Why Openness Matters for Products
A quotient map transfers a topology from one space to another by identifying points. Taking a product with another space is a natural way to make the same identifications in each slice. But quotient maps need not remain quotient after taking products. An openness hypothesis gives a useful sufficient condition: it ensures that the product map is itself open, and hence quotient.
All products below carry the product topology. In particular, a basis for \(X\times Z\) consists of rectangles \(U\times W\), where \(U\) is open in \(X\) and \(W\) is open in \(Z\).
The continuity condition in this definition is essential. Openness and surjectivity alone do not imply continuity, and therefore do not imply that a map is quotient.
Proof. Let \(A\subseteq Y\). If \(A\) is open, then \(f^{-1}(A)\) is open by continuity. Conversely, suppose \(f^{-1}(A)\) is open. Since \(f\) is open, \(f(f^{-1}(A))\) is open in \(Y\). Surjectivity gives \(f(f^{-1}(A))=A\): inclusion in \(A\) follows from the definition of preimage, and every \(y\in A\) equals \(f(x)\) for some \(x\in X\), which then belongs to \(f^{-1}(A)\). Thus \(A\) is open, proving the quotient condition. \(\square\)
This lemma separates the work in the product theorem into two tasks: prove continuity and openness of the product map, and verify that it is surjective. The lemma then supplies the quotient property.
Product with an Identity Map
Proof. First, the map is surjective: for any \((y,z)\in Y\times Z\), surjectivity of \(q\) supplies \(x\in X\) with \(q(x)=y\), and then \((q\times\operatorname{id}_Z)(x,z)=(y,z)\). To prove continuity, take a basis rectangle \(V\times W\) in \(Y\times Z\), with \(V\) open in \(Y\) and \(W\) open in \(Z\). Its preimage is \((q\times\operatorname{id}_Z)^{-1}(V\times W)=q^{-1}(V)\times W\), which is open because \(q\) is continuous. Every open set in the product is a union of basis rectangles, so its preimage is open as well.
For openness, let \(G\) be open in \(X\times Z\). It is a union of basis rectangles \(U_i\times W_i\), with \(U_i\) open in \(X\) and \(W_i\) open in \(Z\). Images commute with unions, and \((q\times\operatorname{id}_Z)(U_i\times W_i)=q(U_i)\times W_i\). Since \(q\) is open, each \(q(U_i)\) is open in \(Y\); hence each image rectangle is open in \(Y\times Z\). Their union, which is the image of \(G\), is open. We have proved that the product map is a continuous, open surjection. The lemma now shows that it is quotient. \(\square\)
The theorem does not say that every product of quotient maps is quotient. Its input requires openness as well as the quotient-map properties. In particular, the proof of openness uses the fact that every \(q(U_i)\) is open.
Worked Example: A Surjective Open Map That Is Not Quotient
This two-point example checks why continuity cannot be omitted when using the lemma. Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\), and let \(Y=\{a,b\}\) have the discrete topology \(\mathcal{P}(Y)\). Define \(f:X\to Y\) by \(f(a)=a\) and \(f(b)=b\).
The map is surjective. It is open because the only open sets in \(X\) are \(\varnothing\) and \(X\), and their images \(\varnothing\) and \(Y\) are open in \(Y\). But \(f\) is not continuous: the singleton \(\{a\}\) is open in \(Y\), while \(f^{-1}(\{a\})=\{a\}\) is not open in \(X\).
It is not quotient either. The set \(\{a\}\subseteq Y\) has open preimage \(\{a\}\) in \(X\) only if that preimage is open; it is not. More directly, the quotient condition fails in the other direction: \(f^{-1}(\varnothing)=\varnothing\) is open, as expected, but for \(A=\{a\}\), \(f^{-1}(A)\) is not open. Thus this example shows the map is not quotient, and that the lemma's continuity assumption cannot be dropped.
Check: the failure of continuity is exactly visible from the open singleton in the target. Surjectivity and openness do not repair the missing preimage condition.
A Careful Check: Endpoint Identification
A familiar identification map illustrates why one should test openness before applying the theorem. Give \(I=[0,1]\) its usual subspace topology from \(\mathbb{R}\), identify \(0\) with \(1\), and leave all other points in singleton equivalence classes. Let \(q:I\to I/{\sim}\) be the quotient map with the quotient topology. By definition, \(A\subseteq I/{\sim}\) is open exactly when \(q^{-1}(A)\) is open in \(I\). This quotient map is not open.
Worked Example: An Open Set Whose Quotient Image Is Not Open
Take \(U=[0,1/3)\), which is open in the subspace \(I\): it is \(I\cap(-1/3,1/3)\). Its image \(q(U)\) contains the equivalence class of both endpoints, since \(0\in U\).
The full preimage of that image is the saturation of \(U\), meaning the union of all equivalence classes that meet \(U\). Here it is \(q^{-1}(q(U))=[0,1/3)\cup\{1\}\). This is not open in \(I\). Indeed, any open neighborhood of \(1\) in \(I\) contains points strictly less than \(1\), whereas this preimage contains no points in \((1/3,1)\). By the quotient-topology definition, \(q(U)\) is therefore not open. So \(q\) is not an open map, and the product theorem cannot be applied to this particular endpoint-identification map.
This calculation makes only a statement about the hypothesis: the map fails to be open. It does not claim that every product involving this quotient map fails to be quotient.
