ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For a monoid action, Yoneda says that an equivariant map from the regular action is determined by the identity element

Example

Let $M$ be a monoid with identity $e$ , viewed as the one-object category $B M$ . A functor $F : B M \to S e t$ is a set $X = F (*)$ with a left $M$ -action. The representable $B M (*, -)$ is the left regular action of $M$ on itself, and Yoneda becomes the bijection

${ϕ : M \to X ∣ ϕ is M -equivariant} \overset{≅}{\to} X, ϕ ⟼ ϕ (e),$

with inverse $x \mapsto (m \mapsto m \cdot x)$ .

Facts & Assumptions

Given: A monoid $(M, \cdot, e)$ , its one-object category $B M$ , and a functor $F : B M \to S e t$ .

[F1]

A monoid has associative multiplication and a two-sided identity (Semigroup and monoid).

[L1]

Every monoid is a one-object category (A monoid is a one-object category, and a group is a one-object category in which every morphism is invertible). This example fixes the resulting data explicitly: $B M (*, *) = M$ , and composition is taken to be $n \circ m = n m$ , the convention that makes the represented functor a left action.

[F2]

Sets and functions form the category $S e t$ (Sets and functions form the large locally small category $S e t$ ).

[L2]

Evaluation at the identity is a bijection $Nat (B M (*, -), F) \to F (*)$ whose inverse sends $x$ to the natural transformation with component $m \mapsto F (m) (x)$ (Evaluation at the identity gives $Nat (C (a, -), F) ≅ F (a)$ and proves that the natural-transformation collection is a set), and it is natural in the represented object and in the target functor (The Yoneda bijection $Nat (C (a, -), F) ≅ F (a)$ is natural in both $a$ and $F$ ).

[F3]

A natural transformation has components commuting with the action of every source morphism (Natural transformation and its components).

Verification

technique · constructive

1.1

Put $m \cdot x = F (m) (x)$ . Functoriality and [L1] give $e \cdot x = x$ and $(n m) \cdot x = n \cdot (m \cdot x)$ , so this is a left action.

givenL1F1F2

1.2

The covariant representable $B M (*, -)$ has value $M$ and sends $n$ to postcomposition $m \mapsto n \circ m = n m$ , so it is the left regular action.

2.1

A natural transformation $B M (*, -) \Rightarrow F$ is a function $ϕ : M \to X$ satisfying $ϕ (n m) = n \cdot ϕ (m)$ for all $m, n$ , exactly equivariance for the two left actions.

step 1.1step 1.2F3

3.1

If $ϕ$ is equivariant, then $ϕ (m) = ϕ (m e) = m \cdot ϕ (e)$ . Conversely, for $x \in X$ , the function $ϕ_{x} (m) = m \cdot x$ satisfies $ϕ_{x} (n m) = (n m) \cdot x = n \cdot ϕ_{x} (m)$ by step 1.1 and has $ϕ_{x} (e) = x$ .

step 1.1step 2.1F1construct

4.1

Step 3.1 proves directly that evaluation at $e$ and $x \mapsto ϕ_{x}$ are inverse. These are precisely the formulas in [L2], whose target-functor naturality says that the bijection commutes with every equivariant map of $M$ -sets.

step 3.1L2discharge-construct∎

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: 36 results over 13 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

Emily Riehl, Category Theory in Context, Proposition 2.2.3 and Theorem 2.2.4 (standard reference, not scraped)