A Finite Continuous Bijection That Is Not a Homeomorphism

A Bijection Need Not Preserve Topology in Both Directions

A bijection matches points perfectly: each point in its codomain comes from exactly one point in its domain. But this set-theoretic fact does not guarantee that the map respects the open sets in both directions. Continuity checks preimages of open sets in the codomain. A homeomorphism requires that check to succeed for the map and its inverse. A finite set with two different topologies gives a particularly direct example of the distinction.

For a map \(f:X\to Y\), continuity means that \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). A homeomorphism is a bijection \(f:X\to Y\) such that both \(f\) and \(f^{-1}\) are continuous. Thus, to disprove that a continuous bijection is a homeomorphism, it suffices to find one open set in the target of the inverse whose preimage is not open in its domain.

Definition: The discrete topology on a set \(X\) is the power set \(\mathcal{P}(X)\), so every subset of \(X\) is open. The indiscrete topology on \(X\) is \(\{\varnothing,X\}\), so the only open sets are the empty set and the whole set.

When the same underlying set \(X\) is given two different topologies, we distinguish the spaces by naming their topologies. Write \(X_{\mathrm{disc}}=(X,\mathcal{P}(X))\) and \(X_{\mathrm{ind}}=(X,\{\varnothing,X\})\). The identity function between them sends each element to itself, but its continuity depends on which topology is used in the domain and which in the codomain.

The Finite Identity Example

Worked Example: Three Points, Two Topologies

Let \(X=\{a,b,c\}\). Give the domain \(X_{\mathrm{disc}}\) the discrete topology, and give the codomain \(X_{\mathrm{ind}}\) the indiscrete topology. Consider the identity map \(i:X_{\mathrm{disc}}\to X_{\mathrm{ind}}\), defined by \(i(x)=x\) for every \(x\in X\).

The map is a bijection: each of \(a,b,c\) is sent to itself, so every codomain point has exactly one preimage. To check continuity, list every open set in the codomain. They are \(\varnothing\) and \(X\). Their preimages are \(i^{-1}(\varnothing)=\varnothing\) and \(i^{-1}(X)=X\). Both are open in the discrete domain. There are no other codomain open sets to test, so \(i\) is continuous.

Now consider the inverse \(i^{-1}:X_{\mathrm{ind}}\to X_{\mathrm{disc}}\). The set \(\{a\}\) is open in its target because that target is discrete. Its preimage under the inverse is \((i^{-1})^{-1}(\{a\})=\{a\}\). But \(\{a\}\) is not open in \(X_{\mathrm{ind}}\): it is neither \(\varnothing\) nor \(X=\{a,b,c\}\). Thus the inverse is not continuous, and \(i\) is not a homeomorphism.

Check: the computation uses the correct direction for each continuity test. The first test takes preimages of open sets in \(X_{\mathrm{ind}}\), and the second takes preimages of open sets in \(X_{\mathrm{disc}}\). In particular, \(\{a\}\) is an open test set for the inverse, not for the forward map.

This example does not depend on having three points. The essential feature is the existence of a nonempty proper subset: such a subset is open in the discrete topology but not in the indiscrete topology. The next result states the full conclusion, including the small cases where no such subset exists.

Theorem: Let \(X\) be a finite set, and let \(i:X_{\mathrm{disc}}\to X_{\mathrm{ind}}\) be the identity map. Then \(i\) is a continuous bijection. It is a homeomorphism if and only if \(X\) has at most one element.

Proof. For every \(X\), the identity is a bijection. The only open sets in \(X_{\mathrm{ind}}\) are \(\varnothing\) and \(X\). Their preimages under \(i\) are respectively \(\varnothing\) and \(X\), both open in \(X_{\mathrm{disc}}\). Hence \(i\) is continuous.

Suppose first that \(X\) has at least two elements. Choose distinct \(x,y\in X\). Then \(\{x\}\) is nonempty and is not all of \(X\), so it is a proper subset. It is open in \(X_{\mathrm{disc}}\). Under the inverse map \(i^{-1}:X_{\mathrm{ind}}\to X_{\mathrm{disc}}\), its preimage is \(\{x\}\), which is not open in \(X_{\mathrm{ind}}\), whose only open sets are \(\varnothing\) and \(X\). Therefore \(i^{-1}\) is not continuous, so \(i\) is not a homeomorphism.

Suppose instead that \(X\) has at most one element. If \(X=\varnothing\), its only subset is \(\varnothing=X\); if \(X\) has one element, its only subsets are \(\varnothing\) and \(X\). In either case, \(\mathcal{P}(X)=\{\varnothing,X\}\), so the discrete and indiscrete topologies are identical. The identity and its inverse are then continuous, since each open set has itself as preimage. Thus \(i\) is a homeomorphism. These cases prove both directions. \(\square\)

Why the Two Continuity Checks Differ

The asymmetry can be understood directly from the preimage definition. For the forward map, the codomain has very few open sets, so there are very few conditions to check. For the inverse, the codomain has every subset as an open set, including sets such as \(\{a\}\). Taking that set's preimage reveals the failure. The map does not fail because it loses or identifies points; it fails because the topology on its source cannot make that preimage open.

