The Quotient Topology

Building a Topology on a Target

A surjection $q:X\to Y$ can be used to transfer the topology of $X$ to the set $Y$. The central idea is to test subsets of $Y$ by pulling them back to $X$: a subset of $Y$ is declared open precisely when its inverse image under $q$ is open in $X$. This construction is especially useful when different points of $X$ are to be treated as one point of $Y$.

Suppose $(X,\mathcal{T})$ is a topological space and $Y$ is a set. Assume throughout that $q:X\to Y$ is surjective. At this stage $Y$ need not have a topology. For $V\subseteq Y$, its inverse image is $q^{-1}(V)=\{x\in X:q(x)\in V\}$. Recall that a map is continuous if the inverse image of every open set in its target is open in its domain.

Definition: The quotient topology induced by $q$ is the collection $$ \mathcal{T}_q=\{V\subseteq Y:q^{-1}(V)\in\mathcal{T}\}. $$ When $Y$ is given this topology, the map $q:(X,\mathcal{T})\to(Y,\mathcal{T}_q)$ is the quotient map associated with the construction.

The definition uses inverse images, not images. This is essential: inverse images preserve unions and intersections, which allows the topology axioms on $Y$ to follow directly from those on $X$. The surjectivity assumption also ensures that $q$ reaches every point of the set being given a topology; it will make the relation between subsets of $Y$ and their preimages faithful.

Why the Definition Gives a Topology

Theorem 1: The collection $\mathcal{T}_q$ is a topology on $Y$.

Proof. We check the topology axioms. Since $q^{-1}(\varnothing)=\varnothing$ and $q^{-1}(Y)=X$, and both $\varnothing$ and $X$ are in $\mathcal{T}$, the sets $\varnothing$ and $Y$ belong to $\mathcal{T}_q$. Let $(V_i)_{i\in J}$ be any family of sets in $\mathcal{T}_q$, including the possibility that $J$ is empty. Then $q^{-1}(\bigcup_{i\in J}V_i)=\bigcup_{i\in J}q^{-1}(V_i)$, which is open in $X$ because each preimage is open and arbitrary unions of open sets are open. For a finite family $V_1,\ldots,V_n$ in $\mathcal{T}_q$, with $n\geq 1$, $q^{-1}(\bigcap_{i=1}^nV_i)=\bigcap_{i=1}^nq^{-1}(V_i)$, which is open in $X$ because finite intersections of open sets are open. Thus the union and finite-intersection axioms hold. The empty finite intersection is $Y$, already checked. Therefore $\mathcal{T}_q$ is a topology on $Y$. $\square$

Theorem 2 (Finest-topology property): The map $q:(X,\mathcal{T})\to(Y,\mathcal{T}_q)$ is continuous. If $\mathcal{S}$ is any topology on $Y$ for which $q:(X,\mathcal{T})\to(Y,\mathcal{S})$ is continuous, then $\mathcal{S}\subseteq\mathcal{T}_q$. Hence $\mathcal{T}_q$ is the finest topology on $Y$ making $q$ continuous.

Proof. For every open set $V\in\mathcal{T}_q$, its inverse image $q^{-1}(V)$ is open in $X$ by the definition of $\mathcal{T}_q$; this is exactly continuity of $q$. Now suppose $\mathcal{S}$ is a topology on $Y$ and $q$ is continuous as a map to $(Y,\mathcal{S})$. For each $V\in\mathcal{S}$, continuity says $q^{-1}(V)\in\mathcal{T}$. The definition of $\mathcal{T}_q$ then gives $V\in\mathcal{T}_q$. Thus every member of $\mathcal{S}$ is a member of $\mathcal{T}_q$, proving $\mathcal{S}\subseteq\mathcal{T}_q$. Since a finer topology contains more open sets, this inclusion proves the stated maximality. $\square$

This property concerns topologies on the same set $Y$, ordered by inclusion. It does not say that every set in $Y$ is open. Rather, it says that among the topologies on $Y$ for which this particular map is continuous, the quotient topology declares as many sets open as possible.

Computing the Topology in Examples

Worked Example: A Surjection Between Finite Sets

Let $X=\{a,b,c,d\}$ have the discrete topology $\mathcal{T}=\mathcal{P}(X)$, let $Y=\{u,v\}$, and define $q(a)=q(b)=q(c)=u$ and $q(d)=v$. Compute $\mathcal{T}_q$.

There are four subsets of $Y$. Their inverse images are $q^{-1}(\varnothing)=\varnothing$, $q^{-1}(\{u\})=\{a,b,c\}$, $q^{-1}(\{v\})=\{d\}$, and $q^{-1}(Y)=X$. Each is open because $X$ has the discrete topology, in which every subset is open. Consequently $\mathcal{T}_q=\{\varnothing,\{u\},\{v\},Y\}=\mathcal{P}(Y)$, the discrete topology on $Y$.

Check: The computation tested every possible subset of $Y$, and each of its preimages is in $\mathcal{T}$. In particular, both singleton subsets of $Y$ are open in the induced topology.

Worked Example: The Absolute-Value Map

Let $X=\mathbb{R}$ with its standard topology, let $Y=[0,\infty)$ as a set, and define $q:\mathbb{R}\to Y$ by $q(x)=|x|$. Determine the quotient topology on $Y$ and compare it with the subspace topology inherited from $\mathbb{R}$.

