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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16
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A surjective local homeomorphism need not be a covering map

Statement refuted

Let E=(0,2)⊔(1,2) and map both summands by inclusion to B=(0,2). This map is a surjective local homeomorphism but is not a covering map.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Every covering map is a surjective local homeomorphism, and each of its fibres is discrete in the subspace topology. (Covering maps are surjective local homeomorphisms with discrete fibres).

[F2]

For a covering p:E→B, the cardinality of p−1(b) is locally constant as a function of b∈B. If B is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).

[F3]

The underlying set. Let I be a set and let Xi be a set for each i∈I. The disjoint union is ⨆i∈IXi  :=  ⋃i∈I(Xi×{i}), whose elements are the pairs (x,i) with i∈I and x∈Xi. For j∈I the j-th canonical injection is κj:Xj→⨆i∈IXi,κj(x):=(x,j). The construction is what makes the word "disjoint" honest. Each κj is injective (def-injection-surjection-bijection), since (x,j)=(x′,j) forces x=x′; the images κj[Xj]=Xj×{j} are pairwise disjoint, since the second coordinate determines j; and their union is the whole set. So no assumption that the Xi are disjoint as sets is needed, and none is made: the tag i separates the copies even when Xi=Xi′ for i≠i′. (The disjoint union (coproduct) ⨆iXi with the final topology of the canonical injections: a set is open exactly when each of its traces is).

[F4]

Throughout, R is the complete ordered field (def-complete-ordered-field, def-ordered-field) with its order (def-real-order). A subset I⊆R is order-convex when x,y∈I and x≤z≤y imply z∈I, and the intervals of R are the nine listed forms, among them [a,b]={x:a≤x≤b} and (a,b)={x:a<x<b}. (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct
1.1givenF3F4

Let the domain be the disjoint union of (0,2) and (1,2) and map both components by inclusion onto the base (0,2).

2.1step 1.1F1F3

The first component makes the map surjective and each inclusion is a local homeomorphism.

3.1step 2.1F1F2F3

The fibre has one point at 1 and two immediately to its right, so local constancy of sheet number rules out a covering; equivalently, the second component supplies only a one-sided partial sheet above every neighbourhood of 1.

4.1step 3.1∎

The preceding construction and implications establish the assertion.

Depends on

Used by

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Dependency tree · two levels

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Sources