Maps That Respect an Identification
A quotient space replaces points that have been declared equivalent by single points. A map from the original space can therefore define a map on the quotient only if it gives the same value to equivalent points. The quotient topology has a further important effect: it makes continuity of the descended map detectable by composing with the quotient map.
Let \((X,\mathcal{T})\) be a topological space. An equivalence relation \(\sim\) on \(X\) is a relation that is reflexive, symmetric, and transitive. The equivalence class of \(x\in X\) is \([x]=\{y\in X:y\sim x\}\), and \(X/{\sim}\) is the set of all equivalence classes. Write \(Q=X/{\sim}\) and \(q:X\to Q\) for the map \(q(x)=[x]\).
The collection defined this way is a topology: preimages preserve the empty set, the whole space, arbitrary unions, and finite intersections, and \(\mathcal{T}\) is a topology. In particular, to test whether a set in \(Q\) is open, it is enough—and necessary—to test its full preimage in \(X\).
Factoring Through the Quotient
Proof. First suppose \(g=f\circ q\) for some \(f:Q\to Y\). If \(x\sim y\), then \([x]=[y]\), so \(q(x)=q(y)\). Therefore \(g(x)=f(q(x))=f(q(y))=g(y)\), and \(g\) is constant on each equivalence class.
Conversely, suppose \(g(x)=g(y)\) whenever \(x\sim y\). Define \(f([x])=g(x)\). If \([x]=[y]\), then \(x\sim y\), so \(g(x)=g(y)\); hence the definition does not depend on which representative is chosen. It gives \(f(q(x))=f([x])=g(x)\) for every \(x\in X\), so \(g=f\circ q\). Finally, if \(f_1\) and \(f_2\) both satisfy \(g=f_i\circ q\), then for any \([x]\in Q\), \(f_1([x])=f_1(q(x))=g(x)=f_2(q(x))=f_2([x])\). Thus the factor is unique. \(\square\)
This proposition is set-theoretic: it does not require a topology on \(Y\), nor continuity of \(g\). When \(Y\) is a topological space and \(g\) is continuous, the quotient topology gives the stronger continuity statement below.
Proof. Suppose first that \(f\) is continuous. The quotient map \(q\) is continuous, and a composite of continuous maps is continuous, so \(g=f\circ q\) is continuous.
For the reverse direction, suppose \(g\) is continuous. Let \(W\subseteq Y\) be open. Using \(g=f\circ q\), we have \(q^{-1}(f^{-1}(W))=g^{-1}(W)\). The right-hand side is open in \(X\), since \(g\) is continuous. By the definition of the quotient topology, this implies \(f^{-1}(W)\) is open in \(Q\). This holds for every open \(W\subseteq Y\), which is exactly the definition of continuity of \(f\). \(\square\)
The equivalence depends on using the quotient topology on \(Q\). Its definition was designed so that openness in \(Q\) can be tested by pulling back along \(q\). The theorem does not say that an arbitrary continuous surjection has this property; it uses the quotient topology specifically.
Worked Examples
Worked Example: A Finite Quotient with the Sierpinski Topology
Let \(X=\{a,b,c,d\}\) with topology \(\mathcal{T}=\{\varnothing,\{a,b\},X\}\). Identify \(a\sim b\) and \(c\sim d\), with no equivalences between these two classes. Give the quotient topology and define a continuous map from the quotient to the Sierpinski space.
Write \(A=[a]=\{a,b\}\) and \(B=[c]=\{c,d\}\), so \(Q=\{A,B\}\). The preimages of the four subsets of \(Q\) are \(q^{-1}(\varnothing)=\varnothing\), \(q^{-1}(\{A\})=\{a,b\}\), \(q^{-1}(\{B\})=\{c,d\}\), and \(q^{-1}(Q)=X\). Of these, \(\varnothing,\{a,b\},X\) are open in \(X\), but \(\{c,d\}\) is not. Therefore the quotient topology is \(\{\varnothing,\{A\},Q\}\).
Let \(Y=\{u,v\}\) with topology \(\{\varnothing,\{u\},Y\}\), and define \(g:X\to Y\) by \(g(a)=g(b)=u\) and \(g(c)=g(d)=v\). It is constant on both equivalence classes, so its factor is \(f(A)=u\), \(f(B)=v\). To check continuity directly, the preimages under \(f\) of the open sets \(\varnothing,\{u\},Y\) are respectively \(\varnothing,\{A\},Q\), all open in \(Q\). The composite \(f\circ q\) is exactly \(g\), as required.
Check: the only nonempty proper open subset of \(Q\) is \(\{A\}\), and \(f^{-1}(\{u\})=\{A\}\). The corresponding preimage in \(X\) is \(\{a,b\}\), which is open. This agrees with the quotient-topology test.
Worked Example: Identifying Opposite Real Numbers
Give \(\mathbb{R}\) its standard topology and define \(x\sim y\) when \(|x|=|y|\). Identify the quotient with \([0,\infty)\), and use the universal property to descend the map \(g(x)=x^2\).
