False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: every continuous bijection of topological spaces is a homeomorphism

Statement

False claim: if $X$ and $Y$ are topological spaces and $f : X \to Y$ is a continuous bijection (Continuity of a map of topological spaces at a point and globally, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), then $f$ is a homeomorphism.

Continuity of $f^{-1}$ is an independent demand, and A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces says exactly what it amounts to: a continuous bijection is a homeomorphism precisely when it is an open map, equivalently a closed map. The claim above asserts that this is automatic, and it is not. The witness below is the smallest possible one — a two-point set carrying two different topologies — and it uses nothing beyond The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies.

Facts & Assumptions

Given: A two-point set $S = \{a,b\}$ with $a \ne b$ , carrying on the one hand the discrete topology $\mathcal{P}(S) = \{\varnothing, \{a\}, \{b\}, S\}$ and on the other the Sierpinski topology $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$ ; and the identity function $\mathrm{id} : (S, \mathcal{P}(S)) \to (S, \mathcal{T}_{\mathrm{Sier}})$ .

[A1]

The discrete topology on $S$ is $\mathcal{P}(S)$ , in which every subset is open; the Sierpinski topology on $\{a,b\}$ is $\{\varnothing, \{b\}, S\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A2]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ , clause (b)).

[L1]

A homeomorphism is a continuous bijection with continuous inverse; an open map carries open sets to open sets (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L2]

A continuous bijection is a homeomorphism if and only if it is an open map (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, claim 1).

Refutation

technique · direct

1.1

$\mathrm{id}$ is a bijection of $S$ onto $S$ , being the identity function of the set $S$ .

given

1.2

$\mathrm{id}$ is continuous: for every $V \in \mathcal{T}_{\mathrm{Sier}}$ the preimage $\mathrm{id}^{-1}[V] = V$ is a subset of $S$ , hence open in the discrete topology.

givenA1A2

1.3

$\{a\}$ is open in the discrete topology on $S$ , and $\{a\} \notin \mathcal{T}_{\mathrm{Sier}}$ , the three members of $\mathcal{T}_{\mathrm{Sier}}$ being $\varnothing$ , $\{b\}$ and $S$ , none of which is $\{a\}$ because $a \ne b$ .

givenA1

2.1

$\mathrm{id}$ is not an open map: by step 1.3 the image $\mathrm{id}[\{a\}] = \{a\}$ of an open set is not open in the target.

step 1.3L1

3.1

By steps 1.1, 1.2 and 2.1, $\mathrm{id}$ is a continuous bijection that is not open, hence not a homeomorphism by [L2]; equivalently, its inverse — again the identity function of $S$ , now read from $(S,\mathcal{T}_{\mathrm{Sier}})$ to $(S,\mathcal{P}(S))$ — is not continuous, because the preimage of the open set $\{a\}$ is $\{a\}$ , which is not open in $\mathcal{T}_{\mathrm{Sier}}$ . So the claim is false.

step 1.1step 1.2step 2.1step 1.3A2L1L2∎

Remarks

The general shape of every witness. Whenever a set carries two distinct comparable topologies $\mathcal{T}_1 \subsetneq \mathcal{T}_2$ , the identity from $(X,\mathcal{T}_2)$ to $(X,\mathcal{T}_1)$ is a continuous bijection and is not open, since a member of $\mathcal{T}_2 \setminus \mathcal{T}_1$ is its own image. The two-point witness above is that construction with the smallest possible $X$ ; the same construction on $\mathbb{R}$ , with the discrete and the usual topologies, is on the companion page (The identity from the discrete topology on $\mathbb{R}$ to the usual topology is a continuous bijection that is not a homeomorphism ↗).
What rescues the claim. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, because it is then a closed map. Neither compactness nor the Hausdorff property is available at this point in the reading order. They are defined later in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right ↗ and Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not ↗, where A continuous bijection from a compact metric space onto a metric space carries open sets to open sets, so its inverse is continuous proves this repair.
Why "bijective and continuous" is the wrong notion of sameness. Sameness of topological spaces is homeomorphy, and by A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces it is an equivalence relation; "there is a continuous bijection $X \to Y$ " is not even symmetric, as the witness above shows, since no continuous bijection runs from Sierpinski space to the two-point discrete space.

Depends on

Used by

The identity from the discrete topology on ℝ to the usual topology is a continuous bijection that is not a homeomorphism Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 45 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Homeomorphism (Wikipedia) (standard reference, not scraped)
Sierpinski space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §18 (standard reference, not scraped)