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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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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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Eckmann–Hilton: two unital operations satisfying interchange coincide and are commutative

Statement

Let a set XX carry two unital binary operations ∘\circ and ∗* with the same unit ee, and suppose

(a∘b)∗(c∘d)=(a∗c)∘(b∗d)(a\circ b)*(c\circ d)=(a*c)\circ(b*d)

for all a,b,c,da,b,c,d. Then the operations coincide and their common operation is commutative.

Facts & Assumptions

Given: The two operations, common unit, and interchange identity in the Statement.

[L1]

A unital associative operation is a monoid operation (Semigroup and monoid); only the unit laws and interchange are needed for the calculation below.

Proof

technique · direct
1.1

For a,b∈Xa,b\in X, interchange gives a∗b=(a∘e)∗(e∘b)=(a∗e)∘(e∗b)=a∘ba*b=(a\circ e)*(e\circ b)=(a*e)\circ(e*b)=a\circ b, so the operations coincide.

givenL1
2.1

A second use gives a∘b=(e∗a)∘(b∗e)=(e∘b)∗(a∘e)=b∗aa\circ b=(e*a)\circ(b*e)=(e\circ b)*(a\circ e)=b*a.

step 1.1L1
3.1

By step 1.1, b∗a=b∘ab*a=b\circ a, so step 2.1 says a∘b=b∘aa\circ b=b\circ a; the common operation is commutative.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 6 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