ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Evaluation of functions is dinatural in its argument set

Example

Fix a set $Y$ and let $S : {S e t}^{o p} \times S e t \to S e t$ be

$S (B_{1}, B_{2}) : = Y^{B_{1}} \times B_{2},$

contravariant in $B_{1}$ by precomposition and covariant in $B_{2}$ (The set $B^{A}$ of all functions $A \to B$ , The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ , Sets and functions form the large locally small category $S e t$ ). The evaluation family

${ev}_{B} : Y^{B} \times B ⟶ Y, {ev}_{B} (g, b) = g (b),$

is a dinatural transformation from $S$ to the constant functor at $Y$ (Dinatural transformation between functors on $C^{o p} \times C$ ), that is, a cowedge under $S$ with vertex $Y$ (Wedges and cowedges, and the categories they form).

Facts & Assumptions

Given: A set $Y$ , the functor $S$ displayed above, and the family of evaluation functions.

[F3]

The functions $A \to B$ form the set $B^{A}$ , and Thus $f \in B^{A}$ holds if and only if $f : A \to B$ . (The set $B^{A}$ of all functions $A \to B$ ).

[F5]

The elements of $A \times B$ are exactly the ordered pairs: Thus $z \in A \times B$ holds if and only if $z = (a, b)$ for some $a \in A$ and some $b \in B$ . (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[F4]

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

[L1]

For every set $A$ the product functor $- \times A$ is left adjoint to $(-)^{A}$ , with $S e t (X \times A, Y) ≅ S e t (X, Y^{A})$ naturally in $X$ and $Y$ ; the bijection sends $k$ to $\overset{ˇ}{k} (x, a) = k (x) (a)$ (Currying gives the adjunction $- \times A ⊣ (-)^{A}$ in $S e t$ ).

[F1]

A dinatural transformation $α : P \to Q$ is a family $α_{c} : P (c, c) \to Q (c, c)$ such that every $f : c \to c^{'}$ satisfies $Q (1_{c}, f) \circ α_{c} \circ P (f, 1_{c}) = Q (f, 1_{c^{'}}) \circ α_{c^{'}} \circ P (1_{c^{'}}, f)$ , the equation displayed by the hexagon (Dinatural transformation between functors on $C^{o p} \times C$ ).

[F2]

A cowedge from $T$ to $d$ is a dinatural transformation from $T$ to a constant functor: a family $ρ_{c} : T (c, c) \to d$ with $ρ_{c} \circ T (f, 1_{c}) = ρ_{c^{'}} \circ T (1_{c^{'}}, f)$ for every $f : c \to c^{'}$ (Wedges and cowedges, and the categories they form).

Verification

technique · direct

1.1

The assignment $S$ is a functor: for $f : B \to B^{'}$ the contravariant slot acts by $g \mapsto g \circ f$ from $Y^{B^{'}}$ to $Y^{B}$ and the covariant slot by $f$ itself, and both actions preserve identities and composites because composition of functions does. The family ${ev}_{B}$ is the transpose $\overset{ˇ}{k}$ of the identity of $Y^{B}$ under the bijection of [L1] with $A = B$ and $X = Y^{B}$ , so it is the counit of that adjunction at $Y$ .

F3F4F5L1

2.1

Fix $f : B \to B^{'}$ and chase an arbitrary element $(g, b)$ of $S (B^{'}, B) = Y^{B^{'}} \times B$ through both legs of the hexagon. Since the target is the constant functor at $Y$ , both outer actions on the target side are identities, and the two legs are ${ev}_{B} \circ S (f, 1_{B})$ and ${ev}_{B^{'}} \circ S (1_{B^{'}}, f)$ . The first sends $(g, b)$ to ${ev}_{B} (g \circ f, b) = g (f (b))$ and the second sends it to ${ev}_{B^{'}} (g, f (b)) = g (f (b))$ . The two values agree for every $(g, b)$ , so the two legs are equal.

F1F5step 1.1

3.1

Since the target is the constant functor at $Y$ , the equation verified in step 2.1 is exactly the cowedge equation of [F2], so the evaluation family is a cowedge under $S$ with vertex $Y$ , and in particular a dinatural transformation.

F2step 2.1∎

Remarks

The displayed evaluation family supplies only diagonal components. There is no canonical evaluation map $Y^{B_{1}} \times B_{2} \to Y$ for unrelated $B_{1}, B_{2}$ , and no natural family of such maps extending evaluation in general; special cases such as singleton $Y$ may admit constant maps. What the family always has is one component per object on the diagonal, tied together by the equation checked above, which is precisely the shape dinaturality was defined to capture.

The chase uses nothing about $Y$ . If $Y$ is empty then $Y^{B}$ is empty unless $B$ is, and the two legs are then functions with empty domain, which are equal for that reason; the computation above covers that case without a separate argument, since it verifies the two legs agree at every element of the domain.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

F. Loregian, (Co)end Calculus (arXiv:1501.02503v7), Remark 1.1.3 (standard reference, not scraped)
B. Richter, From Categories to Homotopy Theory (author's draft), Examples 4.4.3 (standard reference, not scraped)