When Are Two Spaces Topologically the Same?
A continuous map can preserve some structure while losing other information. In particular, a continuous bijection need not have a continuous inverse: the map may carry open sets to sets that are not open. A homeomorphism rules out this loss. It is a bijection that is continuous in both directions, so each space can be described in terms of the other without changing which sets are open.
The distinction between a continuous bijection and a homeomorphism is important whenever we want to say that two spaces have the same topology. The formula for a map may be simple, and it may be one-to-one and onto, but those facts alone do not ensure that it respects the open-set structure when reversed. We will give an exact definition, establish useful equivalent tests, and see how a homeomorphism lets us transfer topological properties between spaces.
The notation \(f^{-1}\) here denotes the inverse function, which exists because \(f\) is bijective. For a subset \(V\subseteq Y\), the notation \(f^{-1}(V)\) denotes the preimage of that subset. Continuity of the inverse means that \(f^{-1}(V)\) is open in \(X\) whenever \(V\) is open in \(Y\). Equivalently, the original map sends every open set in \(X\) to an open set in \(Y\).
Equivalent Tests for a Homeomorphism
Once a map is known to be a bijection, its inverse is determined. This means that continuity of the inverse can be checked without first writing a formula for that inverse: it is equivalent to the original map being open. There is also a closed-set version.
Proof. Suppose first that \(f\) is a homeomorphism. Let \(U\) be open in \(X\). Continuity of \(f^{-1}:Y\to X\) says that the preimage of \(U\) under \(f^{-1}\) is open in \(Y\). That preimage is exactly \(f[U]\), because $$ (f^{-1})^{-1}(U)=\{y\in Y:f^{-1}(y)\in U\}=f[U]. $$ Thus \(f\) is open. If \(F\) is closed in \(X\), the closed-set characterization of continuity applied to \(f^{-1}\) shows that \((f^{-1})^{-1}(F)=f[F]\) is closed in \(Y\). Thus \(f\) is closed as well.
Now suppose \(f\) is continuous and open. For every open \(U\subseteq X\), bijectivity gives \(f^{-1}(f[U])=U\), where \(f^{-1}:Y\to X\) is the inverse function. More directly, the preimage of \(U\) under that inverse function is \(f[U]\), which is open by the openness of \(f\). The open-set criterion for continuity therefore shows that \(f^{-1}\) is continuous. Hence \(f\) is a homeomorphism.
Finally, suppose \(f\) is continuous and closed. If \(F\subseteq X\) is closed, then the preimage of \(F\) under the inverse function \(f^{-1}:Y\to X\) is \(f[F]\), which is closed in \(Y\) by assumption. The closed-set criterion for continuity shows that \(f^{-1}\) is continuous. Thus \(f\) is a homeomorphism. This proves all three conditions equivalent. \(\square\)
In particular, openness alone is not enough: the theorem requires a bijection and continuity as well. The open-map test is useful when it is easier to understand the images of open sets than to work directly with the inverse. The closed-map test can be more convenient when closed sets have a simpler description.
Worked Example: An Exponential Homeomorphism
Give \(\mathbb{R}\) and \((0,\infty)\) their usual topologies, and define \(f:\mathbb{R}\to(0,\infty)\) by \(f(x)=e^x\). Since \(e^x>0\) for every real \(x\), the map takes values in its stated target. The exponential function is strictly increasing, so \(f\) is one-to-one. For every \(y>0\), \(x=\ln y\) satisfies \(e^x=y\), so \(f\) is onto.
Its inverse is \(g:(0,\infty)\to\mathbb{R}\), \(g(y)=\ln y\). The identities verifying the two inverse equations are $$ g(f(x))=\ln(e^x)=x \qquad (x\in\mathbb{R}), $$ and $$ f(g(y))=e^{\ln y}=y \qquad (y>0). $$ Both exponential and logarithm are continuous on their respective domains. Thus \(f\) is a homeomorphism. This proves that the whole real line and the positive half-line have the same topology, even though their elements are different sets.
Worked Example: A Continuous Bijection That Is Not a Homeomorphism
Let \(X=\{a,b\}\) have the discrete topology, in which every subset is open. Let \(Y=\{a,b\}\) have the indiscrete topology, whose only open sets are \(\varnothing\) and \(Y\). Define \(f:X\to Y\) by \(f(a)=a\) and \(f(b)=b\). This is a bijection.
The map \(f\) is continuous: the preimage of either open set \(\varnothing\) or \(Y\) in the target is respectively \(\varnothing\) or \(X\), both open in \(X\). However, its inverse is not continuous. The set \(\{a\}\) is open in \(X\), but its preimage under \(f^{-1}:Y\to X\) is \(\{a\}\), which is not open in the indiscrete space \(Y\). So \(f\) is a continuous bijection but not a homeomorphism. The bijection matches the points, but it does not preserve the open-set structure in both directions.
