Why Check What a Bijection Does to Open Sets?
A continuous bijection matches points one-to-one and respects the topologies in the forward direction. But these facts alone do not ensure that the inverse respects the topologies. One way to test the inverse is to ask whether the original map carries open sets to open sets; another is to ask whether it carries closed sets to closed sets. Either test, together with continuity and bijectivity, guarantees a homeomorphism.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. A function \(f:X\to Y\) is continuous if \(f^{-1}(V)\in\mathcal{T}_X\) for every \(V\in\mathcal{T}_Y\). It is an open map if \(f(U)\in\mathcal{T}_Y\) for every \(U\in\mathcal{T}_X\), and a closed map if \(f(C)\) is closed in \(Y\) for every closed \(C\subseteq X\). These are separate conditions: continuity concerns preimages, whereas openness and closedness concern images.
The Inverse-Continuity Criteria
Proof. Write \(g=f^{-1}:Y\to X\). By the definition of continuity, \(g\) is continuous exactly when \(g^{-1}(U)\) is open in \(Y\) for every open \(U\subseteq X\). Since \(g=f^{-1}\) as a function, \(g^{-1}(U)=f(U)\). Thus this condition says exactly that \(f\) maps every open subset of \(X\) to an open subset of \(Y\).
For the closed-set version, a function is continuous if and only if preimages of closed sets are closed: this follows by taking complements in the open-set definition. Hence \(g\) is continuous exactly when \(g^{-1}(C)\) is closed in \(Y\) for every closed \(C\subseteq X\). Again \(g^{-1}(C)=f(C)\), so this is exactly the condition that \(f\) is a closed map. \(\square\)
Proof. Suppose first that \(f\) is a continuous, open bijection. The proposition shows that \(f^{-1}\) is continuous. The map \(f\) is therefore a bijection continuous in both directions, which is the definition of a homeomorphism. If instead \(f\) is a continuous, closed bijection, the closed-set part of the proposition shows that \(f^{-1}\) is continuous, and the same definition gives that \(f\) is a homeomorphism. These arguments also cover the empty spaces: a bijection involving an empty space has both spaces empty, and the continuity conditions hold vacuously. \(\square\)
The bijection hypothesis matters because it gives an inverse function and identifies its preimages with images under \(f\). The theorem does not say that every continuous bijection is a homeomorphism. The additional open-map or closed-map condition is what certifies continuity of the inverse.
An Open Bijection Between Open Intervals
Worked Example: An Affine Map Between Open Intervals
Give \((2,5)\) and \((-3,3)\) their subspace topologies from the standard topology on \(\mathbb{R}\). Consider \(f:(2,5)\to(-3,3)\) defined by \(f(x)=2x-7\).
Bijection. If \(x\in(2,5)\), then \(2x-7\in(-3,3)\), since \(x>2\) gives \(2x-7>-3\) and \(x<5\) gives \(2x-7<3\). For any \(y\in(-3,3)\), the equation \(y=2x-7\) has the unique solution \(x=(y+7)/2\). The inequalities \(-3<y<3\) give \(2<(y+7)/2<5\). Thus \(f\) is a bijection, with inverse \(f^{-1}(y)=(y+7)/2\).
Continuity. For \(x,x'\in(2,5)\), \(|f(x)-f(x')|=|2x-2x'|=2|x-x'|\). Given \(\varepsilon>0\), taking \(\delta=\varepsilon/2\) ensures that \(|x-x'|<\delta\) implies \(|f(x)-f(x')|<\varepsilon\). This verifies continuity in the subspace metrics.
Openness. Let \(U\) be open in \((2,5)\), and choose any \(y\in f(U)\). There is an \(x\in U\) with \(y=f(x)\). Relative openness gives some \(\delta>0\) such that \((x-\delta,x+\delta)\cap(2,5)\subseteq U\). If \(y'\in(y-2\delta,y+2\delta)\cap(-3,3)\), then \(x'=(y'+7)/2\) lies in \((2,5)\), and \(|x'-x|=|y'-y|/2<\delta\). Consequently \(x'\in U\) and \(y'=f(x')\in f(U)\). We have exhibited an open neighborhood of each \(y\in f(U)\) contained in \(f(U)\), so \(f(U)\) is open in \((-3,3)\).
Conclusion and check. The theorem now gives that \(f\) is a homeomorphism. The inverse formula also checks the result directly: for \(y,y'\in(-3,3)\), \(|f^{-1}(y)-f^{-1}(y')|=|(y-y')/2|=|y-y'|/2\), so the inverse is continuous. Substituting \(y=2x-7\) into the inverse gives \(((2x-7)+7)/2=x\), as required.
A Closed Bijection Between Closed Intervals
Worked Example: A Closed-Map Test on Closed Intervals
Give \([0,1]\) and \([1,4]\) their subspace topologies from \(\mathbb{R}\), and let \(f:[0,1]\to[1,4]\) be \(f(x)=1+3x\).
Bijection and continuity. The endpoint values are \(f(0)=1\) and \(f(1)=4\). For \(x\in[0,1]\), the value \(1+3x\) lies in \([1,4]\); conversely, each \(y\in[1,4]\) has the unique preimage \(x=(y-1)/3\in[0,1]\). For \(x,x'\in[0,1]\), \(|f(x)-f(x')|=3|x-x'|\). Given \(\varepsilon>0\), choosing \(\delta=\varepsilon/3\) proves continuity.
