Reading Open and Closed Sets in a Quotient
A quotient space replaces each equivalence class in a space by a single point. Its topology is designed so that a subset of the quotient is open precisely when its inverse image in the original space is open. This gives a practical test, but it also explains a common surprise: an equivalence class need not be open, and the image of an open set need not be open if that set is not built from whole equivalence classes.
Let \((X,\mathcal{T})\) be a topological space, let \(\sim\) be an equivalence relation on \(X\), and let \(Q=X/{\sim}\) be the set of equivalence classes. Write \(q:X\to Q\) for the map that sends each point to its class. Equip \(Q\) with the quotient topology: a subset \(V\subseteq Q\) is open exactly when \(q^{-1}(V)\) is open in \(X\). The closed-set test follows by taking complements: \(F\subseteq Q\) is closed exactly when \(q^{-1}(F)\) is closed in \(X\).
The last assertion follows directly from the definition of an equivalence class: if \(x\in q^{-1}(V)\) and \(y\sim x\), then \(q(y)=q(x)\in V\), so \(y\in q^{-1}(V)\). Saturation matters when we start with a subset of \(X\) and want to take its image in \(Q\). A set that contains only some members of an equivalence class does not determine an independent portion of that class in the quotient.
The Tests and Their Consequences
Proof. The statement about open sets is the definition of the quotient topology. For the closed-set statement, \(V\) is closed in \(Q\) if and only if \(Q\setminus V\) is open in \(Q\). By the definition of the quotient topology, this is equivalent to \(q^{-1}(Q\setminus V)\) being open in \(X\). Since preimages preserve complements, \(q^{-1}(Q\setminus V)=X\setminus q^{-1}(V)\). This complement is open exactly when \(q^{-1}(V)\) is closed. \(\square\)
Proof. If \(x\in A\), then \(q(x)\in q(A)\), so \(x\in q^{-1}(q(A))\). This proves \(A\subseteq q^{-1}(q(A))\). Conversely, if \(x\in q^{-1}(q(A))\), then \(q(x)=q(a)\) for some \(a\in A\). Equality of these classes means \(x\sim a\), and saturation of \(A\) gives \(x\in A\). Thus \(q^{-1}(q(A))=A\). Apply the open-set test to \(q(A)\): it is open in \(Q\) exactly when its inverse image \(A\) is open in \(X\). The closed-set test gives the corresponding equivalence for closed sets. \(\square\)
In particular, for any equivalence class \([x]\in Q\), the one-point set \(\{[x]\}\) is open in \(Q\) exactly when the class \(q^{-1}(\{[x]\})\) is open in \(X\). It is closed in \(Q\) exactly when that class is closed in \(X\). The class, not just a chosen representative, is what must be tested.
Worked Example: A Class That Is Neither Open Nor Closed
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,c\},X\}\). This is a topology: the listed sets are nested apart from the empty set, so unions and finite intersections of listed sets are still listed. Define the equivalence classes to be \(P=\{a,b\}\) and \(C=\{c\}\). Thus \(Q=\{P,C\}\).
We find the quotient opens by listing the subsets of \(Q\) and their inverse images. The empty set has empty inverse image, and \(Q\) has inverse image \(X\), both open. The other two subsets have \(q^{-1}(\{P\})=\{a,b\}\) and \(q^{-1}(\{C\})=\{c\}\). Neither of these is on the list of open subsets of \(X\). Therefore the quotient topology is \(\{\varnothing,Q\}\), the indiscrete topology.
The class \(P\), viewed as the quotient point \(\{P\}\), is not open because its inverse image \(\{a,b\}\) is not open. It is not closed either: its inverse image \(\{a,b\}\) is not closed, since its complement \(\{c\}\) is not open. Thus this quotient point is neither open nor closed.
Check the nonsaturated-set issue. The set \(\{a\}\) is open in \(X\), but it is not saturated: it contains \(a\) and omits the equivalent point \(b\). Its image is \(q(\{a\})=\{P\}\), which is not open in \(Q\). In contrast, the saturated open set \(X\) has open image \(Q\), as the proposition predicts.
Worked Example: A Quotient with One Open Class and One Closed Class
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\), and again use the classes \(P=\{a,b\}\) and \(C=\{c\}\). These four sets form a topology because they are nested, with \(\varnothing\) and \(X\) included.
