Why a Continuous Bijection May Not Be a Homeomorphism
A continuous bijection \(f:X\to Y\) has an inverse as a function, but that inverse need not be continuous. The useful extra assumptions are compactness of the domain and the Hausdorff property of the codomain. The proof connects them in two steps: continuous images of compact spaces are compact, and compact subsets of Hausdorff spaces are closed. It then shows that \(f\) carries closed sets to closed sets, which is exactly what is needed to prove continuity of \(f^{-1}\).
Recall that a space is compact if every cover of it by open sets has a finite subcover. A space \(Y\) is Hausdorff if for every pair of distinct points \(y,z\in Y\), there are disjoint open neighborhoods \(V\) of \(y\) and \(W\) of \(z\). A map \(f:X\to Y\) is continuous if the preimage of every open subset of \(Y\) is open in \(X\). A homeomorphism is a bijection that is continuous and whose inverse is also continuous.
Two Facts Behind the Argument
Proof. Let \(\{V_i:i\in J\}\) be any cover of \(g(K)\) by sets open in the subspace \(g(K)\). For each \(i\), there is an open set \(W_i\subseteq Y\) such that \(V_i=W_i\cap g(K)\). The sets \(g^{-1}(W_i)\) are open in \(K\) by continuity. They cover \(K\): if \(x\in K\), then \(g(x)\in g(K)\), so some \(V_i\) contains \(g(x)\), and therefore \(x\in g^{-1}(W_i)\). Compactness gives finitely many indices \(i_1,\ldots,i_n\) such that \(K\subseteq \bigcup_{j=1}^n g^{-1}(W_{i_j})\). Applying \(g\), the corresponding \(V_{i_j}\) cover \(g(K)\). Thus every open cover of \(g(K)\) has a finite subcover. \(\square\)
Proof. If \(K=\varnothing\), it is closed. Otherwise, fix any \(y\in Y\setminus K\). For each \(k\in K\), the points \(k\) and \(y\) are distinct. Since \(Y\) is Hausdorff, choose disjoint open sets \(U_k,V_k\subseteq Y\) with \(k\in U_k\) and \(y\in V_k\). The sets \(U_k\) cover \(K\). By compactness, finitely many of them, say \(U_{k_1},\ldots,U_{k_n}\), cover \(K\). The set \(V=\bigcap_{j=1}^n V_{k_j}\) is an open neighborhood of \(y\). It is disjoint from \(K\): if \(z\in V\cap K\), then \(z\in U_{k_j}\) for some \(j\), while \(z\in V\subseteq V_{k_j}\), contradicting \(U_{k_j}\cap V_{k_j}=\varnothing\). Thus every \(y\in Y\setminus K\) has an open neighborhood contained in \(Y\setminus K\). The complement \(Y\setminus K\) is open, so \(K\) is closed. \(\square\)
The finite choice in the second proof is essential to its structure: each point \(k\) has a separating neighborhood pair, but compactness reduces the neighborhoods around the points of \(K\) to finitely many. Intersecting the corresponding finitely many neighborhoods of \(y\) still gives an open neighborhood.
The Compact-to-Hausdorff Theorem
Proof. We prove that the inverse \(f^{-1}:Y\to X\) is continuous. Let \(F\) be any closed subset of \(X\). A closed subset of a compact space is compact: an open cover of \(F\) in its subspace topology can be represented by open sets of \(X\); adjoining the open set \(X\setminus F\) gives an open cover of \(X\), from which a finite subcover yields a finite subcover of \(F\). This also holds when \(F=\varnothing\), which is compact. Proposition 1 now shows that \(f(F)\) is compact in \(Y\), and Proposition 2 shows that \(f(F)\) is closed in \(Y\). Since \(f\) is bijective, \((f^{-1})^{-1}(F)=f(F)\). Therefore the preimage under \(f^{-1}\) of every closed subset \(F\) of \(X\) is closed in \(Y\). A map is continuous if and only if preimages of closed sets are closed, because taking complements converts this condition into the open-set definition of continuity. Hence \(f^{-1}\) is continuous, and \(f\) is a homeomorphism. If \(X=\varnothing\), bijectivity forces \(Y=\varnothing\), and the unique map between them and its inverse are continuous; thus this case is included as well. \(\square\)
Equivalently, the proof shows that \(f\) is a closed map, meaning that it sends closed sets to closed sets. For a bijection, this is precisely the property that makes the inverse continuous. Continuity of \(f\) itself is still a separate hypothesis: the closed-map argument proves continuity of the inverse, not of \(f\).
Worked Example: The Cubing Map on a Compact Interval
Let \(I=[0,1]\) have the subspace topology from the standard topology on \(\mathbb{R}\), and define \(f:I\to I\) by \(f(x)=x^3\). The interval \(I\) is compact by the Heine–Borel theorem, and \(I\) is Hausdorff as a subspace of \(\mathbb{R}\). The formula is continuous because it is the restriction of a polynomial.
To check injectivity, suppose \(0\le x<y\le1\). Then \(y^3-x^3=(y-x)(y^2+xy+x^2)>0\), since \(y-x>0\) and \(y^2+xy+x^2>0\). Thus \(f(x)<f(y)\), so \(f\) is injective. For surjectivity, let \(t\in[0,1]\). The function \(x\mapsto x^3\) is continuous, takes value \(0\) at \(0\) and value \(1\) at \(1\), so the intermediate value theorem gives \(x\in[0,1]\) with \(x^3=t\). Hence \(f\) is a continuous bijection and the theorem implies that it is a homeomorphism.
