Why a Continuous Bijection May Not Be Enough
A continuous map controls preimages of open sets: if \(V\) is open in the codomain, then its preimage is open in the domain. A homeomorphism requires more: its inverse must also be continuous. For a bijection, that amounts to controlling images of open sets, or equivalently images of closed sets. Compactness and the Hausdorff condition provide exactly this additional control.
Throughout, \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) are topological spaces. A space is compact if every open cover has a finite subcover. A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open \(V\subseteq Y\). A map is closed if it sends every closed subset of its domain to a closed subset of its codomain. A homeomorphism is a bijection that is continuous and whose inverse is continuous.
The proof combines three facts. We include the first two briefly to make the chain of reasoning explicit. First, a closed subset \(C\) of a compact space \(X\) is compact: given a cover of \(C\) by sets open in the subspace \(C\), write each as \(C\cap U\) for an open \(U\subseteq X\), and add \(X\setminus C\) to obtain an open cover of \(X\). A finite subcover of \(X\) then gives a finite subcover of \(C\). Second, a continuous image of a compact space is compact: pull an open cover of the image back to an open cover of the domain, take a finite subcover there, and use the corresponding finite collection of original sets.
We also use the standard result proved in the earlier material on compact subsets of Hausdorff spaces: every compact subset of a Hausdorff space is closed. Its hypotheses matter here: compactness applies to the subset, and the Hausdorff condition applies to the space containing it.
The Closed-Set Argument
Proof. Assume \(X\) is compact, \(Y\) is Hausdorff, and \(f:X\to Y\) is a continuous bijection. We first show that \(f\) is closed. Let \(C\subseteq X\) be closed. By the closed-subspace property just described, \(C\) is compact. Since \(f\) is continuous, \(f(C)\) is compact. Because \(f(C)\subseteq Y\) and \(Y\) is Hausdorff, the compact-subsets-are-closed result gives that \(f(C)\) is closed in \(Y\). Thus \(f\) maps every closed subset of \(X\) to a closed subset of \(Y\).
It remains to prove that \(f^{-1}:Y\to X\) is continuous. Let \(C\) be any closed subset of \(X\). The preimage of \(C\) under the inverse map is \( (f^{-1})^{-1}(C)=f(C) \). We have just proved that \(f(C)\) is closed in \(Y\). Therefore the inverse map has closed preimages of closed sets, so it is continuous: indeed, taking complements shows that the preimage of every open set is open. Since \(f\) is already a continuous bijection, it is a homeomorphism. \(\square\)
Proof. Suppose first that \(f^{-1}\) is continuous. If \(C\subseteq X\) is closed, then \(X\setminus C\) is open. Continuity of \(f^{-1}\) gives that \( (f^{-1})^{-1}(X\setminus C)=Y\setminus f(C) \) is open in \(Y\), so \(f(C)\) is closed. Conversely, suppose \(f\) is closed. For every closed \(C\subseteq X\), the identity \( (f^{-1})^{-1}(C)=f(C) \) shows that the preimage of \(C\) under \(f^{-1}\) is closed. Taking complements shows that preimages of open sets under \(f^{-1}\) are open. Hence \(f^{-1}\) is continuous. \(\square\)
Applying the Theorem to Interval Maps
Worked Example: A Nonlinear Bijection of a Compact Interval
Let \(X=Y=[-1,1]\) with their usual subspace topologies from \(\mathbb{R}\), and define \(f:X\to Y\) by \(f(x)=x^3\). We verify the hypotheses and use the theorem to conclude that \(f\) is a homeomorphism.
The interval \([-1,1]\) is compact, and \(Y\), as a subspace of the Hausdorff space \(\mathbb{R}\), is Hausdorff. The polynomial \(x^3\) is continuous. It is injective: if \(x_1<x_2\), then \(x_2^3-x_1^3=(x_2-x_1)(x_2^2+x_1x_2+x_1^2)>0\). The first factor is positive; the second is positive unless both \(x_1\) and \(x_2\) are zero, which is impossible when \(x_1<x_2\). It is surjective onto \([-1,1]\): for each \(y\in[-1,1]\), the real cube root \(x=\sqrt[3]{y}\) belongs to \([-1,1]\) and satisfies \(x^3=y\).
Thus \(f\) is a continuous bijection from a compact space to a Hausdorff space, so its inverse is continuous. For a concrete closed-set check, take \(C=[-1,0]\), closed in \(X\). Since \(x\leq 0\) exactly when \(x^3\leq 0\), its image is \(f(C)=[-1,0]\), closed in \(Y\). The inverse formula is \(f^{-1}(y)=\sqrt[3]{y}\); substituting gives \((\sqrt[3]{y})^3=y\), and applying the formula to \(x^3\) gives \(\sqrt[3]{x^3}=x\). These computations check both inverse identities.