For the four subsets of \(Q=\{P,C\}\), the inverse images are \(q^{-1}(\varnothing)=\varnothing\), \(q^{-1}(\{P\})=\{a,b\}\), \(q^{-1}(\{C\})=\{c\}\), and \(q^{-1}(Q)=X\). Of these, exactly \(\varnothing,\{a,b\},X\) are open in \(X\). Hence the quotient topology is \(\{\varnothing,\{P\},Q\}\).
So \(\{P\}\) is open. It is not closed because its complement \(\{C\}\) is not open. The point \(\{C\}\) is closed because its complement \(\{P\}\) is open, but it is not open. The same answers follow from the class test: \(P=\{a,b\}\) is open but not closed in \(X\), while \(C=\{c\}\) is closed but not open. For these checks, the closed subsets of \(X\), obtained by taking complements of the four open sets, are \(X,\{b,c\},\{c\},\varnothing\).
The open set \(\{a\}\) illustrates a different outcome from the preceding example: it is not saturated, but its image is \(\{P\}\), which is open here. Thus being saturated is a sufficient condition for the image test in the proposition; it is not a necessary condition for every individual image to be open.
Worked Example: Quotienting a Discrete Space
Let \(X=\{1,2,3,4\}\) have the discrete topology, in which every subset is open. Use the equivalence classes \(A=\{1,3\}\) and \(B=\{2,4\}\). The quotient set is \(Q=\{A,B\}\).
There are four subsets of \(Q\). Their inverse images are \(q^{-1}(\varnothing)=\varnothing\), \(q^{-1}(\{A\})=\{1,3\}\), \(q^{-1}(\{B\})=\{2,4\}\), and \(q^{-1}(Q)=X\). Each inverse image is open because \(X\) is discrete. Therefore every subset of \(Q\) is open, and the quotient topology is discrete. Every subset of a discrete space is also closed, since its complement is open, so every subset of \(Q\) is closed as well.
For instance, the quotient point \(\{A\}\) is open and closed: its inverse image \(\{1,3\}\) is both open and closed in \(X\). This example confirms that a quotient class can be open and closed, but the first two examples show that this depends on the topology on the original space.
A Reliable Procedure and a Common Mistake
To test a proposed subset \(V\) of a quotient, compute its full inverse image first. To test the image \(q(A)\) of a subset \(A\subseteq X\), check whether \(A\) is saturated before using the proposition: if it is, then \(q^{-1}(q(A))=A\), and the open and closed tests apply directly. If \(A\) is not saturated, the proposition does not justify deciding whether \(q(A)\) is open or closed merely from the status of \(A\).
Write down \(q^{-1}(V)\) by collecting every point in every class represented by \(V\).
Use the given topology on \(X\): \(V\) is open exactly when the inverse image is open, and closed exactly when the inverse image is closed.
Check saturation. If \(A\) is saturated, test \(A\) itself; if it is not, compute \(q(A)\) and then its full inverse image rather than assuming \(q^{-1}(q(A))=A\).
Common pitfall. A tempting but invalid calculation is: “\(\{a\}\) is open, so \(q(\{a\})\) is open.” In the first worked example, \(\{a\}\) is open but \(q(\{a\})=\{P\}\) is not open. The direct check exposes the error: \(q^{-1}(\{P\})=\{a,b\}\), and \(\{a,b\}\) is not open in the specified topology. An open set in \(X\) may cut through an equivalence class; the quotient topology tests the full inverse image, not the original set whose image was taken.
| Question | What to check |
|---|---|
| Is \(V\subseteq Q\) open? | Is \(q^{-1}(V)\) open in \(X\)? |
| Is \(V\subseteq Q\) closed? | Is \(q^{-1}(V)\) closed in \(X\)? |
| Is \(q(A)\) open for saturated \(A\subseteq X\)? | Is \(A\) open in \(X\)? |
| Is a quotient point \(\{[x]\}\) open or closed? | Is its class \(q^{-1}(\{[x]\})\) open or closed in \(X\)? |
These tests make the quotient topology concrete: openness and closedness downstairs are determined by the corresponding properties of whole collections of equivalence classes upstairs. Checking saturation prevents a partial class from being mistaken for a subset that the quotient can distinguish.
Check Your Understanding
For each question, keep track of whether the set being tested is in \(X\) or in the quotient \(Q\).
- In the first worked example, why is the inverse image of \(\{P\}\) not open?
- In the second worked example, which quotient points are open, and which are closed?
- Prove that \(q^{-1}(V)\) is saturated for every \(V\subseteq Q\).
- For a saturated \(A\subseteq X\), explain why \(q^{-1}(q(A))=A\) can fail if saturation is removed.
- Give the inverse-image test for a quotient subset to be closed, and justify it using complements.