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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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The comparison functor for the free-group adjunction

Example

For the free-group adjunction, the comparison sends a group G to the algebra whose carrier is UG and whose structure map evaluates a reduced word in elements of G to its product. It sends each group homomorphism to its underlying function.

For C2=Z/2Z, the word [1ˉ,1ˉ,1ˉ] evaluates to 1ˉ, while every adjacent inverse pair and every occurrence of the identity may be removed before evaluation.

Facts & Assumptions

Given: The free-group adjunction and a group G.

[L1]

The Eilenberg–Moore category of the free-group monad is isomorphic over Set to the category of groups (The free-group monad has groups as its Eilenberg–Moore algebras).

[L2]

The comparison sends d to (Ud,Uεd) and sends a morphism to its image under U (The comparison functor to the Eilenberg–Moore category exists and is unique).

[L3]

The quotient group (Z,+)/2Z is (Z/2,+) with addition of congruence classes (For every nN, the congruence-class group (Z/n,+) is the quotient group (Z,+)/nZ).

[L4]

Choosing a free group (F(X),iX) on every set X makes F left adjoint to the underlying-set functor U, the adjunction bijection sending φ:F(X)G to U(φ)iX (The free-group functor is left adjoint to the underlying-set functor).

[L5]

An algebra homomorphism is a map commuting with the two algebra structure maps (Algebra and algebra homomorphism for a monad).

Verification

technique · direct
1.1

By [L2], the structure map is the underlying counit UεG of the free-group adjunction. Under [L4] the counit corresponds to the identity of UG, so εG is the unique group homomorphism F(UG)G carrying each basis element g to g; a homomorphism out of a free group is determined by its values on the basis, so it evaluates a reduced word in UG to its product in G.

L2L4
1.2

By [L5] the algebra-homomorphism equation says that a function commutes with evaluation of every word. Evaluating the words [x,y], the empty word and [x1] makes such a function preserve product, identity and inverse, and conversely a group homomorphism preserves the value of every word; so the comparison morphisms are exactly group homomorphisms, in agreement with [L1] and with K(h)=U(h) in [L2].

L1L2L5
2.1

Evaluating a one-letter word returns its letter, and evaluating after substitution of words agrees with evaluating the flattened word by associativity of group multiplication. These are the algebra unit and multiplication laws.

step 1.1algebra
3.1

In the group [L3], one has 1ˉ+1ˉ=0ˉ and 0ˉ+1ˉ=1ˉ, so the displayed three-letter word evaluates to 1ˉ.

step 1.1L3construct

Depends on

Used by

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Sources