The Direction of the Inclusion
A continuous surjection need not be a quotient map. To understand how it can fail, compare the topology already given on its target with the topology induced by the map. This comparison also imposes an important restriction: the target topology cannot be finer than the induced quotient topology. Continuity forces the opposite inclusion. Thus a request for an example with a strictly finer target topology is impossible under the standard definitions; the examples below compute the possible strict inclusion and identify exactly where the quotient-map test fails.
Let \(f:X\to Y\) be a surjection, where \(X\) has topology \(\mathcal{T}_X\) and \(Y\) has topology \(\mathcal{T}_Y\). Recall that \(f\) is continuous if the preimage of every open subset of \(Y\) is open in \(X\). The quotient topology induced by \(f\) is the topology on the underlying set of \(Y\) defined by whether preimages are open.
The defining condition for continuity gives one of the two implications in the quotient-map test. Failure occurs when the other implication fails: some subset of \(Y\) has open preimage but is not open in the given target topology.
Proof. Take any \(V\in\mathcal{T}_Y\). Continuity says that \(f^{-1}(V)\) is open in \(X\). By the definition of \(\mathcal{T}_q\), this means \(V\in\mathcal{T}_q\). Thus \(\mathcal{T}_Y\subseteq\mathcal{T}_q\). If \(f\) is not a quotient map, the quotient-map criterion says that \(\mathcal{T}_Y\ne\mathcal{T}_q\); together with the inclusion, this gives \(\mathcal{T}_Y\subsetneq\mathcal{T}_q\). \(\square\)
This argument also explains why a continuous surjection cannot have a target topology strictly finer than its quotient topology. “Finer” means having more open sets, so it would require \(\mathcal{T}_q\subsetneq\mathcal{T}_Y\), contradicting the inclusion just proved.
How to Locate the Failure
For a continuous surjection, the only possible failure of the quotient-map criterion is a subset \(V\subseteq Y\) for which \(f^{-1}(V)\) is open but \(V\) is not open. It is often useful to locate this failure in \(X\). A subset \(A\subseteq X\) is saturated with respect to \(f\) if it is a union of whole fibers: whenever \(x\in A\) and \(f(x')=f(x)\), then \(x'\in A\). For a surjection, every saturated subset has the form \(f^{-1}(V)\) for some \(V\subseteq Y\).
Proof. Suppose first that \(f\) is not a quotient map. Then there is a subset \(V\subseteq Y\) such that \(f^{-1}(V)\) is open but \(V\) is not open. Put \(A=f^{-1}(V)\). This set is saturated: if \(x\in A\) and \(f(x')=f(x)\), then \(f(x)\in V\), so \(f(x')\in V\) and \(x'\in A\). Since \(f\) is surjective, \(f(f^{-1}(V))=V\), so \(f(A)=V\) is not open.
Conversely, suppose \(A\) is open and saturated, and \(f(A)\) is not open in \(Y\). We have \(A=f^{-1}(f(A))\): inclusion from left to right follows from the definition of image; for the reverse inclusion, if \(f(x)\in f(A)\), some \(a\in A\) satisfies \(f(x)=f(a)\), and saturation gives \(x\in A\). Thus the preimage of \(f(A)\) is open although \(f(A)\) is not open. The quotient-map criterion fails. \(\square\)
In examples, this proposition gives a practical test: find an open set that is a union of fibers, then check whether its image is open in the target. Saturation matters because a set that cuts across a fiber cannot be the full preimage of any target subset.
Worked Example: The Identity from a Discrete Space
Let \(X=Y=\{a,b\}\). Give \(X\) the discrete topology \(\mathcal{T}_X=\mathcal{P}(X)\), and give \(Y\) the indiscrete topology \(\mathcal{T}_Y=\{\varnothing,Y\}\). Let \(f:X\to Y\) be the identity map.
The map is surjective. It is continuous because the only open subsets of \(Y\) are \(\varnothing\) and \(Y\), whose preimages are \(\varnothing\) and \(X\), both open in \(X\). For every \(V\subseteq Y\), \(f^{-1}(V)=V\), and every subset of \(X\) is open. Therefore the induced quotient topology is \(\mathcal{T}_q=\mathcal{P}(Y)=\{\varnothing,\{a\},\{b\},Y\}\).
The target topology is strictly coarser: \(\mathcal{T}_Y\subsetneq\mathcal{T}_q\), since, for example, \(\{a\}\in\mathcal{T}_q\) but \(\{a\}\notin\mathcal{T}_Y\). Thus \(f\) is not quotient. The specific failure is that \(f^{-1}(\{a\})=\{a\}\) is open in \(X\), while \(\{a\}\) is not open in \(Y\). As a check, both topologies contain \(\varnothing\) and \(Y\); the two singleton subsets are precisely the additional quotient-open sets.
