Why Openness Gives the Quotient Test
A quotient map describes a topology on a target by recording exactly which subsets have open preimages. The defining test can be difficult to verify directly: one must consider every subset of the target. A useful sufficient condition is that the map is both open and surjective. Openness lets us push an open preimage forward, while surjectivity ensures that the result is the original target subset.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\in\mathcal{T}_X\) for every \(V\in\mathcal{T}_Y\). It is open if \(f(U)\in\mathcal{T}_Y\) for every \(U\in\mathcal{T}_X\). It is surjective if every point of \(Y\) is \(f(x)\) for some \(x\in X\).
The Open Surjection Theorem
Proof. Let \(V\subseteq Y\). If \(V\) is open, continuity gives that \(f^{-1}(V)\) is open in \(X\). Conversely, suppose \(f^{-1}(V)\) is open in \(X\). Since \(f\) is open, its image \(f(f^{-1}(V))\) is open in \(Y\). Surjectivity gives \[ f(f^{-1}(V))=V. \] Indeed, an image point of \(f(f^{-1}(V))\) has the form \(f(x)\) with \(f(x)\in V\), so it belongs to \(V\). For the reverse inclusion, if \(y\in V\), surjectivity supplies \(x\in X\) with \(f(x)=y\); then \(x\in f^{-1}(V)\), so \(y\in f(f^{-1}(V))\). Thus \(V\) is open. The defining test for a quotient map holds for every \(V\subseteq Y\), and \(f\) is already surjective. \(\square\)
The image identity in the proof is where surjectivity matters. Without it, \(f(f^{-1}(V))\) records only the points of \(V\) that lie in the image of \(f\). Continuity handles one direction of the quotient test; openness and surjectivity together establish the other.
Worked Example: A Projection from a Product
Let \(X\) and \(Y\) have their given topologies, with \(Y\ne\varnothing\), and give \(X\times Y\) the product topology. Consider the first projection \(\pi_1:X\times Y\to X\), defined by \(\pi_1(x,y)=x\). We verify that it is a quotient map.
Surjectivity. For any \(x\in X\), choose \(y_0\in Y\), which is possible because \(Y\ne\varnothing\). Then \(\pi_1(x,y_0)=x\). If \(X=\varnothing\), the map is also surjective, since its target is empty.
Continuity. For every open \(U\subseteq X\), \[ \pi_1^{-1}(U)=U\times Y. \] This is open in the product topology: it is a basic open rectangle \(U\times Y\), since \(Y\) is open in itself. Thus \(\pi_1\) is continuous.
Openness. Every open set \(W\subseteq X\times Y\) is a union of basic open rectangles \(U_\alpha\times V_\alpha\), where \(U_\alpha\) is open in \(X\) and \(V_\alpha\) is open in \(Y\). Projection commutes with unions, and \[ \pi_1(U_\alpha\times V_\alpha)= \begin{cases} U_\alpha,&V_\alpha\ne\varnothing,\\ \varnothing,&V_\alpha=\varnothing. \end{cases} \] Both possible images are open in \(X\). Therefore \(\pi_1(W)\), a union of open subsets of \(X\), is open. The theorem now shows that \(\pi_1\) is a quotient map.
Check with a particular open set. In \(\mathbb{R}\times\mathbb{R}\) with the product of the standard topologies, take \(W=(({-1},1)\times(2,4))\cup((3,5)\times({-1},1))\). The first rectangle projects to \((-1,1)\) because \((2,4)\ne\varnothing\), and the second projects to \((3,5)\) because \((-1,1)\ne\varnothing\). Hence \(\pi_1(W)=(-1,1)\cup(3,5)\), an open subset of \(\mathbb{R}\). This computation illustrates the general rectangle argument.
An Identification by Absolute Value
A quotient map need not be one-to-one. Its fibers can record which points are being identified. For a map \(q:X\to Y\), the fiber over \(y\in Y\) is \(q^{-1}(\{y\})\); points in the same fiber have the same image.
Worked Example: Identifying Opposite Points
Give \(\mathbb{R}\) its standard topology and give \([0,\infty)\) the subspace topology inherited from \(\mathbb{R}\). Define \(q:\mathbb{R}\to[0,\infty)\) by \(q(x)=|x|\). We prove directly that \(q\) is an open, continuous surjection, hence a quotient map.
Surjectivity and fibers. If \(y\in[0,\infty)\), then \(q(y)=y\), so \(q\) is surjective. For \(y=0\), the fiber is \(\{0\}\). For \(y>0\), it is \(\{-y,y\}\), since \(|x|=y\) holds exactly when \(x=-y\) or \(x=y\). Thus the map identifies each pair of opposite nonzero real numbers.
