Why Bijectivity Changes the Quotient Condition
A continuous bijection need not be a homeomorphism: continuity controls preimages of open sets, but a homeomorphism also requires the inverse to be continuous. A quotient map has a stronger open-set condition. When a quotient map is bijective, that condition forces images of open sets to be open, which is exactly what continuity of the inverse requires.
Throughout, a topological space is a pair \((X,\mathcal{T})\), where \(\mathcal{T}\) is its topology. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). It is a homeomorphism if it is bijective, continuous, and its inverse \(f^{-1}:Y\to X\) is continuous.
For a general surjection, different points of \(X\) may map to the same point of \(Y\). For a bijection, there are no such identifications: the preimage of the image of a set \(U\subseteq X\) is exactly \(U\). That fact is the key to the theorem.
The Main Theorem
Proof. Since \(q\) is a quotient map, it is continuous. It remains to prove that \(q^{-1}:Y\to X\) is continuous. Let \(U\) be any open subset of \(X\). Bijectivity gives \(q^{-1}(q(U))=U\): a point of \(X\) maps into \(q(U)\) exactly when it is the unique point mapping to some point \(q(u)\), where \(u\in U\), and hence it equals \(u\). Therefore \(q^{-1}(q(U))\) is open in \(X\). Applying the quotient condition to the subset \(q(U)\subseteq Y\), we conclude that \(q(U)\) is open in \(Y\). But \(q(U)=(q^{-1})^{-1}(U)\), so the preimage under \(q^{-1}\) of every open subset \(U\) of \(X\) is open in \(Y\). Thus \(q^{-1}\) is continuous, and \(q\) is a homeomorphism. This argument also covers the empty space: if \(q\) is a bijection and one space is empty, both are empty, and the same open-set argument applies. \(\square\)
The proof identifies exactly what bijectivity contributes. It makes \(q^{-1}(q(U))=U\), so the quotient test can be applied to turn openness in \(X\) into openness of the corresponding set in \(Y\). Without injectivity, that equality can fail: \(q^{-1}(q(U))\) can include points outside \(U\).
Worked Example: A Bijective Quotient Map Between Finite Spaces
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\), and let \(Y=\{1,2,3\}\) have topology \(\mathcal{U}=\{\varnothing,\{2\},\{2,3\},Y\}\). These are topologies: in each list, arbitrary unions and finite intersections remain in the list. Define \(q:X\to Y\) by \(q(a)=2\), \(q(b)=3\), and \(q(c)=1\).
The map is bijective, and the images of the open sets of \(X\) are \(q(\varnothing)=\varnothing\), \(q(\{a\})=\{2\}\), \(q(\{a,b\})=\{2,3\}\), and \(q(X)=Y\). These are precisely the open sets of \(Y\). For any subset \(V\subseteq Y\), its preimage is open in \(X\) exactly when \(V\) is one of these four images: bijectivity ensures that \(q^{-1}(V)\) is open exactly when \(V=q(q^{-1}(V))\) is the image of an open set. Hence \(q\) is a quotient map.
In particular, the inverse sends \(2\) to \(a\), \(3\) to \(b\), and \(1\) to \(c\). The preimages under the inverse of the open sets \(\varnothing,\{a\},\{a,b\},X\) are respectively \(\varnothing,\{2\},\{2,3\},Y\), all open in \(Y\). This checks the theorem directly. Also, \(q^{-1}(\{2,3\})=\{a,b\}\), so applying \(q\) back gives \(q(\{a,b\})=\{2,3\}\), as required.
A Criterion for Continuous Bijections
Proof. Suppose first that \(f\) is a quotient map. The theorem just proved shows that \(f^{-1}\) is continuous. Conversely, suppose \(f^{-1}\) is continuous. For any subset \(V\subseteq Y\), if \(V\) is open, continuity of \(f\) gives that \(f^{-1}(V)\) is open in \(X\). If \(f^{-1}(V)\) is open in \(X\), continuity of \(f^{-1}\) gives that \((f^{-1})^{-1}(f^{-1}(V))\) is open in \(Y\). Since \(f\) is bijective, \((f^{-1})^{-1}(f^{-1}(V))=V\). Thus \(V\) is open exactly when \(f^{-1}(V)\) is open, and \(f\) is a quotient map. The final statement follows from the definition of homeomorphism and the theorem. \(\square\)
This criterion is useful because it separates two questions. Continuity of \(f\) tests preimages of open sets in \(Y\); continuity of its inverse tests images of open sets in \(X\). A bijection alone does not make these tests equivalent. The quotient condition does.
Worked Example: A Continuous Bijection That Is Not a Homeomorphism
Let \(X=\mathbb{R}\) with its standard topology. Let \(Y=\mathbb{R}\) with the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(\mathbb{R}\) are finite. Consider the identity function \(f:X\to Y\), \(f(x)=x\).