For a concrete closed-set check, take \(F=[1/4,1]\), which is closed in \(I\). Since cubing is increasing on \(I\), its image is \(f(F)=[(1/4)^3,1^3]=[1/64,1]\), closed in \(I\). The endpoint computation is \((1/4)^3=1/64\) and \(1^3=1\). The inverse is \(f^{-1}(t)=\sqrt[3]{t}\): substituting gives \((\sqrt[3]{t})^3=t\) and \(\sqrt[3]{x^3}=x\) for \(t,x\in[0,1]\). The theorem establishes that this inverse is continuous. These checks agree with the conclusion that \(f\) is a homeomorphism.
Why Both Hypotheses Matter
The assumptions concern different spaces and serve different purposes. Compactness is required of the domain so that closed subsets of the domain are compact. Hausdorffness is required of the codomain so that their compact images are closed. Removing either condition can break the conclusion, even when the map remains a continuous bijection.
Worked Example: Removing Compactness
Let \(X=\mathbb{R}\) with the discrete topology, in which every subset is open, and let \(Y=\mathbb{R}\) with the standard topology. Both spaces are Hausdorff: in \(Y\), distinct real numbers have disjoint open intervals around them; in the discrete topology, distinct points have disjoint singleton neighborhoods. Define \(f:X\to Y\) by \(f(x)=x\).
The map is bijective. It is continuous because the preimage of any open set \(U\subseteq Y\) is the same subset \(U\) of \(X\), and every subset of the discrete space \(X\) is open. But \(f^{-1}:Y\to X\) is not continuous: the singleton \(\{0\}\) is open in \(X\), while its preimage under \(f^{-1}\) is \(\{0\}\), which is not open in the standard topology on \(\mathbb{R}\). Also, \(X\) is not compact. The open cover \(\{\{x\}:x\in\mathbb{R}\}\) has no finite subcover, since a finite union of singletons cannot contain every real number. Thus the failure of compactness is real, and the continuous bijection is not a homeomorphism.
Check: the same set \(\{0\}\) is closed in the discrete \(X\), but its image under \(f\) is not closed in \(Y\)? In fact \(\{0\}\) is closed in standard \(\mathbb{R}\), so this set does not witness failure of closedness. Instead, take \(F=(0,1)\), which is closed in the discrete \(X\). Its image \((0,1)\) is not closed in standard \(\mathbb{R}\), since every open neighborhood of \(0\) meets \((0,1)\) and \(0\notin(0,1)\). This explicitly shows why the closed-map route is unavailable.
Worked Example: Removing Hausdorffness
Let \(X=\{a,b\}\) have the discrete topology \(\{\varnothing,\{a\},\{b\},X\}\), and let \(Y=\{a,b\}\) have the topology \(\{\varnothing,\{a\},Y\}\). The finite space \(X\) is compact: from any open cover, choose one member containing \(a\) and one member containing \(b\); these at most two sets cover \(X\). Define \(f:X\to Y\) by \(f(a)=a\) and \(f(b)=b\).
The map is a bijection. It is continuous because every open subset of \(Y\) is a subset of \(X\), and every subset of the discrete \(X\) is open. However, \(f^{-1}:Y\to X\) is not continuous. The set \(\{b\}\) is open in \(X\), but its preimage under \(f^{-1}\) is \(\{b\}\), which is not open in \(Y\). The codomain \(Y\) is not Hausdorff: the only open neighborhood of \(b\) is \(Y\), so \(a\) and \(b\) have no disjoint open neighborhoods. The compact subset \(\{a\}\) of \(Y\) is not closed, because its complement \(\{b\}\) is not open. Thus the compact-subset-is-closed step fails exactly where Hausdorffness is absent.
A Common Pitfall and a Proof Checklist
Common pitfall. A continuous bijection is not automatically a homeomorphism. In the discrete-to-standard example, the calculation “\(f\) is bijective and continuous, therefore \(f^{-1}\) is continuous” is invalid: the singleton \(\{0\}\) is open in the discrete codomain of \(f^{-1}\), but its preimage under \(f^{-1}\) is not open in the standard real line. The correction is to establish the missing inverse continuity, either directly or by proving that the original map is closed. The compact-to-Hausdorff theorem supplies the latter method under its stated hypotheses.
When applying the argument, keep track of which space supplies each property:
| Step | Property used | What it gives |
|---|---|---|
| A closed \(F\subseteq X\) is compact | \(X\) compact | Compactness passes to closed subsets |
| \(f(F)\) is compact | \(f\) continuous | Continuous images preserve compactness |
| \(f(F)\) is closed in \(Y\) | \(Y\) Hausdorff | Compact subsets of \(Y\) are closed |
| \(f^{-1}\) is continuous | \(f\) bijective and closed | Closed-set preimages under \(f^{-1}\) are closed |
The conclusion is stronger than a statement about the sets \(X\) and \(Y\) having the same points: it says their topologies match under \(f\). Compactness controls images of closed sets, while Hausdorffness converts the resulting compactness into closedness. Together they prevent a continuous bijection from losing topological information in its inverse.
Check Your Understanding
For each question, identify the space and hypothesis used at the relevant step of the argument.
- In the theorem, why is \(f(F)\) compact when \(F\) is closed in \(X\)? State both facts needed.
- Where is Hausdorffness used in the proof, and what specific conclusion does it provide?
- In the discrete-to-standard example, give an open set in the codomain of \(f^{-1}\) whose preimage is not open.
- For the two-point example, list the open neighborhoods of \(b\) in \(Y\) and explain why \(Y\) is not Hausdorff.
- Suppose \(f:X\to Y\) is a continuous bijection and is also known to be a closed map. Prove that \(f\) is a homeomorphism, without assuming compactness or Hausdorffness.