False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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FALSE: the two-set van Kampen conclusion needs no path-connectedness hypothesis on the overlap

Statement

False claim: let $X = U \cup V$ with $U, V$ open and path-connected, and let $x_{0} \in U \cap V$ . Even when $U \cap V$ is not path-connected, if $C$ is its path component containing $x_{0}$ , then $π_{1} (X, x_{0})$ is the pushout of

$π_{1} (U, x_{0}) ⟵ π_{1} (C, x_{0}) ⟶ π_{1} (V, x_{0}) .$

Facts & Assumptions

Given: The quotient circle $Q = R / Z$ , its quotient map $p$ , the open arcs $U = p ((- 1 / 8, 5 / 8))$ and $V = p ((3 / 8, 9 / 8))$ , and the basepoint $[0]$ .

[F1]

The quotient map is open, and every interval shorter than one maps homeomorphically to its image in $Q$ (The quotient map is open, and every interval shorter than one embeds in $R / Z$ ).

[F2]

Every nonempty convex subset of a Euclidean space is simply connected (Every nonempty convex subset of $R^{n}$ is simply connected).

[F3]

The degree map is an isomorphism $π_{1} (Q, [0]) ≅ (Z, +)$ ( $Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism).

[F4]

A pointed homeomorphism induces a fundamental-group isomorphism (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

Refutation

technique · direct

1.1

Both defining intervals have length $3 / 4 < 1$ , so [F1] makes $U$ and $V$ open arcs. Their displayed lifts show $U \cup V = Q$ , while $U \cap V = p ((- 1 / 8, 1 / 8)) ⊔ p ((3 / 8, 5 / 8)),$ a disjoint union of two nonempty open arcs. The basepoint lies in the first component $C = p ((- 1 / 8, 1 / 8))$ .

F1algebra

2.1

The three arcs $U, V, C$ are homeomorphic to open intervals, which are nonempty and convex. Hence [F2] and [F4] make all three fundamental groups trivial.

step 1.1F2F4

3.1

The pushout of the two homomorphisms from the trivial group to the two trivial factor groups is itself the trivial group: for every target group there is exactly one compatible pair of homomorphisms and exactly one homomorphism from the trivial group.

step 2.1

4.1

The actual group $π_{1} (Q, [0])$ is isomorphic to $Z$ by [F3], so it is nontrivial and cannot be the pushout computed in step 3.1. Thus the false claim fails for this cover, and path-connectedness of the full overlap cannot be omitted from the two-set theorem.

step 3.1F3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, discussion preceding Theorem 1.20 (standard reference, not scraped)