The map is surjective: for each $y\geq 0$, $q(y)=y$. We show that $V\subseteq Y$ has open preimage in $\mathbb{R}$ exactly when it is open in the subspace topology on $Y$.

First suppose $V$ is open in the subspace topology. For each $y\in V$, there is an open interval $(y-\varepsilon,y+\varepsilon)$ in $\mathbb{R}$ such that $(y-\varepsilon,y+\varepsilon)\cap Y\subseteq V$, for some $\varepsilon>0$. If $x\in q^{-1}(V)$, then $y=|x|\in V$. Whenever $|x'-x|<\varepsilon$, the triangle inequality gives $\big||x'|-|x|\big|\leq |x'-x|<\varepsilon$. Thus $|x'|\in(y-\varepsilon,y+\varepsilon)\cap Y\subseteq V$. So the interval $(x-\varepsilon,x+\varepsilon)$ lies in $q^{-1}(V)$. This holds around each $x\in q^{-1}(V)$, proving that $q^{-1}(V)$ is open.

Conversely, suppose $q^{-1}(V)$ is open, and take $y\in V$. If $y>0$, then $y\in q^{-1}(V)$, so some interval $(y-\delta,y+\delta)$ is contained in $q^{-1}(V)$. Set $\eta=\min(\delta/2,y/2)>0$. If $z\in Y$ and $|z-y|<\eta$, then $z\in(y-\delta,y+\delta)$, so $z=q(z)\in q^{-1}(V)$, which means $z\in V$. If $y=0$, openness of $q^{-1}(V)$ gives a $\delta>0$ with $(-\delta,\delta)\subseteq q^{-1}(V)$. Every $z\in Y$ satisfying $0\leq z<\delta$ is in this interval, so $[0,\delta)\subseteq V$. In both cases, $y$ has a neighborhood in $Y$ contained in $V$; therefore $V$ is subspace-open.

Conclusion and check: the quotient topology induced by $q$ is exactly the standard subspace topology on $[0,\infty)$. The proof checked both implications, including the boundary point $0$, where neighborhoods in $Y$ have the form $[0,\delta)$.

Worked Example: A Finite Target with an Indiscrete Quotient Topology

Let $X=\{a,b,c\}$ have topology $\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}$. Let $Y=\{0,1\}$, and define the surjection $q(a)=q(b)=0$, $q(c)=1$. Compute the quotient topology.

The four subsets of $Y$ have preimages $q^{-1}(\varnothing)=\varnothing$, $q^{-1}(\{0\})=\{a,b\}$, $q^{-1}(\{1\})=\{c\}$, and $q^{-1}(Y)=X$. The sets $\varnothing$, $\{a,b\}$, and $X$ are open in $X$, while $\{c\}$ is not in the listed topology. Hence $\mathcal{T}_q=\{\varnothing,\{0\},Y\}$.

This is a topology: it contains the empty set and $Y$, and unions and finite intersections of its sets remain in the collection. The map is continuous because the preimages of its three open sets are $\varnothing,\{a,b\},X$, all open in $X$. Notice that the quotient topology here is not discrete: $\{1\}$ is not open.

A Common Pitfall: Confusing a Set with Its Preimage

There is no general rule that a subset $V\subseteq Y$ is open just because it looks like a small or simple subset of $Y$. Before the quotient topology has been defined, $Y$ has no designated open sets at all. Afterwards, openness is decided by checking whether $q^{-1}(V)$ is open in $X$.

Wrong calculation: In the preceding finite example, one might say, “$\{1\}$ is a singleton, so it is open in $Y$.” This incorrectly imports the discrete topology onto $Y$. The defining test gives $q^{-1}(\{1\})=\{c\}$, and $\{c\}\notin\mathcal{T}$. Therefore $\{1\}\notin\mathcal{T}_q$. The correction is to compute the preimage and check it against the specified topology on $X$, rather than assume that singletons are open.

Surjectivity gives one further useful check. If $q$ is surjective and $q^{-1}(V)=q^{-1}(W)$ for subsets $V,W\subseteq Y$, then $V=W$: for any $y\in V$, choose $x\in X$ with $q(x)=y$; membership of $x$ in the equal preimages implies $y\in W$, and the reverse inclusion follows in the same way. Thus distinct subsets of $Y$ have distinct preimages. This does not replace the openness test, but it explains why the preimage records the target subset without losing information.

The quotient topology is therefore a controlled way to put a topology on a target set: the topology on $X$ supplies the test, continuity is built into the definition, and the finest-topology theorem specifies exactly how much openness that test permits.

Check Your Understanding

For each question, begin by writing the relevant inverse image and then apply the definition.

For the surjection in the first worked example, what is the quotient topology if $X$ is given the indiscrete topology instead?
In the third worked example, which subsets of $Y$ are not open in the quotient topology, and what are their preimages?
Explain why the inverse image of an arbitrary union is a union of inverse images, and why this matters for Theorem 1.
Can a topology on $Y$ making $q$ continuous contain an open set that is not in $\mathcal{T}_q$? Justify your answer.
For $q(x)=|x|$, describe the form of the preimage of a subset $V\subseteq[0,\infty)$ in terms of positive and zero elements of $V$.

What You'll Learn