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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces

Statement

Let $X$ and $Y$ be topological spaces.

Let $f : X \to Y$ $f : X \to Y$ be 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). The following are equivalent:
- (a) $f$ is a homeomorphism;
- (b) $f$ is an open map;
- (c) $f$ is a closed map.
Homeomorphy is an equivalence relation: $X \cong X$ ; if $X \cong Y$ then $Y \cong X$ ; and if $X \cong Y$ and $Y \cong Z$ then $X \cong Z$ .

Continuity is a genuine hypothesis in claim 1: a bijection that is open and closed but not continuous exists as soon as two comparable topologies differ, for instance the identity from the coarser to the finer of two distinct topologies on one set.

Facts & Assumptions

Given: Topological spaces $X$ , $Y$ , $Z$ and a continuous bijection $f : X \to Y$ with two-sided inverse $g := f^{-1} : Y \to X$ . For a bijection $f$ and any $A \subseteq X$ one has $f[A] = g^{-1}[A]$ , and $f[X \setminus A] = Y \setminus f[A]$ .

[A1]

$f$ is a homeomorphism when it is a continuous bijection and $f^{-1}$ is continuous; $f$ is an open map when images of open sets are open, and a closed map when images of closed sets are closed (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[A2]

A bijection has a unique two-sided inverse, which is itself a bijection, and the inverse of $f^{-1}$ is $f$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, Continuity of a map of topological spaces at a point and globally).

[L1]

A map is continuous exactly when preimages of open sets are open, and exactly when preimages of closed sets are closed (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 $f(\overline{A}) \subseteq \overline{f(A)}$ , clauses (b) and (c)).

[L2]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 1).

Proof

technique · direct

1.1

Since $f$ is a bijection with inverse $g$ , for every $A \subseteq X$ the image $f[A]$ coincides with the preimage $g^{-1}[A]$ : $y \in g^{-1}[A]$ means $g(y) \in A$ , and applying $f$ gives $y \in f[A]$ , while the converse substitution is the same computation read backwards.

givenA2

1.2

Since $f$ is a bijection, $f[X \setminus A] = Y \setminus f[A]$ for every $A \subseteq X$ : surjectivity gives $\supseteq$ and injectivity gives $\subseteq$ .

given

1.3

The identity map of a space is a continuous bijection whose inverse is itself, hence a homeomorphism.

A1L1

1.4

If $h : X \to Y$ is a homeomorphism then so is $h^{-1} : Y \to X$ : it is a bijection, it is continuous by hypothesis, and its own inverse is $h$ , which is continuous.

A1A2

2.1

(a) is equivalent to (b): by step 1.1, $g$ is continuous exactly when $g^{-1}[U] = f[U]$ is open for every open $U \subseteq X$ , that is exactly when $f$ is an open map.

step 1.1A1L1

2.2

(b) is equivalent to (c): by step 1.2, $f$ carries the complement of $A$ to the complement of $f[A]$ , so images of open sets are open exactly when images of closed sets are closed, the two families being exchanged by complementation.

step 1.2A1L1

2.3

If $h : X \to Y$ and $k : Y \to Z$ are homeomorphisms then $k \circ h$ is a homeomorphism: it is a bijection with inverse $h^{-1} \circ k^{-1}$ , and both $k \circ h$ and $h^{-1} \circ k^{-1}$ are continuous as composites of continuous maps.

step 1.4A1A2L2

3.1

Steps 2.1 and 2.2 prove claim 1, and steps 1.3, 1.4 and 2.3 give reflexivity, symmetry and transitivity of $\cong$ , which is claim 2.

step 1.3step 1.4step 2.1step 2.2step 2.3∎

Remarks

The lemma is how homeomorphy is verified in practice. Producing a continuous inverse directly usually means writing a formula and checking continuity a second time; checking instead that the map carries open sets to open sets, or closed sets to closed sets, uses only the map itself.
A continuous bijection that is not a homeomorphism. Take any set carrying two distinct comparable topologies and let $f$ be the identity from the finer to the coarser: it is a continuous bijection, and it is not open, because an open set of the finer topology that is not open in the coarser one is its own image. Both an explicit two-point instance and an instance on $\mathbb{R}$ appear on this page and on the companion page.
What claim 2 licenses. Because $\cong$ is an equivalence relation, "a topological property" is well defined as a property constant on $\cong$ -classes (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), and statements of the form " $X$ is not homeomorphic to $Y$ " can be proved by exhibiting one topological property on which they differ.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 9 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)
Open and closed maps (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §18 (standard reference, not scraped)