Worked Example: A Finite Surjection with a Nontrivial Fiber
Let \(X=\{a,b,c\}\) have the discrete topology. Let \(Y=\{u,v\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{u\},Y\}\). Define \(f:X\to Y\) by \(f(a)=u\) and \(f(b)=f(c)=v\). The map is surjective, with fibers \(f^{-1}(\{u\})=\{a\}\) and \(f^{-1}(\{v\})=\{b,c\}\).
To check continuity, compute the preimages of every target-open set: \(f^{-1}(\varnothing)=\varnothing\), \(f^{-1}(\{u\})=\{a\}\), and \(f^{-1}(Y)=X\). Each is open in the discrete topology on \(X\). For every subset \(V\subseteq Y\), its preimage is also open, since every subset of \(X\) is open. Hence \(\mathcal{T}_q=\mathcal{P}(Y)=\{\varnothing,\{u\},\{v\},Y\}\), strictly larger than \(\mathcal{T}_Y\).
The quotient criterion fails at \(V=\{v\}\): its preimage \(\{b,c\}\) is open, but \(\{v\}\notin\mathcal{T}_Y\). Equivalently, \(A=\{b,c\}\) is an open saturated subset of \(X\), and \(f(A)=\{v\}\) is not target-open. The fiber computation verifies saturation directly: \(A\) contains the whole fiber over \(v\), and contains none of the fiber over \(u\).
A Real-Line Example and a Common Reversal
Worked Example: The Identity to the Finite-Complement Topology
Let \(X=\mathbb{R}\) with its standard topology, and let \(Y=\mathbb{R}\) have the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(\mathbb{R}\) are finite. Let \(f:\mathbb{R}\to\mathbb{R}\) be the identity on the underlying set.
The map is surjective. It is continuous because the preimage of any target-open set is that same set, and every cofinite subset of \(\mathbb{R}\) is standard-open: its complement is finite, each singleton in \(\mathbb{R}\) is closed, and a finite union of closed sets is closed. The preimage of \(\varnothing\) is open as well.
Since \(f^{-1}(V)=V\), the induced quotient topology consists exactly of the standard-open subsets of \(\mathbb{R}\). It is strictly finer than the target topology: the standard-open interval \((0,1)\) is not cofinite, because its complement contains, for example, every integer \(n\ge 2\), and so is not open in \(Y\). Therefore \(f^{-1}((0,1))=(0,1)\) is open in \(X\) while \((0,1)\) is not open in \(Y\), confirming that \(f\) is not quotient.
This example checks both the inclusion and the failure directly: every finite-complement-open set is standard-open, but not every standard-open set is finite-complement-open. The target topology is strictly coarser than the quotient topology.
Common pitfall. A mistaken argument may say, “The map is continuous, so every set whose preimage is open is target-open; hence \(\mathcal{T}_q\subseteq\mathcal{T}_Y\).” That implication is not the definition of continuity. Continuity starts with a target-open set and concludes that its preimage is open: \(V\in\mathcal{T}_Y\Rightarrow f^{-1}(V)\in\mathcal{T}_X\). By the definition of \(\mathcal{T}_q\), this gives \(\mathcal{T}_Y\subseteq\mathcal{T}_q\), the reverse of the mistaken inclusion. In the finite example, the proposed inclusion fails concretely: \(\{v\}\in\mathcal{T}_q\) but \(\{v\}\notin\mathcal{T}_Y\).
The distinction matters whenever a map is used to regard points of a target as identified images of points in a source. Continuity alone controls preimages of target-open sets; it does not ensure that every subset with open preimage is target-open. A quotient map has exactly that additional property, and the open saturated set in the proposition pinpoints its absence.
| Map condition | Open-set implication | Consequence |
|---|---|---|
| Continuous surjection | \(V\in\mathcal{T}_Y\Rightarrow f^{-1}(V)\) is open | \(\mathcal{T}_Y\subseteq\mathcal{T}_q\) |
| Quotient map | \(V\in\mathcal{T}_Y\) if and only if \(f^{-1}(V)\) is open | \(\mathcal{T}_Y=\mathcal{T}_q\) |
| Continuous surjection that is not quotient | Some \(f^{-1}(V)\) is open while \(V\) is not open | \(\mathcal{T}_Y\subsetneq\mathcal{T}_q\) |
Check Your Understanding
For each question, state the topology on each space before applying the quotient-map test.
- Can a continuous surjection have a target topology strictly finer than its induced quotient topology? Justify your answer using the open-set definitions.
- Let \(X=\{0,1\}\) be discrete, let \(Y=\{0,1\}\) be indiscrete, and let \(f\) be the identity. Identify a subset witnessing that \(f\) is not quotient.
- In the finite surjection example, list both fibers and explain why \(\{b,c\}\) is saturated.
- For the identity from standard \(\mathbb{R}\) to \(\mathbb{R}\) with the finite-complement topology, name a standard-open set that is not target-open.
- Prove that if \(f\) is a continuous surjection and \(\mathcal{T}_Y=\mathcal{T}_q\), then \(f\) is a quotient map.