The Gold Standard of Topology | A Continuous Bijection with a Discontinuous Inverse

A bijection is one-to-one and onto. Each target point has exactly one source point. This makes a set-theoretic inverse possible, but says nothing by itself about how nearby points behave under that inverse.

A homeomorphism requires a bijection whose forward map and inverse are both continuous. It preserves the open-set structure in both directions. That is a stronger claim than merely providing a unique label for each point.

Before using the definition, name the spaces and their topologies. The same underlying set can behave differently with a different choice of open sets. Here all interval and circle topologies are inherited from ordinary Euclidean spaces.

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Let X = [0,1), including zero but excluding one, with its usual subspace topology from the real line. Let S¹ be the unit circle in the plane, with its usual subspace topology from the plane.

Define f(t) = (cos(2πt), sin(2πt)). The number t records a fraction of one counterclockwise turn starting at the point p = (1,0). Thus f(0) = p, f(0.25) = (0,1), and f(0.5) = (−1,0).

The trigonometric functions are continuous, so f is continuous on X. Restricting a continuous real-variable formula to a subspace does not introduce a discontinuity. The trouble, if there is one, must occur in the reverse interpretation.

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Every point of the circle has an angle represented by exactly one fraction of a turn in [0,1). This gives surjectivity. For injectivity, two parameters with the same sine and cosine differ by an integer number of turns.

Two members of [0,1) differ by less than one in absolute value. Their difference can therefore be an integer only if it is zero. The parameters are equal, establishing injectivity.

Excluding the endpoint 1 matters. If we used [0,1], then f(0) = f(1), so the map would no longer be one-to-one. Removing that duplicate produces a bijection. It does not automatically produce a homeomorphism.

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Let g be the inverse that assigns each circle point its unique parameter in [0,1). Approach p from the side corresponding to parameters just below one. Those circle points get arbitrarily close to p, while their assigned parameters stay close to one.

The circle distances are chord lengths, calculated as 2 sin(π(1−t)). They shrink towards zero. The parameter distances do not shrink towards zero, since g(p) = 0. A unique label can jump even while the point being labelled moves smoothly.

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For n ≥ 2, take tn = 1 − 1/n. Continuity of sine and cosine gives f(tn) → p as n grows. Applying g returns tn, which does not converge to g(p) = 0. For example, every tn is at least 1/2 away from zero.

If g were continuous at p, every sequence converging to p would have images converging to g(p). This sequence violates that necessary condition. Therefore g is discontinuous at p, and f is not a homeomorphism.

It is fine to say tn tends to 1 as a sequence in the real line. But 1 is not a point of X. Do not describe it as convergence to a missing point within X, and do not mistake a limit near one for convergence to zero.

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Take U = [0,1/4). It is open in X because U = X ∩ (−1,1/4), the intersection of X with an open real interval. Being open in a subspace does not require being open in the whole real line.

For g to be continuous, the inverse image g⁻¹(U) must be open in the circle. Because g is the inverse bijection, this set is f(U): the quarter-circle arc starting at p, including p and excluding (0,1).

That arc is not open in the circle. Every sufficiently small circle neighbourhood of p also contains points just below p, with parameters near one, which are outside the arc. Thus g fails the open-set definition of continuity.

This also pinpoints a common logical slip: continuity of f controls preimages of open sets under f. It does not say that f sends every open set to an open set.

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A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The circle is Hausdorff because it is a metric space. Our domain [0,1), however, is not compact.

One direct check uses the relative open sets Un = [0,1−1/n), for integers n ≥ 2. Their union covers X. Any finite selection has a largest upper endpoint below one, leaving some points of X uncovered. There is no finite subcover.

The theorem therefore cannot be applied to this domain. Its failure is consistent with the theorem, not a counterexample to it. Allen Hatcher’s point-set topology notes, especially the continuity and Hausdorff-space sections, provide the formal definitions and compact-domain result.

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In the third construction, identify 0 and 1 as one equivalence class. The other points remain separate. The quotient topology records that the two ends now meet. The induced map from this quotient to the circle is continuous and bijective.

The quotient is compact because it is a continuous image of the compact interval [0,1]. The circle is Hausdorff, so the theorem now applies. This repairs the topology by changing the domain space, rather than by pretending the old inverse had been continuous.

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A direction can move smoothly through the chosen zero angle while its reported angle jumps from nearly one full turn to zero. The discontinuity can belong to the chosen numerical representation rather than to the direction itself.

Moving the seam to another direction moves the problem. On an arc that stays away from a seam, one can choose a continuous local angle coordinate. Multiple overlapping coordinate descriptions can therefore work where a single global interval-valued inverse fails.

Be precise about the claim. The circle certainly has continuous real-valued functions, including constant functions and its Cartesian coordinates. What fails here is the unique whole-circle angle inverse taking values in this particular half-open interval with its ordinary topology.

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Doesn’t a continuous bijection preserve nearby points?

The forward map respects continuity. A homeomorphism must respect it in both directions. Our sequence puts the two requirements to separate tests.

Can numerical testing prove continuity?

A finite table can suggest the mechanism but cannot check every neighbourhood. The sequence and open-set arguments establish the failure beyond the displayed values.

Does this show the interval and circle are never homeomorphic?

For X = [0,1), yes, another quick obstruction is compactness: the circle is compact and X is not. The failed map already explains why this particular attractive pairing cannot establish equivalence. A proposed map failing alone would not generally exclude every other map between two spaces.

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Use h(t) = (−cos(2πt), −sin(2πt)) on the same domain. It starts at (−1,0). Predict where its inverse fails, then use tn = 1−1/n to prove it. The failure moves to (−1,0); the same parameter sequence returns values far from zero.

Next restrict f to [0,0.75]. Its image is a closed three-quarter-circle arc. This restriction is a continuous bijection from a compact interval to a Hausdorff subspace, so its inverse is continuous. The problematic approach through parameters near one is no longer present in that image.

Finally, explain why adding 1 back to the original domain solves one issue but creates another: compactness returns, while injectivity fails. A theorem needs all its hypotheses at the same time.

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For the central language, return to Topology and How Mathematics Works: Topology. Read their continuity, compactness and quotient-space sections alongside the example, rather than treating a picture of wrapping as a proof.

For formal foundations and further exercises, use Hatcher’s introductory point-set topology notes. The forward and inverse tests above can be checked independently using their definitions; the decimal table is only an illustration.

For the wider learning purpose, continue through How X Works and The Gold Standard of Education. The habit that transfers is to check every direction of an equivalence and every hypothesis of the theorem used to justify it.

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