Why Restricting a Quotient Map Requires Care
A quotient map identifies points of a space, and its target topology records exactly which subsets have open preimages. If we keep only part of the domain, we may keep some points of a fiber while discarding others. The restricted map can then lose the information that made the original map quotient. Two conditions prevent this problem: the chosen domain subset must be open, and it must contain whole fibers.
Let \(q:X\to Y\) be a map. For \(A\subseteq Y\), the preimage is \(q^{-1}(A)=\{x\in X:q(x)\in A\}\), and for \(U\subseteq X\), the image is \(q(U)=\{q(x):x\in U\}\). A map is a quotient map if it is surjective and a subset \(A\subseteq Y\) is open in \(Y\) exactly when \(q^{-1}(A)\) is open in \(X\). A subset of \(X\) is given the subspace topology when its open sets are intersections with open sets of \(X\).
The equality in the definition is useful as a direct test: \(q^{-1}(q(U))\) includes every point of \(X\) equivalent, under \(q\), to a point of \(U\). Thus \(U\) is saturated precisely when this operation adds no points. We will prove that if \(q\) is quotient and \(U\) is both open and saturated, then \(q|_U\) is quotient.
The Restriction Theorem
Proof. Write \(f=q|_U\). For every \(y\in q(U)\), the definition of \(q(U)\) gives an \(x\in U\) with \(q(x)=y\). Thus \(f\) is surjective onto \(q(U)\). The map \(f\) is continuous: if \(V\) is open in \(q(U)\), there is an open set \(W\subseteq Y\) such that \(V=W\cap q(U)\). Then \(f^{-1}(V)=U\cap q^{-1}(W)\), which is open in \(U\) because \(q\) is continuous.
It remains to verify the quotient condition. Let \(A\subseteq q(U)\). By saturation, \(q^{-1}(q(U))=U\), so \(q^{-1}(A)\subseteq U\), and consequently \(f^{-1}(A)=q^{-1}(A)\). Suppose first that \(f^{-1}(A)\) is open in \(U\). Since \(U\) is open in \(X\), every set open in \(U\) is open in \(X\): such a set is \(U\cap W\) for some open \(W\subseteq X\). Therefore \(q^{-1}(A)\) is open in \(X\). The quotient property of \(q\) implies that \(A\) is open in \(Y\), hence open in the subspace \(q(U)\).
Conversely, suppose \(A\) is open in \(q(U)\). We first show \(q(U)\) is open in \(Y\): its preimage is \(q^{-1}(q(U))=U\), which is open, so the quotient property of \(q\) gives that \(q(U)\) is open in \(Y\). Since \(A\) is open relative to this open subspace, \(A\) is open in \(Y\). Thus \(q^{-1}(A)\) is open in \(X\), again by the quotient property of \(q\); hence \(f^{-1}(A)=q^{-1}(A)\) is open in \(U\). We have proved both directions of the quotient condition, so \(f\) is quotient. \(\square\)
The proof uses the hypotheses at distinct points. Openness of \(U\) turns a set open in \(U\) into a set open in \(X\). Saturation ensures that preimages of subsets of \(q(U)\) have no additional points outside \(U\). These facts let the quotient criterion for \(q\) transfer to the restriction.
Proof. The quotient criterion says that \(q(U)\) is open in \(Y\) if and only if its preimage \(q^{-1}(q(U))\) is open in \(X\). Saturation identifies that preimage with \(U\). \(\square\)
Checking the Hypotheses in Familiar Quotients
Worked Example: Integer Orbits on the Real Line
Let \(X=\mathbb{R}\) with its standard topology, and let \(Y=\mathbb{R}/\mathbb{Z}\) have the quotient topology for the map \(q:\mathbb{R}\to Y\) given by \(q(x)=x+\mathbb{Z}\). Thus \(q(x)=q(y)\) exactly when \(x-y\in\mathbb{Z}\), and \(q\) is quotient by the definition of the quotient topology. Consider \(U=\bigcup_{n\in\mathbb{Z}}(n,n+\tfrac12)\).
This set is open, since it is a union of open intervals. It is saturated: if \(x\in(n,n+\tfrac12)\) and \(q(y)=q(x)\), then \(y=x+k\) for an integer \(k\). Hence \(y\in(n+k,n+k+\tfrac12)\subseteq U\). Conversely, \(U\subseteq q^{-1}(q(U))\) holds for every set \(U\), so these inclusions give \(q^{-1}(q(U))=U\).
The image is \(q(U)=\{q(t):0<t<\tfrac12\}\). To check this description, if \(x\in(n,n+\tfrac12)\), then \(t=x-n\) satisfies \(0<t<\tfrac12\) and \(q(x)=q(t)\); each such \(t\) belongs to \(U\). Since \(U\) is open and saturated, the restriction \(q|_U:U\to q(U)\) is quotient by the theorem. In particular, the target subspace is open in \(Y\), as its preimage is \(U\). This example has many points in each fiber retained by \(U\): for each \(t\in(0,\tfrac12)\), all \(t+n\), \(n\in\mathbb{Z}\), lie in \(U\).
