Why Closed Surjections Give Quotient Maps
A quotient map is a surjection that determines the topology of its target entirely through preimages: a subset of the target is open exactly when its preimage is open. This condition can be hard to check directly, because it quantifies over every subset of the target. The closed-surjection criterion provides a useful alternative. It does not require the domain to be compact, but it does require all three of continuity, closedness, and surjectivity.
Let \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) be topological spaces. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open \(V\subseteq Y\). It is closed if \(f(C)\) is closed in \(Y\) for every closed \(C\subseteq X\). A map is surjective if every point of \(Y\) is the image of some point of \(X\).
Continuity supplies one direction of the quotient condition: if \(V\) is open in \(Y\), then its preimage is open in \(X\). The extra direction asks whether openness of a preimage forces openness of the original set. For a surjection, it is enough to establish the corresponding statement for closed sets.
Proof. Suppose first that \(q\) is a quotient map and \(q^{-1}(F)\) is closed in \(X\). Then \(q^{-1}(Y\setminus F)=X\setminus q^{-1}(F)\) is open. By the quotient condition, \(Y\setminus F\) is open, so \(F\) is closed. Conversely, suppose the stated closed-set implication holds. Let \(V\subseteq Y\) have open preimage. Then \(q^{-1}(Y\setminus V)=X\setminus q^{-1}(V)\) is closed. The implication gives that \(Y\setminus V\) is closed, so \(V\) is open. The reverse direction, that open \(V\) has open preimage, follows from continuity. Together these are exactly the quotient condition. \(\square\)
The Closed-Surjection Theorem
Proof. Let \(V\subseteq Y\) and suppose \(f^{-1}(V)\) is open in \(X\). Then \(X\setminus f^{-1}(V)\) is closed in \(X\), so its image is closed in \(Y\) because \(f\) is a closed map. Surjectivity gives the exact equality $$ f\bigl(X\setminus f^{-1}(V)\bigr)=Y\setminus V. $$ Indeed, if \(y=f(x)\) and \(x\notin f^{-1}(V)\), then \(f(x)\notin V\), so \(y\in Y\setminus V\). Conversely, if \(y\in Y\setminus V\), surjectivity gives \(x\in X\) with \(f(x)=y\); this \(x\) cannot belong to \(f^{-1}(V)\), so \(y\) belongs to the image on the left. Therefore \(Y\setminus V\) is closed and \(V\) is open. The other direction follows from continuity. Thus \(V\) is open if and only if \(f^{-1}(V)\) is open, and \(f\) is a quotient map. \(\square\)
The proof depends on all three hypotheses in distinct ways. Continuity gives one direction of the quotient condition. Closedness turns the closed preimage complement into a closed image. Surjectivity ensures that this image is the entire complement \(Y\setminus V\), rather than merely a part of it.
Worked Example: The Square Map
Give \(X=\mathbb{R}\) and \(Y=[0,\infty)\) their standard (for \(Y\), subspace) topologies. Consider \(f:X\to Y\) defined by \(f(x)=x^2\). We verify directly that it is a quotient map, without assuming its domain is compact.
The map is surjective because for each \(y\in[0,\infty)\), the real number \(\sqrt y\) satisfies \(f(\sqrt y)=y\). It is continuous because the polynomial \(x\mapsto x^2\) is continuous as a real-valued map, and its values lie in \(Y\).
To prove that \(f\) is closed, let \(C\subseteq\mathbb{R}\) be closed. If \(C=\varnothing\), then \(f(C)=\varnothing\), which is closed. Otherwise, take any convergent sequence \((y_n)\) in \(f(C)\), with limit \(y\in[0,\infty)\). For each \(n\), choose \(x_n\in C\) such that \(x_n^2=y_n\). A convergent real sequence \((y_n)\) is bounded, so there is \(M\geq 0\) with \(0\leq y_n\leq M\) for every \(n\). Hence \(|x_n|=\sqrt{y_n}\leq\sqrt M\). The bounded sequence \((x_n)\) has a convergent subsequence \((x_{n_k})\), with limit \(x\). Since \(C\) is closed, \(x\in C\); continuity gives \(x^2=\lim_k x_{n_k}^2=\lim_k y_{n_k}=y\). Thus \(y\in f(C)\).
In a metric space, a set containing the limit of every convergent sequence of its points is closed: if a point lies in its closure but not in the set, choose a point of the set in each ball of radius \(1/n\) around it to obtain a sequence converging to that point. Since \(Y\) is a metric space, the argument above proves that \(f(C)\) is closed. Therefore \(f\) is a closed surjection, and the theorem shows that it is a quotient map.
For a direct check of the criterion on particular sets, \(V=[0,1)\) is open in the subspace \(Y\), and \(f^{-1}(V)=(-1,1)\) is open in \(\mathbb{R}\). In contrast, \(V=\{0\}\) is not open in \(Y\), and \(f^{-1}(V)=\{0\}\) is not open in \(\mathbb{R}\). The theorem guarantees that this agreement between openness and preimage openness holds for every subset \(V\subseteq Y\), not just these two.
Other Noncompact Examples
Worked Example: Projection with a Finite Fiber
Let \(X=\mathbb{R}\times\{0,1\}\) have the product topology, where \(\mathbb{R}\) has its standard topology and \(\{0,1\}\) has the discrete topology. Give \(Y=\mathbb{R}\) its standard topology, and define \(f(x,i)=x\). The map is surjective. It is continuous because for every open \(V\subseteq\mathbb{R}\), the preimage \(V\times\{0,1\}\) is open in the product topology.