Continuity. The triangle inequality gives \(\bigl||x|-|z|\bigr|\le |x-z|\) for all \(x,z\in\mathbb{R}\). Given \(x\in\mathbb{R}\) and \(\varepsilon>0\), if \(|z-x|<\varepsilon\), then \(\bigl||z|-|x|\bigr|<\varepsilon\). This proves continuity at every \(x\), so \(q\) is continuous.
Openness. Let \(U\subseteq\mathbb{R}\) be open and take \(y\in q(U)\). There is \(x\in U\) with \(|x|=y\). If \(y>0\), choose \(\delta>0\) so that \((x-\delta,x+\delta)\subseteq U\) and \(\delta<y\). When \(x=y\), the absolute values of points in this interval include \((y-\delta,y+\delta)\); when \(x=-y\), they include the same interval. Thus \(q(U)\) contains a relatively open neighborhood of \(y\) in \([0,\infty)\). If \(y=0\), then \(x=0\), and openness of \(U\) supplies \(\delta>0\) with \((-\delta,\delta)\subseteq U\). Consequently \([0,\delta)\subseteq q(U)\), which is a neighborhood of \(0\) open in the subspace \([0,\infty)\). Every point of \(q(U)\) therefore has a relative neighborhood contained in \(q(U)\), proving that \(q(U)\) is open in \([0,\infty)\).
All three hypotheses of the theorem have been checked, so \(q\) is a quotient map. For an explicit test, \(V=[0,1)\) is open in \([0,\infty)\), and \(q^{-1}(V)=(-1,1)\), which is open in \(\mathbb{R}\). In contrast, \(V=[0,1]\) is not open in \([0,\infty)\), and \(q^{-1}(V)=[-1,1]\) is not open in \(\mathbb{R}\). For the latter claim, every open interval around \(1\) contains points greater than \(1\), so no open neighborhood of \(1\) lies inside \([-1,1]\). These tests agree with the quotient criterion.
Worked Example: A Finite Identification
Let \(X=\{a,b,c\}\) have the discrete topology, and let \(Y=\{u,v\}\) also have the discrete topology. Define \(q:X\to Y\) by \(q(a)=q(b)=u\) and \(q(c)=v\). The map is surjective, and its fibers are \(\{a,b\}\) and \(\{c\}\): it identifies \(a\) with \(b\), but not with \(c\).
Every subset of \(X\) is open, so \(q\) is continuous: the preimage of any open subset of \(Y\) is a subset of \(X\) and hence open. Every subset of \(X\) is also open, so the image of every open subset is a subset of \(Y\) and hence open. Thus \(q\) is open. For a direct quotient check, the four subsets of \(Y\) are \(\varnothing,\{u\},\{v\},Y\); their preimages are respectively \(\varnothing,\{a,b\},\{c\},X\), all open. Since all four target subsets are open as well, the criterion holds for every subset. Therefore \(q\) is a quotient map.
Check the fibers. The preimage of \(\{u\}\) contains both \(a\) and \(b\), while the preimage of \(\{v\}\) contains exactly \(c\). This confirms that the quotient test applies to the stated identification, not to a one-to-one map.
A Common Mistake: Surjectivity and Continuity Are Not Enough
It is incorrect to conclude that every continuous surjection is a quotient map. For a counterexample, take the same underlying set \(\mathbb{R}\) as both domain and target, give the domain its standard topology, give the target the indiscrete topology \(\{\varnothing,\mathbb{R}\}\), and let \(f\) be the identity function.
This map is surjective. It is continuous because the only open target sets are \(\varnothing\) and \(\mathbb{R}\), whose preimages are both open in the standard topology. But it is not open: the standard open set \((0,1)\) maps to \((0,1)\), which is not open in the indiscrete target topology. It also fails the quotient test directly. For \(V=(0,1)\), the preimage \(f^{-1}(V)=(0,1)\) is open in the domain, while \(V\) is not open in the target. The reverse implication in the quotient criterion fails.
The mistaken argument is: “\(f\) is a continuous surjection, so whenever \(f^{-1}(V)\) is open, \(V\) is open.” Continuity only proves that open target sets have open preimages. The open-surjection theorem supplies the missing direction by making \(f(f^{-1}(V))\) open and using surjectivity to identify that image with \(V\).
Check Your Understanding
For each question, identify which part of the quotient criterion or open-surjection theorem applies.
- In the product projection example, why is the condition \(Y\ne\varnothing\) needed for surjectivity when \(X\ne\varnothing\)?
- For \(q:\mathbb{R}\to[0,\infty)\), \(q(x)=|x|\), find the fiber over \(3\) and the preimage of \([0,2)\).
- Explain why openness of \(f\) gives that \(f(f^{-1}(V))\) is open when \(f^{-1}(V)\) is open.
- For the finite map \(q(a)=q(b)=u,\ q(c)=v\), list the preimages of \(\varnothing,\{u\},\{v\},Y\).
- In the identity-map counterexample, identify a subset \(V\) whose preimage is open but which is not open in the target.