Deletion Data as a Test for Homeomorphism
A homeomorphism preserves the topological structure of a space, so it must preserve more than the number of points. One way to detect that structure is to delete a point and ask how many connected pieces remain. If two spaces have different answers to this question, no homeomorphism can relate them. The key is to track the points as well: a homeomorphism carries the deletion of \(x\) to the deletion of its image.
Throughout, a space is a pair \((X,\mathcal{T})\). A separation of \(X\) is a pair of disjoint, nonempty open sets whose union is \(X\). A space is connected if it has no separation. Equivalently, its only subsets that are both open and closed are \(\varnothing\) and \(X\): in a separation, each of the two open sets is also closed, and any nonempty proper set that is both open and closed gives a separation with its complement.
The component count is defined whether or not \(X\) is connected; the term cut point, as used here, is reserved for a point of a connected space. A homeomorphism is a bijection that is continuous and whose inverse is continuous.
What a Homeomorphism Preserves
Proof. First suppose \(X\) is connected. If \(Y\) had a separation \(U,V\), then \(f^{-1}(U)\) and \(f^{-1}(V)\) would be disjoint, nonempty open sets in \(X\) whose union is \(X\), because \(f\) is continuous and onto. This would separate \(X\), a contradiction. Thus \(Y\) is connected. Applying the same argument to the continuous inverse \(f^{-1}\) proves the converse.
Since \(f\) is a bijection, it maps \(X\setminus\{x\}\) bijectively onto \(Y\setminus\{f(x)\}\). Its restriction is continuous, and its inverse is the restriction of \(f^{-1}\), so this restriction is a homeomorphism. A homeomorphism maps connected sets to connected sets: if the image of a connected set had a separation in its subspace topology, the inverse images of the separating sets would separate the original set. It also maps components to components. Indeed, if \(C\) is a component and \(f(C)\) were contained in a larger connected set \(D\), then \(f^{-1}(D)\) would be connected and contain \(C\); maximality of \(C\) would force \(f^{-1}(D)=C\), and hence \(D=f(C)\). Applying this to the restriction after deletion gives a bijection of components and the equality of their counts. Finally, disconnectedness of the two deletion spaces is equivalent under a homeomorphism, proving the cut-point assertion when the original spaces are connected. \(\square\)
The component argument uses only connectedness and maximality. A component containing a given point can be described as the union of all connected subsets containing that point: such a union is connected, because any separation of the union would separate one of the connected subsets meeting the common point. This also explains why the maximal connected subsets being compared in the proof are well-defined.
Reading the Invariant in Finite Spaces
Finite spaces provide explicit examples where every open set, deletion, and component can be checked directly. In the examples below, connectedness is verified by checking for nontrivial subsets that are both open and closed. For the deletion spaces, that check determines whether the whole deletion is connected; when the deletion is discrete, its singleton subsets give its components.
Worked Example: A Three-Point Space with a Cut Point
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{b\},\{c\},\{b,c\},X\}\). This is a topology: unions of listed sets are listed, and intersections of listed sets are listed. Find \(c_X(x)\) for every point and identify the cut points.
Connectedness of \(X\). The complements of the open sets show that its closed sets are \(\varnothing,\{a\},\{a,c\},\{a,b\},X\). Comparing the open and closed lists, the only sets that are both open and closed are \(\varnothing\) and \(X\). Thus \(X\) is connected.
Delete \(a\). The subspace \(X\setminus\{a\}=\{b,c\}\) has topology \(\{\varnothing,\{b\},\{c\},\{b,c\}\}\), since each of those sets is the intersection of an open set of \(X\) with \(\{b,c\}\). This is the discrete topology. Its two singleton subsets are connected, and no connected subset can contain both points: the two singletons separate \(\{b,c\}\). Therefore its components are \(\{b\}\) and \(\{c\}\), and \(c_X(a)=2\).
Delete \(b\). On \(\{a,c\}\), the subspace topology is \(\{\varnothing,\{c\},\{a,c\}\}\). Its only sets that are both open and closed are the empty set and the whole subspace, so it is connected and has one component. Hence \(c_X(b)=1\).
Delete \(c\). On \(\{a,b\}\), the subspace topology is \(\{\varnothing,\{b\},\{a,b\}\}\). Again the only clopen sets are the empty set and the whole subspace, so this deletion is connected and \(c_X(c)=1\).
Thus \(a\) is the unique cut point of \(X\): the original space is connected, and deleting \(a\) leaves two components, whereas deleting either other point leaves one.
Worked Example: Different Deletion Data Rule Out a Homeomorphism
Let \(Y=\{p,q,r\}\) have the topology \(\mathcal{U}=\{\varnothing,\{r\},\{q,r\},Y\}\). Show that \(X\) from the preceding example and \(Y\) are not homeomorphic.
The listed sets form a topology: their unions and finite intersections remain in the list. The closed sets of \(Y\), obtained by taking complements, are \(\varnothing,\{p\},\{p,q\},Y\). The only sets both open and closed are \(\varnothing\) and \(Y\), so \(Y\) is connected.
