Successive Quotient Maps
A quotient map replaces a space by a space in which points have been identified, while defining the topology on the result through preimages. If we perform one identification and then another, it is natural to ask whether the combined map still has precisely this quotient property. The answer is yes: quotient maps compose. The proof depends on the defining open-set test, not on either map being open or closed.
For a surjection \(q:X\to Y\), the fiber over \(y\in Y\) is \(q^{-1}(\{y\})\). A subset of \(X\) is saturated with respect to \(q\) if it is a union of whole fibers. An equivalence relation on \(X\) gives a typical quotient map: each fiber is one equivalence class. The quotient topology on \(Y\) is defined by testing whether preimages are open in \(X\).
The Composition Theorem
Proof. Since \(q\) and \(r\) are surjective, for every \(z\in Z\) there is \(y\in Y\) with \(r(y)=z\), and there is \(x\in X\) with \(q(x)=y\). Thus \((r\circ q)(x)=z\), so \(r\circ q\) is surjective. Now let \(A\subseteq Z\). The preimage identity \((r\circ q)^{-1}(A)=q^{-1}(r^{-1}(A))\) holds by the definition of composition. If \(A\) is open in \(Z\), continuity of \(r\) makes \(r^{-1}(A)\) open in \(Y\), and continuity of \(q\) makes \(q^{-1}(r^{-1}(A))\) open in \(X\). Conversely, suppose \((r\circ q)^{-1}(A)\) is open in \(X\). By the quotient-map test for \(q\), applied to the subset \(r^{-1}(A)\subseteq Y\), the set \(r^{-1}(A)\) is open in \(Y\). By the quotient-map test for \(r\), \(A\) is open in \(Z\). Hence \(A\) is open exactly when its preimage under \(r\circ q\) is open, and the composite is a quotient map. \(\square\)
The reverse implication in the proof is the key step: openness of the composite preimage first passes through the quotient test for \(q\), then through the quotient test for \(r\). The argument works for arbitrary topological spaces and does not require the maps to be open or closed.
Proof. A point \(x\in X\) belongs to the left-hand side exactly when \(r(q(x))=z\). This holds exactly when \(q(x)=y\) for some \(y\in r^{-1}(\{z\})\), which means \(x\in q^{-1}(\{y\})\) for at least one such \(y\). That is precisely membership in the union on the right. \(\square\)
Worked Example: Two Identifications of a Discrete Space
Let \(X=\{a,b,c,d,e\}\) have the discrete topology, in which every subset is open. Define \(q:X\to Y=\{u,v,w\}\) by \(q(a)=q(b)=u\), \(q(c)=q(d)=v\), and \(q(e)=w\). Give \(Y\) the discrete topology. Then define \(r:Y\to Z=\{s,t\}\) by \(r(u)=r(v)=s\) and \(r(w)=t\), and give \(Z\) the discrete topology.
Both maps are surjective. For any \(V\subseteq Y\), the preimage \(q^{-1}(V)\) is a subset of the discrete space \(X\), hence open; and every \(V\subseteq Y\) is open. Therefore \(q\) is quotient. For any \(A\subseteq Z\), \(r^{-1}(A)\subseteq Y\) is open, and every \(A\subseteq Z\) is open, so \(r\) is quotient.
The composite has \((r\circ q)(a)=(r\circ q)(b)=(r\circ q)(c)=(r\circ q)(d)=s\) and \((r\circ q)(e)=t\). Its fibers are therefore \(\{a,b,c,d\}\) over \(s\) and \(\{e\}\) over \(t\). Since every subset of \(X\) is open, the preimage of every subset of \(Z\) is open. Every subset of \(Z\) is itself open, so the composite passes the quotient test.
Check: the preimages of \(\varnothing,\{s\},\{t\},Z\) are respectively \(\varnothing,\{a,b,c,d\},\{e\},X\). All four are open in \(X\), matching the four open subsets of the discrete space \(Z\).
A Non-Discrete Example: Compute Each Stage
To check quotient maps in a finite non-discrete example, list the open sets and compare them with the preimages of all subsets of the target. This makes it possible to see exactly which open sets survive each identification.
Worked Example: A Two-Stage Quotient with Sierpinski Topologies
Let \(X=\{a,b,c,d,e,f\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{a,b\},\{a,b,c,d\},X\}\). These sets form a nested chain, so their unions and finite intersections are again in the collection; thus this is a topology. Define \(q:X\to Y=\{u,v,w\}\) by sending \(a,b\) to \(u\), \(c,d\) to \(v\), and \(e,f\) to \(w\). Give \(Y\) the topology \(\{\varnothing,\{u\},\{u,v\},Y\}\).