The finite-complement sets form a topology. A finite intersection of such sets has finite complement, since the complement of the intersection is a finite union of finite sets. A union of any collection of open sets is open: if the union contains a nonempty member with finite complement, the complement of the union is a subset of that finite complement; and if all members are empty, the union is empty. The identity \(f\) is a bijection.
To test continuity of \(f\), let \(V\) be open in \(Y\). If \(V=\varnothing\), then \(f^{-1}(V)=\varnothing\), which is open in \(X\). Otherwise \(\mathbb{R}\setminus V\) is finite. Finite subsets of the standard real line are closed, since each singleton is closed and a finite union of closed sets is closed. Therefore \(V\) is open in the standard topology on \(X\), and \(f^{-1}(V)=V\) is open in \(X\). Thus \(f\) is continuous.
Its inverse is not continuous. The set \((0,1)\) is open in the standard topology on the codomain of \(f^{-1}\), but its preimage under \(f^{-1}\) is \((0,1)\) in the finite-complement topology on the domain. This preimage is not open there: its complement \((-\infty,0]\cup[1,\infty)\) is infinite, so \((0,1)\) is neither empty nor a set with finite complement. Hence \(f^{-1}\) is not continuous, and \(f\) is not a homeomorphism.
The same set also verifies that \(f\) is not a quotient map. The subset \(V=(0,1)\) is not open in \(Y\), but \(f^{-1}(V)=(0,1)\) is open in \(X\). The quotient condition requires the two openness statements to be equivalent, so it fails here.
Check: For every finite-complement open set \(V\) in \(Y\), its preimage under \(f\) is open in standard \(\mathbb{R}\), as shown above. The failure is in the reverse implication of the quotient condition: this particular standard-open preimage comes from a set that is not open in \(Y\).
Checking the Topology in the Right Direction
Continuity always refers to open sets in the codomain of the map being tested. For \(f^{-1}:Y\to X\), begin with an open set \(U\) in \(X\), then ask whether \((f^{-1})^{-1}(U)\) is open in \(Y\). In the preceding example, \(X\) has the standard topology, while \(Y\) has the finite-complement topology. Therefore \((0,1)\) is a valid open set to test for continuity of \(f^{-1}\), because it is open in \(X\), the codomain of \(f^{-1}\). Its preimage is a subset of \(Y\), and openness must be judged using \(Y\)'s finite-complement topology.
Common mistake. One might say, “The identity is a bijection and is continuous, so it is a quotient map; therefore its inverse is continuous.” The first implication is false. In the example, \(f^{-1}((0,1))=(0,1)\) is open in \(X\), while \((0,1)\) is not open in \(Y\); this directly violates the quotient condition. The correction is to check the condition in both directions for every subset of the codomain, or, for a continuous bijection, to test inverse continuity separately.
Worked Example: A Two-Point Failure
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), and let \(Y=\{1,2\}\) have the indiscrete topology \(\mathcal{U}=\{\varnothing,Y\}\). Define \(f:X\to Y\) by \(f(a)=1\) and \(f(b)=2\).
The map is bijective. It is continuous because the only open sets in \(Y\) are \(\varnothing\) and \(Y\), whose preimages are \(\varnothing\) and \(X\), both open in \(X\). But its inverse is not continuous: \(\{a\}\) is open in \(X\), the codomain of \(f^{-1}\), and \((f^{-1})^{-1}(\{a\})=\{1\}\) is not open in \(Y\). Thus \(f\) is not a homeomorphism.
It is also not a quotient map. For \(V=\{1\}\subseteq Y\), the preimage \(f^{-1}(V)=\{a\}\) is open in \(X\), whereas \(V\) is not open in \(Y\). This example checks the same failure as the real-line example with only two points and verifies every relevant open-set test directly.
What the Result Does—and Does Not—Say
The theorem is a strong test for maps out of quotient spaces: if such a map is bijective and satisfies the quotient open-set condition, there is no further inverse-continuity question to settle. But bijectivity and continuity by themselves are insufficient. The two counterexamples exhibit the missing condition concretely: some open subset of the domain has an image that is not open in the codomain.
Keep the direction of each test explicit:
- For continuity of \(f:X\to Y\), take an open set in \(Y\) and test its preimage in \(X\).
- For continuity of \(f^{-1}:Y\to X\), take an open set in \(X\) and test its preimage in \(Y\).
- For a quotient map \(f:X\to Y\), test both directions: \(V\) is open in \(Y\) exactly when \(f^{-1}(V)\) is open in \(X\).
Check Your Understanding
For each question, identify the topology in which every claimed open set is being tested.
- In the two-point example, which subset of \(Y\) shows that the continuous bijection is not a quotient map?
- Explain why \(q^{-1}(q(U))=U\) in the theorem's proof depends on injectivity.
- For a continuous bijection \(f:X\to Y\), which continuity condition is equivalent to \(f\) being a quotient map?
- In the real-line example, why is \((0,1)\) open in the domain of \(f\) but not open in its codomain?
- Can a bijective quotient map fail to be a homeomorphism? Justify your answer using the theorem.