Homeomorphisms Preserve Topological Properties
If \(f:X\to Y\) is a homeomorphism, it is continuous, and its inverse is continuous too. The theorem on continuous images of compact sets therefore applies in both directions. The same reasoning applies to connectedness, using the theorem that continuous images of connected spaces are connected. In each case the two directions are essential: continuity of \(f\) transfers the property from \(X\) to \(Y\), and continuity of \(f^{-1}\) transfers it back.
Proof. Let \(f:X\to Y\) be a homeomorphism. If \(X\) is compact, then its continuous image \(f[X]=Y\) is compact by the theorem on continuous images of compact sets. Conversely, if \(Y\) is compact, then \(f^{-1}:Y\to X\) is continuous and has image \(X\), so \(X\) is compact. This proves the compactness equivalence.
If \(X\) is connected, its continuous image \(f[X]=Y\) is connected by the theorem on continuous images of connected spaces. If \(Y\) is connected, its continuous image under \(f^{-1}\) is \(X\), which is therefore connected. This proves the connectedness equivalence. \(\square\)
A property preserved whenever spaces are homeomorphic is called a topological invariant. Compactness and connectedness are two such invariants. This gives a way to prove that spaces are not homeomorphic: it is enough to find a topological invariant that they do not share. Such a test can rule out a homeomorphism without examining every possible bijection.
Worked Example: Compactness Rules Out a Homeomorphism
The interval \([0,1]\) is compact by the Heine–Borel Theorem. The open interval \((0,1)\) is not compact. To verify the latter claim, consider the open sets in \((0,1)\) $$ U_n=(1/n,1), \qquad n\geq 2. $$ Each \(U_n\) is open in \((0,1)\), and their union is \((0,1)\): given \(x\in(0,1)\), choose an integer \(n\geq 2\) with \(n>1/x\), so \(1/n<x<1\) and \(x\in U_n\).
Any finite subfamily has a largest index \(N\), and its union is \(U_N=(1/N,1)\), because the sets increase with \(n\). This finite union misses, for example, \(1/(2N)\), which lies in \((0,1)\) and is less than \(1/N\). Thus the cover has no finite subcover, and \((0,1)\) is not compact. Since compactness is preserved by homeomorphisms, \((0,1)\) and \([0,1]\) cannot be homeomorphic.
Composing Homeomorphisms
Homeomorphisms also fit naturally with composition. If \(f:X\to Y\) and \(g:Y\to Z\) are homeomorphisms, their composite is a continuous bijection by the composition theorem and the elementary fact that a composition of bijections is a bijection. Its inverse is \(f^{-1}\circ g^{-1}\), which is continuous by the same composition theorem. Thus the composite is again a homeomorphism.
Proof. The identity map is a bijection and is its own continuous inverse, so it is a homeomorphism. If \(f:X\to Y\) is a homeomorphism, then \(f^{-1}:Y\to X\) is a continuous bijection whose inverse is \(f\), also continuous; hence \(f^{-1}\) is a homeomorphism.
Now let \(f:X\to Y\) and \(g:Y\to Z\) be homeomorphisms. Their composite \(g\circ f\) is a continuous bijection. Its inverse is \(f^{-1}\circ g^{-1}\): composing in either order gives the appropriate identity map, since \(f^{-1}\circ f=\operatorname{id}_X\) and \(g^{-1}\circ g=\operatorname{id}_Y\), with the corresponding identities in the other order. Both inverse maps are continuous, so their composite is continuous. Therefore \(g\circ f\) is a homeomorphism. Reflexivity, symmetry, and transitivity of the relation “is homeomorphic to” now follow respectively from the identity, inverse, and composition statements. \(\square\)
The definition and the equivalent tests separate three ideas that should not be conflated: a map may be bijective as a function, continuous in one direction, and still fail to be a homeomorphism. To prove homeomorphism directly, exhibit a continuous inverse. To use the open-map or closed-map test, first verify bijectivity and continuity, then check the relevant preservation property. When proving two spaces are not homeomorphic, look for a topological invariant that distinguishes them.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What three conditions must a map satisfy to be a homeomorphism?
- For a bijection \(f:X\to Y\), how can openness of \(f\) be used to check continuity of its inverse?
- Why is the identity map from the discrete two-point space to the indiscrete two-point space continuous but not a homeomorphism?
- Which property rules out a homeomorphism between \((0,1)\) and \([0,1]\), and how is that property verified for \((0,1)\)?
- If \(f:X\to Y\) and \(g:Y\to Z\) are homeomorphisms, what is the inverse of \(g\circ f\)?