Each equivalence class has the form \([x]=\{x,-x\}\), with the two points equal when \(x=0\). Define \(\Phi:Q\to[0,\infty)\) by \(\Phi([x])=|x|\). This is well-defined by the definition of \(\sim\), and it is a bijection: every \(t\geq0\) is \(\Phi([t])\), and if \(|x|=|y|\), then \([x]=[y]\).
Here \([0,\infty)\) has the subspace topology from the standard topology on \(\mathbb{R}\). For \(W\subseteq[0,\infty)\), the quotient-topology condition for \(\Phi^{-1}(W)\) is that \(q^{-1}(\Phi^{-1}(W))=\{x\in\mathbb{R}:|x|\in W\}\) be open in \(\mathbb{R}\). This set is open exactly when \(W\) is open in \([0,\infty)\). In one direction, if \(W\) is relatively open, then each \(t\in W\) has a \(\delta>0\) such that \((t-\delta,t+\delta)\cap[0,\infty)\subseteq W\). For any \(x\) with \(|x|\in W\), the inequality \(||y|-|x||\leq|y-x|\) shows that \(y\) sufficiently near \(x\) still has \(|y|\in W\). Thus the displayed preimage is open. In the other direction, suppose that preimage is open. For \(t>0\) in \(W\), it contains \(t\), so it contains an interval \((t-\delta,t+\delta)\) for some \(\delta>0\); choosing \(\delta<t\) gives \((t-\delta,t+\delta)\subseteq W\). If \(0\in W\), the preimage contains \(0\), so it contains \((-\delta,\delta)\) for some \(\delta>0\), which gives \([0,\delta)\subseteq W\). These neighborhoods at every \(t\in W\) show that \(W\) is relatively open.
Now \(g(x)=x^2\) is continuous and constant on each class, since \(x\sim y\) implies \(x^2=|x|^2=|y|^2=y^2\). Its factor is \(f([x])=x^2\), or, in the coordinate \(\Phi([x])=t\), \(f\) is the map \(t\mapsto t^2\). The theorem guarantees that this map on the quotient is continuous. For example, \([0]\), \([1/2]\), and \([2]\) map respectively to \(0^2=0\), \((1/2)^2=1/4\), and \(2^2=4\). Substituting into the composite gives \(g(0)=0\), \(g(1/2)=1/4\), and \(g(2)=4\), matching the factor at each class.
Worked Example: The Quotient of the Integers by Parity
Give \(X=\mathbb{Z}\) the discrete topology, and identify integers when their difference is even. Find the quotient topology and descend \(g(n)=(-1)^n\) to a map into \(\mathbb{R}\) with its standard topology.
There are two classes, \(E=[0]\), the even integers, and \(O=[1]\), the odd integers. The preimage of each subset of \(Q=\{E,O\}\) is a subset of the discrete space \(\mathbb{Z}\), hence is open. Therefore every subset of \(Q\) is open, so \(Q\) has the discrete topology.
The map \(g\) is constant on each class: \(g(n)=1\) for even \(n\) and \(g(n)=-1\) for odd \(n\). It is continuous because every subset of its discrete domain is open, so every preimage of an open subset of \(\mathbb{R}\) is open. Its factor is \(f(E)=1\) and \(f(O)=-1\). Since \(Q\) is discrete, the preimage under \(f\) of any open subset of \(\mathbb{R}\) is a subset of \(Q\) and therefore open; \(f\) is continuous. For a numerical check, \(f(q(4))=f(E)=1=(-1)^4\), while \(f(q(3))=f(O)=-1=(-1)^3\).
Why the Two Conditions Matter
Common pitfall: forgetting constancy on classes. Continuity of \(g\) alone does not guarantee that it defines a map on the quotient. In the parity quotient above, take \(g(n)=n\) from discrete \(\mathbb{Z}\) to standard \(\mathbb{R}\). This map is continuous because every subset of \(\mathbb{Z}\) is open. But \(0\sim2\) while \(g(0)=0\ne2=g(2)\). A proposed factor would have to assign both \(0\) and \(2\) to the single point \(E=[0]\), which is impossible. The fiber-constancy condition is needed before continuity of a factor can even be discussed.
The universal property is useful because it replaces a continuity check on the quotient by one on the original space. To construct a map from \(Q\), first verify that the intended rule gives the same output for every representative of a class. Then define the map on classes and test continuity after composing with \(q\). This method applies regardless of whether the quotient has a familiar geometric description.
Check Your Understanding
For each question, identify the equivalence classes first, then use the quotient map and its defining preimage test.
- Let \(X=\{1,2,3\}\) have the discrete topology, with \(1\sim2\) and \(3\) in its own class. What is the quotient topology?
- On \(\mathbb{R}\) with its standard topology, identify \(x\sim y\) when \(x^2=y^2\). Does \(g(x)=|x|\) descend to the quotient? If so, give its value on \([x]\).
- In the parity quotient of \(\mathbb{Z}\), explain why \(g(n)=n^2\) cannot define a map on the quotient.
- Suppose \(g:X\to Y\) is constant on equivalence classes and continuous. Which definition of the topology on \(Q\) lets you conclude that its factor \(f:Q\to Y\) is continuous?
- If \(f:Q\to Y\) is continuous, what two continuity facts imply that \(f\circ q\) is continuous?