Closedness. Let \(C\) be closed in \([0,1]\), so \(O=[0,1]\setminus C\) is relatively open. We first check that \(f(O)\) is relatively open in \([1,4]\). For any \(y=f(x)\in f(O)\), relative openness gives \(\delta>0\) such that \((x-\delta,x+\delta)\cap[0,1]\subseteq O\). If \(y'\in(y-3\delta,y+3\delta)\cap[1,4]\), then \(x'=(y'-1)/3\in[0,1]\) and \(|x'-x|=|y'-y|/3<\delta\). Thus \(x'\in O\) and \(y'=f(x')\in f(O)\). So \(f(O)\) is open relative to \([1,4]\).
Since \(f\) is a bijection, \(f(O)=[1,4]\setminus f(C)\): a point is outside \(f(C)\) precisely when its unique preimage is outside \(C\). The complement of \(f(C)\) is therefore open, and \(f(C)\) is closed. This holds for every closed \(C\), so \(f\) is a closed map. By the theorem, \(f\) is a homeomorphism. As a check, its inverse is \(f^{-1}(y)=(y-1)/3\), and \(|f^{-1}(y)-f^{-1}(y')|=|y-y'|/3\), which verifies continuity of the inverse directly.
Finite Spaces: Testing Images and Preimages
Worked Example: A Finite Homeomorphism
Let \(X=\{a,b\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{a\},X\}\), and let \(Y=\{u,v\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{u\},Y\}\). Define \(f(a)=u\) and \(f(b)=v\).
The map is a bijection. To check continuity, test all open sets of \(Y\): their preimages are \(f^{-1}(\varnothing)=\varnothing\), \(f^{-1}(\{u\})=\{a\}\), and \(f^{-1}(Y)=X\), all open in \(X\). To check openness, the open sets of \(X\) have images \(\varnothing\), \(\{u\}\), and \(Y\), all open in \(Y\). Thus \(f\) is a continuous open bijection and hence a homeomorphism.
For a closed-map check, the closed sets of \(X\), obtained by taking complements of its open sets, are \(\varnothing,\{b\},X\). Their images are \(\varnothing,\{v\},Y\), which are closed in \(Y\), since the closed sets of \(Y\) are \(\varnothing,\{v\},Y\). So this same bijection is also closed. Finally \(f^{-1}(\{u\})=\{a\}\) and \(f^{-1}(\{v\})=\{b\}\), in agreement with the inverse function; the inverse is continuous because it is itself a bijection between these matching topologies.
Worked Example: A Finite Continuous Bijection That Fails Both Tests
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{P}(X)\), and let \(Y=\{u,v\}\) have topology \(\{\varnothing,\{u\},Y\}\). Define \(h(a)=u\) and \(h(b)=v\). This example shows why the extra condition in the theorem cannot be omitted.
The map is a bijection. To check continuity, the open sets in \(Y\) are \(\varnothing,\{u\},Y\), and their preimages are \(\varnothing,\{a\},X\), all open in the discrete space \(X\). But \(h\) is not open: \(\{b\}\) is open in \(X\), whereas \(h(\{b\})=\{v\}\) is not open in \(Y\). It is not closed either: \(\{a\}\) is closed in \(X\), whereas \(h(\{a\})=\{u\}\) is not closed in \(Y\), whose closed sets are \(\varnothing,\{v\},Y\).
The inverse is not continuous: the set \(\{a\}\) is open in \(X\), but its preimage under \(h^{-1}:Y\to X\) is \(\{u\}\), which is not open in \(Y\). Notice the distinction in the continuity test for \(h\): \(\{u\}\) is open in \(Y\), so its preimage \(\{a\}\) must be checked and is open; \(\{v\}\) is not open in \(Y\), so its preimage is not a required continuity test. The failure of openness, by contrast, is detected using the image of the open source set \(\{b\}\).
A Common Pitfall and a Practical Test
Common pitfall. A tempting but incorrect calculation is: “\(f\) is a continuous bijection, so \(f(U)\) is open whenever \(U\) is open.” Continuity does not say this; it says that preimages of open target sets are open. In the finite example above, \(h\) is continuous and bijective, but the open set \(\{b\}\subseteq X\) has image \(\{v\}\), which is not open in \(Y\). The correction is to test images separately when proving a map is open. Alternatively, to use the closed-map criterion, test images of closed source sets. Neither test can be replaced by checking continuity alone.
For a proposed homeomorphism proof, first establish that the map is a bijection and continuous. Then choose one inverse-continuity test: show that images of all open sets are open, or show that images of all closed sets are closed. In small finite spaces, list the open or closed sets and check each one. For interval maps, use the relative neighborhoods of the interval topologies, as in the examples.
Check Your Understanding
For each question, specify which images or preimages must be checked and why.
- State the two inverse-continuity criteria for a bijection \(f:X\to Y\).
- For \(f:[0,1]\to[2,5]\) defined by \(f(x)=2+3x\), give its inverse and verify continuity of the inverse using a distance calculation.
- In the finite counterexample, which open set of \(X\) demonstrates that the map is not open?
- For that same example, list the open subsets of \(Y\) and their preimages under \(h\). Why is \(\{v\}\) not part of this continuity check?
- Explain why showing that a bijection is closed proves continuity of its inverse, even though the inverse-continuity definition is stated using open sets.