For the eight subsets of \(Y\), the preimages are \(\varnothing,\{a,b\},\{c,d\},\{e,f\},\{a,b,c,d\},\{a,b,e,f\},\{c,d,e,f\},X\), corresponding in order to \(\varnothing,\{u\},\{v\},\{w\},\{u,v\},\{u,w\},\{v,w\},Y\). Exactly the preimages \(\varnothing,\{a,b\},\{a,b,c,d\},X\) are open in \(X\). These correspond exactly to the stated open sets of \(Y\), so \(q\) is quotient.
Now define \(r:Y\to Z=\{s,t\}\) by \(r(u)=r(v)=s\) and \(r(w)=t\). The preimages of \(\varnothing,\{s\},\{t\},Z\) are \(\varnothing,\{u,v\},\{w\},Y\). Among these, exactly \(\varnothing,\{u,v\},Y\) are open in \(Y\). Consequently the quotient topology on \(Z\) is \(\{\varnothing,\{s\},Z\}\), and \(r\) is quotient.
The composite fibers are \(\{a,b,c,d\}\) over \(s\) and \(\{e,f\}\) over \(t\). The preimages of \(\varnothing,\{s\},\{t\},Z\) under the composite are \(\varnothing,\{a,b,c,d\},\{e,f\},X\). Exactly \(\varnothing,\{a,b,c,d\},X\) are open in \(X\). Those correspond exactly to \(\varnothing,\{s\},Z\), the open sets of \(Z\).
Check: the only nonempty proper open set of \(Z\) is \(\{s\}\), and its composite preimage is \(\{a,b,c,d\}\), which is listed in \(\mathcal{T}_X\). The other nonempty proper subset \(\{t\}\) is not open, and its preimage \(\{e,f\}\) is not in \(\mathcal{T}_X\). This verifies the quotient test in both directions.
Quotient Does Not Mean Open
An open map sends every open subset of its domain to an open subset of its codomain. That is a different condition from being a quotient map, which tests openness by taking preimages of subsets of the codomain. The composition theorem uses the latter test; it does not claim that quotient maps, or their composites, must be open.
Worked Example: A Quotient Map That Is Not Open
Let \(X=\{a,b,c,d\}\) have topology \(\{\varnothing,\{a,d\},X\}\), and let \(Y=\{u,v,w\}\) have the indiscrete topology \(\{\varnothing,Y\}\). Define \(q:X\to Y\) by \(q(a)=q(b)=u\), \(q(c)=v\), and \(q(d)=w\).
The preimages of the eight subsets \(\varnothing,\{u\},\{v\},\{w\},\{u,v\},\{u,w\},\{v,w\},Y\) are, respectively, \(\varnothing,\{a,b\},\{c\},\{d\},\{a,b,c\},\{a,b,d\},\{c,d\},X\). Among these, exactly \(\varnothing\) and \(X\) are open in \(X\). These correspond exactly to the open subsets of the indiscrete space \(Y\), so the surjection \(q\) is quotient.
However, \(\{a,d\}\) is open in \(X\), and \(q(\{a,d\})=\{u,w\}\). The set \(\{u,w\}\) is a nonempty proper subset of \(Y\), so it is not open in the indiscrete topology. Therefore \(q\) is not an open map.
Check: the preimage of \(\{u,w\}\) is \(\{a,b,d\}\), which is not open in \(X\). Thus its failure to be open in \(Y\) is consistent with the quotient-map test, while the image \(q(\{a,d\})=\{u,w\}\) demonstrates separately that \(q\) is not an open map.
Common Pitfall and Why the Theorem Matters
Common pitfall. A frequent incorrect argument is: “The composite is quotient because each map is open.” Quotient maps need not be open, as the preceding calculation shows. This argument assumes an unnecessary property and does not establish the quotient condition. The correct argument starts with an arbitrary \(A\subseteq Z\) and uses \((r\circ q)^{-1}(A)=q^{-1}(r^{-1}(A))\). If this set is open, the quotient test for \(q\) makes \(r^{-1}(A)\) open, and the quotient test for \(r\) makes \(A\) open. The converse follows from continuity of both quotient maps.
This result is useful whenever a construction is performed in stages. Instead of rebuilding the final topology from scratch, one can verify that each stage is a quotient and then apply the theorem. The fiber formula tells which original points are identified in the end; the quotient test confirms that the topology on the final space is the one dictated by the composite. Neither calculation should be replaced by a picture or by an assumption that images of open sets remain open.
Check Your Understanding
Use the quotient-map definition in each question, distinguishing preimages from images.
- If \(q:X\to Y\) and \(r:Y\to Z\) are quotient maps, identify the two quotient tests used in the proof that \(r\circ q\) is quotient.
- For the two-stage example on \(X=\{a,b,c,d,e,f\}\), list the composite fibers over \(s\) and \(t\).
- In the non-discrete example, why is \(\{t\}\) not open in \(Z\), while \(\{s\}\) is open?
- Explain why the fact that a quotient map is continuous does not imply that it is an open map.
- For a composite of quotient maps, write the preimage of an arbitrary subset \(A\subseteq Z\) in terms of \(q\) and \(r\).