Let \(C\) be closed in \(X\), and define \(C_i=\{x\in\mathbb{R}:(x,i)\in C\}\) for \(i=0,1\). The map \(j_i:\mathbb{R}\to X\) given by \(j_i(x)=(x,i)\) is continuous: the preimage of a basic open set \(U\times D\), with \(U\) open in \(\mathbb{R}\) and \(D\subseteq\{0,1\}\), is \(U\) if \(i\in D\) and \(\varnothing\) otherwise. Thus \(C_i=j_i^{-1}(C)\) is closed in \(\mathbb{R}\). Now \(f(C)=C_0\cup C_1\), a union of two closed sets, so \(f(C)\) is closed. The theorem applies, and \(f\) is a quotient map.
The domain is not compact: the open sets \((-n,n)\times\{0,1\}\), for \(n\in\mathbb{N}\), cover \(X\), but no finite subcollection covers \(X\). In this example, checking closedness reduces to checking the two slices of a closed set.
Worked Example: A Noninjective Polynomial Map
Give \(X=Y=\mathbb{R}\) their standard topologies and define \(f(x)=x^3-x\). We check that this is a quotient map by verifying the theorem's hypotheses.
The polynomial is continuous. To see that it is surjective, fix \(t\in\mathbb{R}\). Choose \(M\geq 2\) so large that \(\frac34 M^3>|t|\). For \(M\geq2\), \(M^2-1\geq\frac34M^2\), so $$ f(M)=M(M^2-1)\geq\frac34M^3>|t|, \qquad f(-M)=-M(M^2-1)\leq-\frac34M^3<-|t|. $$ In particular \(f(-M)<t<f(M)\). The intermediate value theorem gives \(x\in[-M,M]\) with \(f(x)=t\). Thus \(f\) is surjective.
To check closedness, let \(C\subseteq\mathbb{R}\) be closed. The empty case gives \(f(C)=\varnothing\), which is closed. Otherwise, take a convergent sequence \((y_n)\) in \(f(C)\), with limit \(y\), and choose \(x_n\in C\) with \(f(x_n)=y_n\). Since \((y_n)\) converges, there is \(L\geq0\) such that \(|y_n|\leq L\) for all \(n\). For \(|x|\geq2\), $$ |f(x)|=|x|\,|x^2-1|\geq\frac34|x|^3. $$ Consequently, if \(|x_n|\geq2\), then \(\frac34|x_n|^3\leq L\). Together with the alternative \(|x_n|<2\), this bounds the whole sequence \((x_n)\). A bounded real sequence has a convergent subsequence \(x_{n_k}\to x\). Closedness gives \(x\in C\), and continuity gives \(f(x)=\lim_k y_{n_k}=y\). The metric-space sequential test for closedness therefore shows that \(f(C)\) is closed.
All three hypotheses hold, so \(f\) is a quotient map. Its fibers need not be singletons: \(f(x)=0\) exactly when \(x(x-1)(x+1)=0\), so \(f^{-1}(\{0\})=\{-1,0,1\}\). As a check on a different preimage, the factorization \(f(x)=x(x-1)(x+1)\) shows that \(f^{-1}((0,\infty))=(-1,0)\cup(1,\infty)\), which is open, as required for the open set \((0,\infty)\).
Why the Hypotheses Matter
Continuity and surjectivity alone do not guarantee that a map is quotient. For a concrete failure, let \(X=\mathbb{R}\) have its standard topology, let \(Y=\mathbb{R}\) have the indiscrete topology \(\{\varnothing,\mathbb{R}\}\), and let \(f:X\to Y\) be the identity function. The map is surjective. It is continuous because the only open subsets of \(Y\) are \(\varnothing\) and \(Y\), whose preimages are open in \(X\). But \(V=(0,1)\) is not open in \(Y\), while \(f^{-1}(V)=(0,1)\) is open in \(X\). Thus \(f\) is not a quotient map. It is also not closed: \(\{0\}\) is closed in \(X\), but its image is not closed in the indiscrete space \(Y\), whose only closed sets are \(\varnothing\) and \(Y\).
Common pitfall. The incorrect inference is “continuous surjection, therefore quotient.” In the counterexample, the preimage of every open set of \(Y\) is open, so continuity holds; the failure is the converse direction of the quotient test. To use the closed-surjection theorem, check closedness as well as continuity and surjectivity. Conversely, do not replace the theorem's hypotheses with compactness: compactness can help establish closedness in other settings, but it is not needed when closedness has been verified directly, as in the examples above.
Check Your Understanding
For each question, distinguish the hypotheses that have been verified from the conclusion that follows.
- In the proof of the closed-surjection theorem, where is surjectivity used to identify the image of the complement?
- For \(f(x)=x^2\) from \(\mathbb{R}\) to \([0,\infty)\), compute \(f^{-1}([0,4))\) and state whether the target set is open in the subspace topology.
- For the projection \(\mathbb{R}\times\{0,1\}\to\mathbb{R}\), explain why the image of a closed set is the union of two closed subsets of \(\mathbb{R}\).
- In the polynomial example, factor \(x^3-x\) and determine its zeros.
- For the identity map from the standard real line to the indiscrete real line, give a subset whose preimage is open but which is not open in the target.