From Equivalence Classes to Open Sets
A quotient identifies points of a space and treats each resulting equivalence class as one point. The underlying set of the quotient is easy to list; its topology requires a separate calculation. For a finite space, the calculation can be made explicit: inspect the open sets in the original space, retain those that contain either all or none of each equivalence class, and translate each retained set into a set of quotient points.
Let \((X,\mathcal{T})\) be a topological space, and let \(\sim\) be an equivalence relation on \(X\). Write \([x]=\{y\in X:y\sim x\}\) for the equivalence class of \(x\), and let \(Y=X/{\sim}\) be the set of all equivalence classes. The quotient map is \(q:X\to Y\), where \(q(x)=[x]\). In particular, \(q\) is surjective, including when \(X=\varnothing\), in which case \(Y=\varnothing\).
The preimage \(q^{-1}(V)\) is always saturated: if it contains \(x\), then \([x]\in V\), so every \(y\sim x\) also belongs to \(q^{-1}(V)\). Conversely, every saturated \(U\subseteq X\) is the preimage of exactly one subset of \(Y\), namely \(q(U)=\{[x]:x\in U\}\). Indeed, saturation gives \(q^{-1}(q(U))=U\). These facts are what make the finite calculation work.
Proof. First, \(\mathcal{T}_q\) is a topology. The preimage of \(\varnothing\) is \(\varnothing\), and the preimage of \(Y\) is \(X\), both open. For any family \((V_i)_{i\in I}\) of subsets of \(Y\), \(q^{-1}(\bigcup_{i\in I}V_i)=\bigcup_{i\in I}q^{-1}(V_i)\), so a union of members of \(\mathcal{T}_q\) belongs to \(\mathcal{T}_q\). For \(V_1,V_2\in\mathcal{T}_q\), \(q^{-1}(V_1\cap V_2)=q^{-1}(V_1)\cap q^{-1}(V_2)\), which is open; thus finite intersections belong to \(\mathcal{T}_q\). Now suppose \(V\in\mathcal{T}_q\). Its preimage \(U=q^{-1}(V)\) is open and saturated, and surjectivity gives \(q(U)=V\). Conversely, if \(U\) is open and saturated, then \(q^{-1}(q(U))=U\) is open, so \(q(U)\in\mathcal{T}_q\). This proves both directions of the test. If \(X=\varnothing\), then \(Y=\varnothing\) and the same statements give the topology \(\{\varnothing\}\). \(\square\)
A Table-Based Procedure
Suppose the equivalence classes are \(C_1,\ldots,C_k\). For each open \(U\in\mathcal{T}\), check every class: if \(U\cap C_j\) is nonempty, does \(U\) contain all of \(C_j\)? The open set passes precisely when, for every \(j\), either \(U\cap C_j=\varnothing\) or \(C_j\subseteq U\). Each passing set is a union of classes, and hence specifies an open subset of \(Y\). If several open sets specify the same subset of classes, list that quotient open set only once.
Use the equivalence-class table to record each class as a subset of \(X\).
Use the given topology \(\mathcal{T}\), not the power set unless \(X\) is discrete.
Keep an open set only if it contains each equivalence class it meets in its entirety.
Replace each retained union of classes by the corresponding subset of quotient points.
The theorem also verifies the result: the translated sets are exactly the open sets in the quotient topology. In particular, one must not decide that a subset of quotient points is open merely because it is the image of some open subset of \(X\). Its full preimage, not an arbitrary set mapping onto it, must be open.
Worked Example: Two Classes in a Three-Point Space
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{b\},\{a,b\},X\}\). This is a topology: unions and intersections of the listed sets are again listed. Identify \(a\sim b\), leaving \(c\) in its own class. Find the quotient topology.
The classes are \(A=\{a,b\}\) and \(C=\{c\}\), so the quotient set is \(Y=\{A,C\}\). A saturated subset must be one of \(\varnothing,A,C,Y\). Testing the five source opens gives:
| Open \(U\subseteq X\) | Saturated? | Quotient set \(q(U)\), if saturated |
|---|---|---|
| \(\varnothing\) | Yes | \(\varnothing\) |
| \(\{a\}\) | No: it contains only part of \(A\) | — |
| \(\{b\}\) | No: it contains only part of \(A\) | — |
| \(\{a,b\}\) | Yes | \(\{A\}\) |
| \(X\) | Yes | \(\{A,C\}\) |
Therefore \(\mathcal{T}_q=\{\varnothing,\{A\},Y\}\). The quotient has three open sets, not all four subsets of its two-point underlying set. As a check, \(q^{-1}(\{A\})=\{a,b\}\in\mathcal{T}\), while \(q^{-1}(\{C\})=\{c\}\notin\mathcal{T}\). Hence \(\{A\}\) is open and \(\{C\}\) is not.
