The Open-Set Test
A continuous map carries information about open sets from its target back to its domain: the inverse image of each open set is open. A quotient map is more precisely controlled. For a surjective quotient map, this implication also works in reverse: a subset of the target is open whenever its full inverse image is open. This gives a test that can recognize quotient maps without trying to describe every open set of the target in advance.
Throughout, \(X\) and \(Y\) are topological spaces, with their topologies understood, and \(f:X\to Y\) is a map. For \(V\subseteq Y\), the full inverse image is \(f^{-1}(V)=\{x\in X:f(x)\in V\}\). It includes every point of every fiber over \(V\), not just a chosen representative.
The equivalence has two directions. The implication “\(V\) open implies \(f^{-1}(V)\) open” is exactly the definition of continuity. The reverse implication is the additional requirement that distinguishes a quotient map from a continuous surjection. Surjectivity is also part of the definition: the open-set equivalence alone does not force a map to be onto.
Proof. Suppose first that \(f\) is a quotient map. By definition, it is surjective, and its target topology is precisely the collection of subsets \(V\subseteq Y\) for which \(f^{-1}(V)\) is open in \(X\). Therefore, for every \(V\subseteq Y\), \(V\) is open if and only if its full inverse image is open.
Conversely, suppose \(f\) is surjective and satisfies the stated equivalence for every subset \(V\subseteq Y\). Then the open subsets of \(Y\) are exactly those subsets whose full inverse images are open in \(X\). This is the defining condition for the topology on \(Y\) to be the quotient topology induced by \(f\). Together with surjectivity, it says that \(f\) is a quotient map. \(\square\)
One can also see directly that the sets \(\{V\subseteq Y:f^{-1}(V)\text{ is open in }X\}\) form a topology: inverse images preserve the empty set and the whole space, arbitrary unions, and finite intersections. The quotient-map test says that this induced topology must agree with the topology already given on \(Y\).
Applying the Test
For a finite target, the test can be carried out by checking each subset. To prove that a surjection is quotient, check both directions for every subset: every open target set must have open inverse image, and every subset with open inverse image must be open in the target. To disprove the quotient property, it is enough to find one subset \(V\) for which exactly one of those two sets is open.
Worked Example: A Quotient Map with a Nontrivial Target Topology
Let \(X=\{a,b,c\}\) have topology \(\{\varnothing,\{a,b\},X\}\), and let \(Y=\{u,v\}\) have topology \(\{\varnothing,\{u\},Y\}\). Define \(f:X\to Y\) by \(f(a)=u\), \(f(b)=u\), and \(f(c)=v\). Determine whether \(f\) is a quotient map.
Both displayed collections are topologies: they contain the empty set and the whole space, and their only nontrivial open set is preserved under arbitrary unions and finite intersections. The map is surjective because \(f(a)=u\) and \(f(c)=v\). To apply the test, list every target subset and its full inverse image:
The open subsets of \(Y\) are \(\varnothing,\{u\},Y\). Their inverse images are, respectively, \(\varnothing,\{a,b\},X\), all open in \(X\). The one subset of \(Y\) that is not open is \(\{v\}\), and its inverse image \(\{c\}\) is not open in \(X\), since \(\{c\}\) is absent from the listed topology. Thus, among all four subsets, openness of the subset agrees exactly with openness of its inverse image. The map is a quotient map.
Check: the fiber over \(u\) is \(\{a,b\}\), and the fiber over \(v\) is \(\{c\}\). In particular, the inverse image of the open set \(\{u\}\) contains both points of its fiber, as required for a full inverse image.
Worked Example: A Quotient Map That Is Not Open
Let \(X=\{a,b,c\}\) have topology \(\{\varnothing,\{a\},X\}\), let \(Y=\{u,v\}\) have the indiscrete topology \(\{\varnothing,Y\}\), and define \(f(a)=u\), \(f(b)=u\), \(f(c)=v\). Determine whether \(f\) is quotient, and whether it is an open map. An open map is a map that sends every open subset of its domain to an open subset of its target.