Periodic Identifications and Their Products
There are quotient maps identifying points that differ by integers for which the openness hypothesis does hold. Define \(x\sim y\) on \(\mathbb{R}\) when \(x-y\in\mathbb{Z}\), let \(Y=\mathbb{R}/{\sim}\) have the quotient topology, and write \(q:\mathbb{R}\to Y\) for the map sending a real number to its equivalence class. This is a periodic identification: every class consists of the integer translates of any one of its members. We use only this quotient description here.
Worked Example: Verifying Openness of the Periodic Quotient Map
Let \(U\) be open in \(\mathbb{R}\). The preimage of its image is the saturation \[ \bigcup_{n\in\mathbb{Z}}(U+n), \] where \(U+n=\{u+n:u\in U\}\). Each translate \(U+n\) is open, since translation by \(n\) preserves open intervals, and a union of open sets is open. The quotient-topology definition now implies that \(q(U)\) is open in \(Y\).
The map \(q\) is surjective by the definition of equivalence classes. It is continuous because \(Y\) has the quotient topology: if \(A\) is open in \(Y\), then \(q^{-1}(A)\) is open in \(\mathbb{R}\). Thus \(q\) is continuous, open, and surjective; by the lemma it is quotient. Applying the theorem with \(Z=\mathbb{R}\), the map \(q\times\operatorname{id}_{\mathbb{R}}:\mathbb{R}\times\mathbb{R}\to Y\times\mathbb{R}\) is quotient.
For a direct check of the image calculation, take \(U=(-1/4,1/4)\). Its saturation is \(\bigcup_{n\in\mathbb{Z}}(n-1/4,n+1/4)\), an open union of intervals. Therefore \(q(U)\) is open in \(Y\). The image of the open rectangle \(U\times(1,2)\) under the product map is \(q(U)\times(1,2)\), which is open in \(Y\times\mathbb{R}\).
Check: the first coordinate of a point in \(q(U)\) has a representative in \((-1/4,1/4)\). Its representatives are precisely the integer translates of that interval, matching the computed saturation.
The result also gives a quotient map for the simultaneous periodic identification in both coordinates. First apply the theorem to \(q\times\operatorname{id}_{\mathbb{R}}\). Then apply its coordinate-swapped version to \(\operatorname{id}_{Y}\times q\). That version follows by the same rectangle proof, or by swapping the two product coordinates, which carries basis rectangles to basis rectangles. The composition is \[ (\operatorname{id}_{Y}\times q)\circ(q\times\operatorname{id}_{\mathbb{R}})=q\times q. \] The composition of quotient maps is quotient: if \(f:X\to Y\) and \(g:Y\to Z\) are quotient maps, then for any \(A\subseteq Z\), \((g\circ f)^{-1}(A)=f^{-1}(g^{-1}(A))\) is open exactly when \(g^{-1}(A)\) is open, exactly when \(A\) is open. Thus \(q\times q:\mathbb{R}^2\to Y\times Y\) is quotient.
Worked Example: The Two-Coordinate Identification
Under \(q\times q\), the point \((x,y)\) maps to \(([x],[y])\), where brackets denote equivalence classes modulo integer differences. Its preimage class is \[ \{(x+m,y+n):m,n\in\mathbb{Z}\}. \] For example, the class of \((1/3,2/5)\) contains \((1/3+m,2/5+n)\) for all integers \(m,n\). No other points lie in this class: equality of first-coordinate classes requires a difference in \(\mathbb{Z}\), and equality of second-coordinate classes requires a difference in \(\mathbb{Z}\) as well.
If \(U=(-1/4,1/4)\) and \(V=(1/3,2/3)\), then \[ (q\times q)(U\times V)=q(U)\times q(V). \] The preimage of this image is \[ \left(\bigcup_{m\in\mathbb{Z}}(U+m)\right)\times \left(\bigcup_{n\in\mathbb{Z}}(V+n)\right), \] an open subset of \(\mathbb{R}^2\). The product of quotient maps established above guarantees that this image is open in \(Y\times Y\).
Check: the first-coordinate intervals are \((m-1/4,m+1/4)\); the second-coordinate intervals are \((n+1/3,n+2/3)\). The displayed preimage consists exactly of the rectangles formed by choosing one interval of each kind, as required by the independent integer translations in the two coordinates.
Common Pitfall and Takeaway
A common error is to write “an open surjection is quotient” without checking continuity. The two-point example shows why that implication is invalid. A related error is to assume that every quotient map is open; the endpoint-identification example disproves that assumption. For the product theorem, the useful chain of reasoning is specific: establish continuity, openness, and surjectivity of \(q\times\operatorname{id}_Z\); then invoke the lemma to conclude that it is quotient.
The identity factor can be any topological space \(Z\); no separation or compactness assumption on \(Z\) is needed. The openness of \(q\), however, is the feature that makes images of basis rectangles open. This is why the theorem is particularly effective for periodic identifications, whose saturations are unions of open translates.
Check Your Understanding
For each question, identify which definition or step of the product theorem applies.
- In the proof that \(q\times\operatorname{id}_Z\) is continuous, what is the preimage of a basis rectangle \(V\times W\)?
- Why does openness of \(q\) imply that the image of an arbitrary open subset of \(X\times Z\) is open?
- For endpoint identification on \(I\), compute the saturation of \([0,1/3)\) and explain why it is not open in \(I\).
- For the periodic quotient \(q:\mathbb{R}\to\mathbb{R}/{\sim}\), describe the saturation of an open set \(U\subseteq\mathbb{R}\).
- Explain why continuity is required in the lemma that a continuous, open surjection is quotient.