When Two Metric Spaces Have the Same Topology
An isometry preserves every distance, so it preserves the metric geometry of its domain exactly. A homeomorphism asks for less: it preserves which sets are open, and thus the continuity structure of the spaces, without requiring distances to remain unchanged. For example, a space can be stretched or compressed in a way that changes distances but does not alter its open sets in the essential sense.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A bijection is important here because it allows every point of either space to correspond to exactly one point of the other. But bijectivity and continuity in just one direction do not suffice: the inverse map must also be continuous.
A homeomorphism identifies the points of two spaces in a way that is reversible and continuous in both directions. An isometry is a homeomorphism, because it is bijective when it is a surjective isometry, and its inverse also preserves distances. By the Lipschitz Maps Theorem, distance-preserving maps are continuous; the same argument applies to the inverse. The converse does not hold: homeomorphisms can change distances substantially.
Open and Closed Sets Under a Homeomorphism
Continuity can be expressed in terms of inverse images of open sets: a map is continuous if the inverse image of every open set is open. For a bijection, this gives a useful way to understand why continuity of the inverse is a separate condition. The inverse image under \(f^{-1}\) of a set \(U\subseteq X\) is exactly \(f(U)\).
Proof. Suppose first that \(f\) is a homeomorphism. Fix an open set \(U\subseteq X\). Since \(f^{-1}:Y\to X\) is continuous, the inverse image of \(U\) under \(f^{-1}\) is open in \(Y\). That inverse image is $$ (f^{-1})^{-1}(U)=\{y\in Y:f^{-1}(y)\in U\}=f(U), $$ so \(f(U)\) is open.
Conversely, suppose \(f\) is a continuous bijection and maps open sets in \(X\) to open sets in \(Y\). For every open \(U\subseteq X\), the same identity gives \((f^{-1})^{-1}(U)=f(U)\), which is open by assumption. Thus the inverse image under \(f^{-1}\) of every open subset of \(X\) is open, so \(f^{-1}\) is continuous. Therefore \(f\) is a homeomorphism. \(\square\)
There is a parallel test using closed sets. A homeomorphism maps closed sets to closed sets, and a continuous bijection that maps closed sets to closed sets has a continuous inverse. Indeed, for a bijection, \(f(X\setminus F)=Y\setminus f(F)\); thus if \(F\) is closed in \(X\), then \(X\setminus F\) is open, and the open-set characterization can be applied to the complements.
Worked Example: A Stretching That Is a Homeomorphism
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=3x-4\), with the usual metric on both spaces. This map is continuous, and it is bijective: for any \(y\in\mathbb{R}\), the equation \(y=3x-4\) has the unique solution \(x=(y+4)/3\). Thus $$ f^{-1}(y)=\frac{y+4}{3}. $$
The inverse is continuous because it is an affine function. Hence \(f\) is a homeomorphism. It is not an isometry: for example, \(d_{\mathbb{R}}(f(1),f(0))=|{-1}-({-4})|=3\), while \(d_{\mathbb{R}}(1,0)=1\). The map changes distances but still gives a reversible continuous correspondence between the spaces.
The open-set test is often easier than checking inverse continuity directly. It also clarifies the distinction between a continuous bijection and a homeomorphism: continuity of \(f\) controls inverse images of open sets, whereas continuity of \(f^{-1}\) controls images of open sets under \(f\).
Sequences Give a Two-Way Test
The Sequential Characterization of Continuity from “Continuity Between Metric Spaces” lets us restate the definition in terms of convergent sequences. A homeomorphism must preserve convergence in both directions. This is particularly useful when an inverse formula is awkward or when a proposed inverse can be shown discontinuous by constructing a sequence.
Proof. Suppose \(f\) is a homeomorphism. If \(x_n\to x\), continuity of \(f\) gives \(f(x_n)\to f(x)\). If \(f(x_n)\to f(x)\), continuity of \(f^{-1}\) gives $$ x_n=f^{-1}(f(x_n))\longrightarrow f^{-1}(f(x))=x. $$ So the equivalence holds.