Proposition: If \(Y\) has the indiscrete topology, every map \(f:Z\to Y\) from any topological space \(Z\) is continuous. If \(X\) has the discrete topology, every map \(g:X\to Z\) to any topological space \(Z\) is continuous.

Proof. For the first claim, let \(V\) be open in \(Y\). Since \(Y\) is indiscrete, either \(V=\varnothing\) or \(V=Y\). Then \(f^{-1}(V)\) is respectively \(\varnothing\) or \(Z\), both open in \(Z\). This proves continuity of \(f\). For the second claim, let \(W\) be any open set in \(Z\). The preimage \(g^{-1}(W)\) is a subset of \(X\), and every subset of a discrete space is open. This proves continuity of \(g\). \(\square\)

These statements explain why both directions of the example behave as they do. The target of the forward identity is indiscrete, so its open-set tests impose little restriction. The source of the inverse is indiscrete while its target is discrete; the open sets in that target include all the subsets that expose the failure.

Worked Example: The Smallest Nontrivial Case

Let \(X=\{0,1\}\), with the discrete topology in the domain and the indiscrete topology in the codomain. The identity \(i:X_{\mathrm{disc}}\to X_{\mathrm{ind}}\) has the following continuity checks. The codomain open sets are only \(\varnothing\) and \(\{0,1\}\), and \(i^{-1}(\varnothing)=\varnothing\), \(i^{-1}(\{0,1\})=\{0,1\}\). Both preimages are open in the discrete domain, so \(i\) is continuous.

The inverse has codomain open sets \(\varnothing,\{0\},\{1\},\{0,1\}\). For the open set \(\{0\}\), its preimage is \((i^{-1})^{-1}(\{0\})=\{0\}\). This is not open in the indiscrete domain. The inverse is therefore not continuous. The same check with \(\{1\}\) also fails, since its preimage is \(\{1\}\), again neither empty nor the whole domain.

Check: the bijection sends \(0\) to \(0\) and \(1\) to \(1\), so its two fibers are \(\{0\}\) and \(\{1\}\). The failure is specifically that neither fiber is open in the indiscrete topology, even though each is open in the discrete topology.

Worked Example: The Degenerate Cases

For \(X=\{p\}\), the power set is \(\mathcal{P}(X)=\{\varnothing,\{p\}\}\). Since \(X=\{p\}\), this is also \(\{\varnothing,X\}\), the indiscrete topology. The identity map and its inverse are the same function. For each open set, the preimage is itself: \(\varnothing\) maps back to \(\varnothing\), and \(\{p\}\) maps back to \(\{p\}\). Both are open, so the identity is a homeomorphism.

For \(X=\varnothing\), the power set is \(\{\varnothing\}\), and \(\{\varnothing,X\}\) is also \(\{\varnothing\}\), since \(X=\varnothing\). The unique function from the empty set to itself is a bijection. Its only open-set preimage is \(\varnothing\), which is open, so the function and its inverse are continuous. It too is a homeomorphism.

Check: the failure in the two-point example required a nonempty proper subset. The singleton has no such subset, and the empty set has none either. These cases confirm why the theorem's condition is “at most one element,” rather than simply “finite.”

Common Pitfall: Testing Images Instead of Preimages

A common incorrect argument says that the identity \(i:X_{\mathrm{disc}}\to X_{\mathrm{ind}}\) is not continuous because the open domain set \(\{a\}\) has image \(i(\{a\})=\{a\}\), which is not open in the codomain. This tests whether images of open sets are open, not whether the map is continuous. Continuity requires preimages of open sets in the codomain to be open in the domain.

For the three-point example, the correct forward test uses \(V=\varnothing\) and \(V=X\), the only open subsets of the codomain. Their preimages are \(\varnothing\) and \(X\), both open in the domain. So the forward map is continuous despite sending the open set \(\{a\}\) to a non-open set. For the inverse, the correct test does use \(\{a\}\) as an open set in its codomain, and its preimage is not open in its domain. Keeping track of the map's direction and the topology on each side prevents the two tests from being confused.

Takeaway: A continuous bijection need not be a homeomorphism. In this finite example, the forward identity is continuous because the codomain has only two open sets to test, while the inverse fails on any nonempty proper subset. For the empty set and a one-point set, the discrete and indiscrete topologies coincide, so the identity is a homeomorphism.

Check Your Understanding

For each question, specify the topology on the domain and codomain before checking preimages.

On \(X=\{r,s,t\}\), list the open sets that must be checked to prove that the identity from the discrete topology to the indiscrete topology is continuous.
For the identity in question 1, choose an open set in the codomain of the inverse that proves the inverse is not continuous.
Why does the argument using a nonempty proper subset not apply when \(X\) has exactly one element?
Is every map from a discrete space continuous? State the preimage argument that justifies your answer.
For the identity from the discrete topology to the indiscrete topology, does a non-open image of a domain open set disprove continuity? Explain which condition continuity actually tests.