More Quotient Topologies from the Same Test
Worked Example: A Three-Point Quotient with Two Branches
Let \(X=\{a,a',b,c\}\) and take \(\mathcal{T}=\{\varnothing,\{a\},\{a,a'\},\{a,a',b\},\{a,a',c\},X\}\). This is a topology: the two incomparable nonempty sets \(\{a,a',b\}\) and \(\{a,a',c\}\) intersect in \(\{a,a'\}\) and have union \(X\); taking unions or intersections with the other listed sets also produces listed sets. Let the classes be \(A=\{a,a'\}\), \(B=\{b\}\), and \(C=\{c\}\).
The saturated subsets of \(X\) are exactly the unions of \(A,B,C\). Check the listed opens in order. The empty set is saturated. The set \(\{a\}\) is not, because it contains only part of \(A\). The set \(\{a,a'\}\) is saturated and translates to \(\{A\}\). The set \(\{a,a',b\}\) is saturated and translates to \(\{A,B\}\). The set \(\{a,a',c\}\) is saturated and translates to \(\{A,C\}\). Finally, \(X\) translates to \(Y=\{A,B,C\}\).
Thus the quotient topology is \(\mathcal{T}_q=\{\varnothing,\{A\},\{A,B\},\{A,C\},Y\}\). The two nontrivial proper opens are different, so neither may be omitted. As a check on the quotient definition, their preimages are respectively \(\{a,a'\}\), \(\{a,a',b\}\), and \(\{a,a',c\}\), all of which are listed source opens.
Worked Example: A Discrete Quotient from a Nondiscrete Space
Let \(X=\{u,v,w,z\}\) with \(\mathcal{T}=\{\varnothing,\{u,v\},\{w,z\},X\}\). This is a topology because the two nonempty proper opens are disjoint and their union is \(X\). The space is not discrete: for instance, \(\{u\}\notin\mathcal{T}\). Use the equivalence classes \(A=\{u,v\}\) and \(B=\{w,z\}\).
Every open set in the displayed topology is a union of whole classes. Their translations are \(\varnothing,\{A\},\{B\},\{A,B\}\). Therefore \(\mathcal{T}_q=\mathcal{P}(Y)\), the discrete topology on the two-point quotient. For a direct check, \(q^{-1}(\{A\})=\{u,v\}\) and \(q^{-1}(\{B\})=\{w,z\}\), and both are open. This example shows that the quotient can be discrete even though the original space is not.
A Common Mistake: Taking Images of Opens
Consider \(X=\{a,b,c\}\) with \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\), and identify \(a\sim c\), with \(b\) alone in its class. This is a topology because its listed sets are nested. Write \(A=\{a,c\}\) and \(B=\{b\}\). The open set \(\{a\}\) has image \(q(\{a\})=\{A\}\). But \(\{A\}\) is not quotient-open: its full preimage is \(q^{-1}(\{A\})=\{a,c\}\), which is not one of the source opens. The saturated open sets are only \(\varnothing\) and \(X\), so the quotient topology is \(\{\varnothing,\{A,B\}\}\), the indiscrete topology.
The incorrect calculation would be: “\(\{a\}\) is open, so its image \(\{A\}\) is open in the quotient.” The correction is to test the full preimage: \(q^{-1}(\{A\})=\{a,c\}\notin\mathcal{T}\), so \(\{A\}\) is not open. Quotient maps are defined by the preimage condition; they are not required to send every open set to an open set.
Check Your Understanding
In each problem, list the saturated source opens before writing down the quotient topology.
- Let \(X=\{p,q,r\}\) have topology \(\{\varnothing,\{p\},\{p,q\},X\}\), with classes \(\{p\}\) and \(\{q,r\}\). Find the quotient topology.
- In a finite discrete space \(X\), prove that every equivalence relation gives the discrete topology on \(X/{\sim}\).
- For the space in the common-mistake example, compute \(q^{-1}(\varnothing)\) and \(q^{-1}(\{A,B\})\), and use them to verify both quotient opens.
- Can a nonempty quotient open set have empty preimage? Explain using the definition of \(q\).
- For an empty space \(X\), state its quotient set and quotient topology under the unique equivalence relation.