Conversely, suppose the stated equivalence holds for every sequence and every \(x\in X\). Its forward implication says that \(f\) is sequentially continuous, and its reverse implication says that \(f^{-1}\) is sequentially continuous: any sequence \((y_n)\) converging to \(y\) in \(Y\) can be written as \(y_n=f(x_n)\), \(y=f(x)\), because \(f\) is bijective. The reverse implication then gives \(x_n\to x\), which is precisely \(f^{-1}(y_n)\to f^{-1}(y)\). By the Sequential Characterization of Continuity, both \(f\) and \(f^{-1}\) are continuous. Thus \(f\) is a homeomorphism. \(\square\)
Worked Example: A Homeomorphism from an Interval to the Positive Reals
Let \(f:(0,1)\to(0,\infty)\) be defined by \(f(x)=x/(1-x)\). The denominator is positive on \((0,1)\), so \(f(x)>0\). To find an inverse, solve \(y=x/(1-x)\): $$ y(1-x)=x,\qquad y=x(1+y),\qquad x=\frac{y}{1+y}. $$ For every \(y>0\), \(y/(1+y)\) lies in \((0,1)\), since it is positive and \(y<1+y\). Substitution verifies both inverse identities: $$ f\left(\frac{y}{1+y}\right) =\frac{y/(1+y)}{1-y/(1+y)} =\frac{y/(1+y)}{1/(1+y)} =y, $$ and $$ f^{-1}(f(x)) =\frac{x/(1-x)}{1+x/(1-x)} =\frac{x/(1-x)}{1/(1-x)} =x. $$
The function \(f\) is continuous on \((0,1)\), and its inverse \(f^{-1}(y)=y/(1+y)\) is continuous on \((0,\infty)\), since their denominators do not vanish on their domains. Therefore \(f\) is a homeomorphism. The interval and the positive real axis have different shapes as subsets of the real line, but the map gives a continuous correspondence with a continuous inverse.
A Continuous Bijection Can Have a Discontinuous Inverse
The requirement that the inverse be continuous cannot be omitted. A continuous bijection can send points that are far apart in the domain to points that approach one another in the target. In that situation, the inverse fails to preserve the corresponding sequence convergence.
Worked Example: A Continuous Bijection That Is Not a Homeomorphism
Let \(X=[0,2\pi)\) with the usual metric, and let \(Y\) be the unit circle in \(\mathbb{R}^2\), equipped with the restricted Euclidean metric. Define $$ f(t)=(\cos t,\sin t). $$ The coordinate functions are continuous, so \(f\) is continuous. It maps into \(Y\), since \(\cos^2t+\sin^2t=1\). It is bijective: each point on the unit circle has exactly one angle in \([0,2\pi)\).
For integers \(n\geq1\), put \(t_n=2\pi-1/n\). Then \(t_n\in X\), and $$ f(t_n)=\left(\cos(2\pi-1/n),\sin(2\pi-1/n)\right)\longrightarrow(1,0)=f(0). $$ But \(t_n\) does not converge to \(0\) in \(X\); in fact, \(t_n\to2\pi\) as real numbers, and \(|t_n-0|=2\pi-1/n\) does not approach zero. If \(f^{-1}\) were continuous, the convergence \(f(t_n)\to f(0)\) would imply \(t_n\to0\), a contradiction. Thus \(f^{-1}\) is not continuous, and \(f\) is not a homeomorphism.
The circle closes up at \((1,0)\), but the interval \([0,2\pi)\) has only one endpoint there: angles near \(2\pi\) are close on the circle to the image of \(0\), while they are not close to \(0\) in the domain. This is precisely the mismatch that inverse continuity detects.
What a Homeomorphism Preserves
A homeomorphism preserves open sets in both directions. If \(U\) is open in \(X\), then \(f(U)\) is open in \(Y\). Conversely, if \(V\) is open in \(Y\), then \(f^{-1}(V)\) is open in \(X\) by continuity. Since \(f\) is bijective, this says that the open-set structure of one space determines the open-set structure of the other. The same conclusion holds for closed sets by taking complements.
These facts do not say that distances, lengths, or angles are preserved. For example, the affine homeomorphism above multiplies every distance by \(3\), while more general homeomorphisms can change different distances by different amounts. Nor does a homeomorphism necessarily extend to boundary points missing from either space, as the circle example illustrates.
Show that each target point has exactly one preimage.
Use the metric definition, a continuity theorem, or a known continuous formula.
Find an inverse formula and prove it continuous, or use the open-set or sequential characterization.
The principal pitfall is to stop after verifying that a map is continuous and bijective. Those properties alone do not guarantee that nearby target points correspond to nearby domain points. To establish a homeomorphism, one must also verify this reverse control, either directly through the inverse or by an equivalent criterion.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What three properties must a map have to be a homeomorphism?
- Why does a homeomorphism map every open subset of its domain to an open subset of its target?
- State the sequential test for a bijection to be a homeomorphism.
- For \(f(x)=x/(1-x)\) on \((0,1)\), what is the inverse formula, and what is its range?
- In the circle example, which sequence shows that the inverse is not continuous?