \subseteq X$, $f(U)$ is closed.

Let $X$ be a topological space and let $A$ be a set. Let $p: X \to A$ be a surjective map. Then there is exactly one topology, the quotient topology, that can be endowed on $A$ to make $p$ a quotient map.

The definition of a quotient map makes the quotient topology unique. Indeed, with our previous definition in mind, our topology on $A$ must be such that $U$ is in the topology if and only if $p^{-1}(U)$ is open in $X$. Of course, the sets $\emptyset$ and $A$ are open in the quotient topology, since $p^{-1}(\emptyset) = \emptyset$ and $p^{-1}(A) = X$, which are open sets in $X$. The conditions on arbitrary unions and finite intersections come by how preimages work:

and so the quotient topology is a valid topology. Moreover, it is uniquely determined by the topology on $X$, since the only subsets of $A$ that are open are the ones such that $p^{-1}(U)$ is open. Adding an additional set to the quotient topology would make $p$ no longer continuous, whereas omitting an existing set in the quotient topology would make $p$ no longer a quotient map.

Let $X$ be a topological space, and let $X^*$ be a partition of $X$ (a set of disjoint subsets of $X$ whose union is $X$). Let $p: X \to X^*$ be an identification map that takes elements of $X$ to the subset they belong to in $X^*$. In the quotient topology induced by $p$, $X^*$ is called a quotient space of $X$.

Recall the idea behind the canonical set decomposition in $\sf{Set}$ and its representation as the First Isomorphism Theorem in $\sf{Grp}$.

For any set $X$ and some map out of it, say $f: X \to Y$, the image of the map induces an equivalence relation that partitions $X$. Call this partition $X^*$. We define $\ker(f) = {\sim}$ to be such that $x_1 \sim x_2 \iff f(x_1) = f(x_2)$. Then two elements of $X$ belong to the same subset of $X^*$ if their image is the same under $f$. Essentially, $X^*$ changes the notion of equivalence, where two points of $X$ are indistinguishable in $X^*$ if they are “equivalent” to eachother, where equivalence is defined by having the same image under $f$.

We can decompose the action of $f$ into three steps. First, we can define a map $\pi: X \to X^*$ that sends an element in $X$ to its equivalence class in $X^*$. Then, since $X^*$ is defined specifically so that elements of its equivalence classes have the same image, we can define a map $\tilde{f}: X^* \to \im(f)$ that sends each equivalence class to its image. Finally, we can define an inclusion map $\iota: \im(f) \to Y$ that embeds the image of $f$ into its codomain. This is represented by this commutative diagram.

Notice that $\tilde f$ is a bijection. Equating points of $X$ whose image is equal under $f$ resolves the non-injectivity of $f$, since the many points that may map to an element in the image collapse to one point under $\sim$. Mapping into $\im(f)$ instead of $Y$ automatically resolves the non-surjectivity of $f$. If $f$ is either injective or surjective, then these processes collapse trivially—namely, if $f$ is injective, then $X^*$ is a set of singleton subsets of $X$, and if $f$ is surjective, then $\im(f) = Y$.

Let $X$ be a topological space and let $X^*$ be a partition of $X$ that is a quotient space relative to the identification map $p: X \to X^*$. Recall a set $U$ is open in the quotient topology endowed on $X^*$ is $p^{-1}(U)$ is open. A subset $U \subseteq X^*$ is a set of equivalence classes of $X$, and its preimage under $p$ is simply the union of each equivalence class. Thus $U$ is open if the union of equivalence classes it contains is an open subset of $X$.

Suppose $p: X \to Y$ is a quotient map, and let $A \subseteq X$ be a subspace. Then let $p|_A = q: A \to p(A)$ be the restriction.

1. If $A$ is an open or closed subset of $X$, then $q$ is a quotient map.

2. If $p$ is an open or closed map, then $q$ is a quotient map.