The listed collection on \(X\) is a topology: unions of its open sets are \(\varnothing,\{a\}\), or \(X\), and finite intersections have one of those same forms. The listed topology on \(Y\) is indiscrete, meaning that its only open sets are \(\varnothing\) and \(Y\). The map is surjective. Its inverse images are
The open target subsets are \(\varnothing\) and \(Y\), and their inverse images \(\varnothing\) and \(X\) are open. The other two target subsets are not open; their inverse images are also not open, since neither \(\{a,b\}\) nor \(\{c\}\) belongs to the topology on \(X\). Therefore the open-set equivalence holds for every subset, so \(f\) is quotient.
But \(\{a\}\) is open in \(X\), and \(f(\{a\})=\{u\}\), which is not open in the indiscrete space \(Y\). Thus \(f\) is not an open map. The quotient condition concerns which target subsets have open full inverse images; it does not require images of open subsets of \(X\) to be open.
Check: the crucial distinction is between \(f^{-1}(\{u\})=\{a,b\}\), the full inverse image, and \(f(\{a\})=\{u\}\), the image of a particular open set. The quotient test uses the former; failure of openness is shown by the latter.
A Continuous Surjection That Fails the Test
A common mistake is to treat continuity and surjectivity as sufficient. Continuity checks only that open sets in the target have open inverse images. The quotient test asks for the converse as well, for every subset of the target.
Worked Example: The Identity Map from Discrete to Indiscrete
Let \(X=Y=\{p,q\}\) as sets. Give \(X\) the discrete topology \(\mathcal{P}(X)\), so every subset is open, and give \(Y\) the indiscrete topology \(\{\varnothing,Y\}\). Let \(f:X\to Y\) be the identity on the underlying set. Is \(f\) quotient?
The map is surjective. It is also continuous: the only open subsets of \(Y\) are \(\varnothing\) and \(Y\), and their inverse images are \(\varnothing\) and \(X\), both open in \(X\). Now test \(V=\{p\}\). This subset is not open in the indiscrete topology on \(Y\), but \(f^{-1}(\{p\})=\{p\}\) is open in the discrete topology on \(X\). The equivalence fails, so \(f\) is not quotient.
Check: continuity holds because both open target sets have open inverse images. The quotient condition fails because the inverse image of the non-open subset \(\{p\}\) is open. Thus the failure is exactly in the direction that continuity does not test.
For a surjective map, the fibers help organize the test. If \(V\subseteq Y\), its full inverse image consists of all points whose images lie in \(V\); in particular it contains either an entire fiber or none of that fiber. A subset of \(X\) that is a union of whole fibers is called saturated. Since every \(f^{-1}(V)\) is saturated, the quotient condition asks whether openness of these saturated inverse images detects openness in \(Y\). It does not say that every open subset of \(X\) must be saturated.
Common Pitfall: Checking Only One Direction
Suppose someone argues, “\(f\) is a continuous surjection, so whenever \(V\) is open in \(Y\), \(f^{-1}(V)\) is open in \(X\); therefore \(f\) is quotient.” The first claim is the definition of continuity, but it proves only one implication. In the identity-map example, \(V=\{p\}\) gives the concrete failure: \(f^{-1}(V)=\{p\}\) is open while \(V\) is not open.
The correction is to keep the two implications separate. Continuity verifies that target-open subsets have open inverse images. To establish the quotient property, one must additionally show that every subset with an open full inverse image is open in the target. For a counterexample, one such subset with a mismatch suffices. For a proof, the reverse implication must hold for all subsets, not merely for a selected collection of examples.
Check Your Understanding
For each question, use the open-set equivalence and distinguish full inverse images from images of subsets.
- State which direction of the quotient-map equivalence follows directly from continuity.
- In the first worked example, list the inverse image of each of the four subsets of \(Y\), and identify which are open.
- Why does the second worked example show that a quotient map need not be an open map?
- In the identity-map example, identify a subset of \(Y\) that proves the map is not quotient.
- Does the open-set equivalence by itself guarantee that a map is surjective? What additional condition is included in the definition of a quotient map?