Check: for \(t=\tfrac13\), its fiber is \(\{\tfrac13+n:n\in\mathbb{Z}\}\). Each member lies in \((n,n+\tfrac12)\), because \(0<\tfrac13<\tfrac12\); so this fiber is wholly contained in \(U\), as saturation requires.
Worked Example: Identifying the Endpoints of an Interval
Let \(X=[0,1]\) with the subspace topology from \(\mathbb{R}\), and let \(q:X\to Y\) identify \(0\) and \(1\) while leaving every other point in a singleton fiber. Give \(Y\) the quotient topology, so \(q\) is quotient. Take \(U=(0,1)\), which is open in \(X\).
The only fiber with more than one point is \(\{0,1\}\). It is disjoint from \(U\), while every other fiber is a singleton. Therefore \(U\) is saturated: adding all points in fibers that meet \(U\) adds no new points. More explicitly, \(q^{-1}(q(U))=(0,1)=U\). The restriction \(q|_U\) is consequently quotient onto \(q(U)\). The omitted point of \(Y\) is \(p=q(0)=q(1)\), so \(q(U)=Y\setminus\{p\}\), and its preimage is precisely \(U\). The quotient criterion then also shows that \(Y\setminus\{p\}\) is open in \(Y\).
This example illustrates why the fiber test must be made for every point of \(U\), not merely for the visibly identified points: all singleton fibers are complete automatically, and the one non-singleton fiber is absent altogether. In contrast, an open subset containing \(0\) but not \(1\) would not be saturated.
What Can Go Wrong Without Saturation?
Worked Example: A Finite Quotient Where the Restriction Fails
Let \(X=\{a,b,c,d\}\) have topology \(\mathcal{T}=\{\varnothing,\{a,b\},\{c\},\{a,b,c\},X\}\). These sets form a topology: arbitrary unions of listed sets are listed, and intersections of any two listed sets are listed. Let \(Y=\{p,r\}\), with topology \(\{\varnothing,\{p\},Y\}\), and define \(q(a)=q(b)=p\) and \(q(c)=q(d)=r\).
We verify that \(q\) is quotient. It is surjective, and the preimages of the four subsets of \(Y\) are \(q^{-1}(\varnothing)=\varnothing\), \(q^{-1}(\{p\})=\{a,b\}\), \(q^{-1}(\{r\})=\{c,d\}\), and \(q^{-1}(Y)=X\). Among these, precisely \(\varnothing,\{a,b\},X\) are open in \(X\), matching precisely the open sets \(\varnothing,\{p\},Y\) in \(Y\). Thus the quotient criterion holds.
Now take \(U=\{a,b,c\}\), which is open in \(X\), and note that \(q(U)=Y\). But \(U\) is not saturated: \(c\in U\), \(q(c)=q(d)=r\), and \(d\notin U\). For the subset \(A=\{r\}\subseteq q(U)\), the restricted preimage is \((q|_U)^{-1}(A)=\{c\}\). This is open in \(U\), since \(\{c\}\) is open in \(X\). Yet \(A=\{r\}\) is not open in \(q(U)=Y\). Therefore \(q|_U\) is not quotient.
Check: the full preimage \(q^{-1}(A)=\{c,d\}\) is not open in \(X\). The restricted preimage kept only \(\{c\}\), which is open, while discarding \(d\) hid the failure of openness in the original space. That is exactly the obstruction saturation prevents.
Common pitfall. One might argue, “\(q\) is quotient and \(U\) is open, so its restriction is quotient.” The finite example disproves this inference: \(U\) is open, but for \(A=\{r\}\), the restricted preimage \(\{c\}\) is open in \(U\) although \(A\) is not open in \(q(U)\). The missing point \(d\) lies in the same fiber as \(c\), so \(U\) is not saturated. The theorem gives a sufficient pair of hypotheses, not a claim that every restriction fails whenever saturation fails; an individual restriction without saturation must be checked by its own quotient criterion.
For a practical test, first list or characterize the fibers of \(q\). Check that every fiber meeting \(U\) is contained in \(U\), and separately check that \(U\) is open in \(X\). Only after both checks can the restriction theorem be applied. An open set is not automatically saturated, and a saturated set is not automatically open.
Check Your Understanding
For each question, state the relevant topology or quotient criterion as part of your reasoning.
- For a map \(q:X\to Y\), prove directly from the definition that \(q^{-1}(q(U))\) is the union of all fibers that meet \(U\).
- For the integer-orbit quotient, determine whether \(\bigcup_{n\in\mathbb{Z}}(n+\tfrac14,n+\tfrac34)\) is open and saturated.
- For the endpoint-identification quotient of \([0,1]\), is \(U=[0,\tfrac12)\) saturated? Explain by checking the fiber of \(0\).
- In the finite example, identify an open subset of \(U=\{a,b,c\}\) that is the restricted preimage of a non-open subset of \(q(U)\).
- Which part of the restriction theorem's proof uses that \(U\) is open, and which part uses that \(U\) is saturated?