PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Canonical associativity, symmetry, and unit maps for smash products

Statement

For based CGWH spaces $X, Y, Z$ , the formulas

$(x \land y) \land z ⟼ x \land (y \land z), x \land y ⟼ y \land x,$

and $S^{0} \land X \to X$ , $1 \land x \mapsto x$ , induce natural based homeomorphisms

$(X \land Y) \land Z ≅ X \land (Y \land Z), X \land Y ≅ Y \land X, S^{0} \land X ≅ X .$

These maps satisfy the usual pentagon, triangle, and symmetry coherence identities.

Facts & Assumptions

[F1]

Every product and quotient is formed with the CGWH convention of Smash product of based spaces.

[F2]

The published product-of-quotients lemma identifies an iterated quotient product with the corresponding quotient of the product (Compact-test exponential law and products of quotient maps).

[F3]

The coordinate reassociations and permutations of k-products are mutually inverse homeomorphisms.

[F4]

Maps constant on quotient classes descend uniquely and continuously (For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map).

Proof

Given: Based CGWH spaces $X, Y, Z$ as in the statement.

1.1F1F2F4

Put both triple smashes over one quotient. Let $W = (X \times Y \times {*_{Z}}) \cup (X \times {*_{Y}} \times Z) \cup ({*_{X}} \times Y \times Z) .$

By [F1] and [F2], both $(X \land Y) \land Z$ and $X \land (Y \land Z)$ are canonically the kified quotient of $X \times_{k} Y \times_{k} Z$ by $W$ . The identity on triples therefore descends by [F4] to the displayed associator, and the same construction in reverse gives its continuous inverse. [F1, F2, F4]

1.2F1F3F4

Descend the twist and unit maps. The transposition $(x, y) \mapsto (y, x)$ takes $X \lor Y$ onto $Y \lor X$ , so [F3, F4] give the symmetry homeomorphism and its inverse. Since the nonbasepoint of $S^{0}$ is an open-and-closed singleton, collapsing $S^{0} \lor X$ leaves precisely the copy ${1} \times X$ , giving $S^{0} \land X ≅ X$ .

1.3F1

Check naturality and coherence. Naturality follows because every map above is induced by the corresponding coordinate map before quotienting. Each pentagon, triangle, or symmetry composite is induced by the same reassociation or permutation of the same product coordinates. Those maps agree before quotienting, hence their descended maps agree.

$□$

Depends on

Dependency tree · 2 levels

Sources

J. P. May, A Concise Course in Algebraic Topology