Worked Example: A Rational Formula on a Compact Interval
Give \(X=[0,1]\) and \(Y=[0,\tfrac12]\) their usual subspace topologies from \(\mathbb{R}\), and define \(f(x)=\frac{x}{1+x}\). The denominator is positive on \(X\), so the formula defines a continuous map there. To check its range and bijectivity, suppose \(0\leq x_1<x_2\leq1\). Then \( f(x_2)-f(x_1)=\frac{x_2-x_1}{(1+x_1)(1+x_2)}>0 \), so \(f\) is strictly increasing. Its endpoint values are \(f(0)=0/(1+0)=0\) and \(f(1)=1/(1+1)=\tfrac12\). Continuity and the intermediate value theorem show that it takes every value between these endpoints; strict increase shows it takes each value only once.
Here \(X\) is compact and \(Y\) is Hausdorff, so the theorem makes \(f\) a homeomorphism. Solving \(y=x/(1+x)\) gives \(y+xy=x\), then \(y=x(1-y)\), and hence \(x=y/(1-y)\). This denominator is positive for \(y\in[0,\tfrac12]\). The inverse is therefore \(f^{-1}(y)=y/(1-y)\). Checking at the endpoints gives \(f^{-1}(0)=0/(1-0)=0\) and \(f^{-1}(\tfrac12)=(\tfrac12)/(\tfrac12)=1\), as required.
For another closed-set check, take \(C=[0,\tfrac12]\subseteq X\). Monotonicity and the endpoint computations give \( f(C)=[f(0),f(\tfrac12)]=[0,\tfrac{1/2}{1+1/2}]=[0,\tfrac13] \), which is closed in \(Y\). The fraction simplifies because \((1/2)/(3/2)=(1/2)(2/3)=1/3\).
Why Both Hypotheses Matter
The theorem is useful because it supplies continuity of the inverse without requiring a direct neighborhood-by-neighborhood analysis. But neither compactness nor the Hausdorff condition is decoration: omitting either can invalidate the conclusion.
Worked Example: Dropping the Hausdorff Condition
Let \(X=Y=\{a,b\}\). Give \(X\) the discrete topology \(\{\varnothing,\{a\},\{b\},X\}\), and give \(Y\) the indiscrete topology \(\{\varnothing,Y\}\). The identity map \(f:X\to Y\) is a bijection. The finite space \(X\) is compact: from any open cover, choose one covering set for each of its two points, giving at most two sets. The map is continuous because the only open sets of \(Y\) have preimages \(\varnothing\) and \(X\), both open in \(X\).
However, \(f^{-1}\) is not continuous. The set \(\{a\}\) is open in \(X\), while its preimage under \(f^{-1}:Y\to X\) is \(\{a\}\), which is not open in the indiscrete topology on \(Y\). The target \(Y\) is not Hausdorff: the only open neighborhood of either point is \(Y\), so distinct points cannot have disjoint open neighborhoods. Thus the Hausdorff hypothesis cannot simply be removed.
Worked Example: Dropping Compactness
Let \(X=\mathbb{R}\) with the discrete topology and \(Y=\mathbb{R}\) with its standard topology. The identity map \(f:X\to Y\) is a bijection and is continuous: every subset of the discrete domain is open, so the preimage of any open subset of \(Y\) is open in \(X\). The codomain \(Y\) is Hausdorff.
The domain is not compact. The open cover \(\{\{x\}:x\in\mathbb{R}\}\) has no finite subcover, since a finite collection of singleton sets covers only finitely many real numbers. The inverse identity map \(f^{-1}:Y\to X\) is not continuous: the singleton \(\{0\}\) is open in the discrete space \(X\), but its preimage under \(f^{-1}\) is \(\{0\}\), which is not open in the standard topology on \(\mathbb{R}\). This shows why compactness is needed in the theorem.
Common pitfall. Continuity does not in general say that images of open sets are open. For example, with the standard topology on \(\mathbb{R}\), the continuous map \(g:\mathbb{R}\to\mathbb{R}\), \(g(x)=x^2\), sends the open set \((-1,1)\) to \([0,1)\). This image is not open: every open interval around \(0\) contains negative numbers, which are not in \([0,1)\). The compact-to-Hausdorff theorem does not claim that every continuous map is open. It gives a closed-map conclusion specifically for a continuous bijection under its stated hypotheses, and that conclusion is what makes the inverse continuous.
Check Your Understanding
For each question, identify which hypotheses are being used before drawing a conclusion.
- In the proof of the theorem, why is \(f(C)\) compact when \(C\) is closed in \(X\)? State the results used in order.
- Show that if \(f:X\to Y\) is a continuous bijection and \(X\) is compact, then \(f(C)\) is compact for every closed \(C\subseteq X\), even if \(Y\) is not Hausdorff.
- For the map \(f(x)=x^3\) on \([-1,1]\), compute the image of the closed set \([-\tfrac12,\tfrac12]\).
- In the finite example, which open set in the discrete space witnesses that the inverse identity is not continuous?
- Explain why a closed bijection has a continuous inverse, using the preimage-of-a-closed-set test.