Deleting \(p\) gives \(\{q,r\}\) with subspace topology \(\{\varnothing,\{r\},\{q,r\}\}\), which is connected because it has no nonempty proper clopen subset. Deleting \(q\) gives \(\{p,r\}\) with topology \(\{\varnothing,\{r\},\{p,r\}\}\), also connected. Deleting \(r\) gives \(\{p,q\}\) with topology \(\{\varnothing,\{q\},\{p,q\}\}\), again connected. Thus \(c_Y(p)=c_Y(q)=c_Y(r)=1\), and \(Y\) has no cut point.
In contrast, the first example gave \(c_X(a)=2\). If a homeomorphism \(f:X\to Y\) existed, the theorem would require \(c_X(a)=c_Y(f(a))\). The left side is \(2\), while every possible value on the right side is \(1\). Therefore no such homeomorphism exists.
Worked Example: Relabeling Preserves the Component Counts
Let \(X\) be the three-point space in the first example, and let \(Z=\{u,v,w\}\) have topology \(\{\varnothing,\{v\},\{w\},\{v,w\},Z\}\). Define \(f:X\to Z\) by \(f(a)=u\), \(f(b)=w\), and \(f(c)=v\). Check that this map is a homeomorphism and compare the deletion counts.
The map is a bijection. It sends the open sets of \(X\) to open sets of \(Z\): \(\{b\}\) maps to \(\{w\}\), \(\{c\}\) to \(\{v\}\), \(\{b,c\}\) to \(\{v,w\}\), and \(\varnothing,X\) to \(\varnothing,Z\). Thus \(f\) is continuous. Since these correspondences also show that every open set of \(Z\) has an open preimage in \(X\), \(f^{-1}\) is continuous too.
The deletion counts in \(X\) are \(c_X(a)=2\), \(c_X(b)=1\), and \(c_X(c)=1\). The theorem, or the same subspace-topology calculations with the labels changed, gives \(c_Z(u)=2\), \(c_Z(w)=1\), and \(c_Z(v)=1\). In particular, the unique cut point \(a\) corresponds to \(u=f(a)\), as required.
Check: the map sends the two components \(\{b\}\) and \(\{c\}\) of \(X\setminus\{a\}\) to \(\{w\}\) and \(\{v\}\), respectively, which are precisely the two components of \(Z\setminus\{u\}\).
Why a Continuous Bijection Is Not Enough
The theorem depends on having a homeomorphism, not merely a continuous bijection. Continuity carries connected sets to connected sets, but without a continuous inverse it need not preserve connectedness in both directions or preserve the components of deleted spaces.
Worked Example: A Continuous Bijection Changes a Deletion Count
Let \(A=\{1,2,3\}\) have the discrete topology, in which every subset is open, and let \(B=\{1,2,3\}\) have the indiscrete topology, in which the only open sets are \(\varnothing\) and \(B\). Consider the identity bijection \(g:A\to B\). Is it a homeomorphism? Compare the component counts after deleting \(1\).
The map \(g\) is continuous because the preimages of the only open sets \(\varnothing\) and \(B\) in \(B\) are \(\varnothing\) and \(A\), both open in \(A\). But its inverse is not continuous: the set \(\{2\}\) is open in \(A\), while its preimage under \(g^{-1}\), viewed as a map from \(B\) to \(A\), is \(\{2\}\), which is not open in \(B\). Thus \(g\) is not a homeomorphism.
In \(A\setminus\{1\}=\{2,3\}\), the subspace topology is discrete, so the two singleton sets are its components and \(c_A(1)=2\). In \(B\setminus\{1\}=\{2,3\}\), the subspace topology is indiscrete. This subspace is connected: its only open sets are the empty set and itself, so it has no separation. It has one component, giving \(c_B(1)=1\).
Correction to the tempting claim: “Every continuous bijection preserves deletion component counts” is false. Here the counts are \(2\) and \(1\). The theorem applies to homeomorphisms, and its inverse-continuity hypothesis is what lets the component correspondence work in both directions.
Using the Test Carefully
Deletion data give a necessary condition for homeomorphism, not a complete classification. If \(X\) and \(Y\) are homeomorphic, their points must match so that corresponding deletion spaces have the same component counts. Consequently, a mismatch proves that no homeomorphism exists. Matching counts alone does not establish a homeomorphism: other topological features may still differ.
When using the test, check the original spaces are connected before calling a point a cut point, and count components in the subspace topology after deletion. A deletion with one component is connected; it does not mean the deletion is the same space as the original. The count may be zero if deleting the only point leaves the empty space. Keeping these cases distinct prevents a component count from being mistaken for a pointwise description of the space.
Check Your Understanding
For each question, use the definition of connectedness and the component correspondence theorem where appropriate.
- Suppose \(f:X\to Y\) is a homeomorphism and \(X\) is connected. Explain why \(Y\) is connected.
- In the three-point space \(X\) with topology \(\{\varnothing,\{b\},\{c\},\{b,c\},X\}\), list the components of \(X\setminus\{a\}\).
- Can a homeomorphism send a cut point of a connected space to a point whose deletion leaves one component? Explain.
- Why does the identity map from a discrete three-point space to an indiscrete three-point space not contradict the theorem?
- If two connected spaces have different numbers of components after deleting every possible point, what conclusion can you draw about a homeomorphism between them?