LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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A wedge on a product index category is exactly a family dinatural in each variable separately

Statement

Let $C$ , $D$ and $E$ be categories and let

$T : (C \times D)^{o p} \times (C \times D) ⟶ E$

be a functor. Reindexing the source as $C^{o p} \times D^{o p} \times C \times D$ (Product category and its projection functors, Opposite category $C^{o p}$ ), write $T (c_{1}, d_{1}, c_{2}, d_{2})$ for its values, contravariant in the first two slots and covariant in the last two.

Let $X$ be an object of $E$ and let $ω_{(c, d)} : X \to T (c, d, c, d)$ be a family indexed by the objects of $C \times D$ . Then a wedge on a product index category is exactly a family dinatural in each variable separately: $ω$ is a wedge from $X$ to $T$ (Wedges and cowedges, and the categories they form) if and only if

for every $f : c \to c^{'}$ in $C$ and every object $d$ of $D$ , $T (1_{c}, 1_{d}, f, 1_{d}) \circ ω_{(c, d)} = T (f, 1_{d}, 1_{c^{'}}, 1_{d}) \circ ω_{(c^{'}, d)}$ , and
for every $g : d \to d^{'}$ in $D$ and every object $c$ of $C$ , $T (1_{c}, 1_{d}, 1_{c}, g) \circ ω_{(c, d)} = T (1_{c}, g, 1_{c}, 1_{d^{'}}) \circ ω_{(c, d^{'})}$ .

The equivalence is asserted for wedges, whose source is the constant functor at $X$ . It is not asserted for a dinatural transformation between two varying functors (Dinatural transformation between functors on $C^{o p} \times C$ ).

Facts & Assumptions

Given: A functor $T$ as displayed, an object $X$ of $E$ , and a family $ω_{(c, d)} : X \to T (c, d, c, d)$ .

[F2]

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

[F3]

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

[F4]

The product category has objects the pairs, componentwise identities, and componentwise composition $(f^{'}, g^{'}) \circ (f, g) = (f^{'} \circ f, g^{'} \circ g)$ (Product category and its projection functors).

[F6]

The opposite category has the same objects and reverses every morphism: $C^{o p} (A, B) = C (B, A)$ (Opposite category $C^{o p}$ ).

[F5]

A functor satisfies $F (1_{A}) = 1_{F A}, F (g \circ f) = F g \circ F f$ (Covariant functor, identity functor, composite functor, and contravariant functor).

Proof

technique · direct

1.1

A morphism $(c, d) \to (c^{'}, d^{'})$ of $C \times D$ is a pair $(f, g)$ with $f : c \to c^{'}$ and $g : d \to d^{'}$ , and under the reindexing the wedge equation of [F2] at that morphism reads $T (1_{c}, 1_{d}, f, g) \circ ω_{(c, d)} = T (f, g, 1_{c^{'}}, 1_{d^{'}}) \circ ω_{(c^{'}, d^{'})}$ , an equation between morphisms $X \to T (c, d, c^{'}, d^{'})$ . The two displayed conditions of the Statement are this equation at $(f, 1_{d})$ and at $(1_{c}, g)$ .

F2F3F4F6

2.1

For the forward direction, if $ω$ is a wedge then the equation of step 1.1 holds at every morphism of $C \times D$ , in particular at $(f, 1_{d})$ and at $(1_{c}, g)$ , which are the two displayed conditions.

F2F3step 1.1

3.1

For the converse direction, assume the two conditions and fix $(f, g)$ . Since $T$ acts independently in its four slots, $T (1_{c}, 1_{d}, f, g) = T (1_{c}, 1_{d}, 1_{c^{'}}, g) \circ T (1_{c}, 1_{d}, f, 1_{d})$ , so the first condition rewrites the left-hand side of step 1.1 as $T (1_{c}, 1_{d}, 1_{c^{'}}, g) \circ T (f, 1_{d}, 1_{c^{'}}, 1_{d}) \circ ω_{(c^{'}, d)} = T (f, 1_{d}, 1_{c^{'}}, g) \circ ω_{(c^{'}, d)}$ . Factoring again as $T (f, 1_{d}, 1_{c^{'}}, g) = T (f, 1_{d}, 1_{c^{'}}, 1_{d^{'}}) \circ T (1_{c^{'}}, 1_{d}, 1_{c^{'}}, g)$ and applying the second condition at $c^{'}$ gives $T (f, 1_{d}, 1_{c^{'}}, 1_{d^{'}}) \circ T (1_{c^{'}}, g, 1_{c^{'}}, 1_{d^{'}}) \circ ω_{(c^{'}, d^{'})} = T (f, g, 1_{c^{'}}, 1_{d^{'}}) \circ ω_{(c^{'}, d^{'})}$ , which is the right-hand side of step 1.1. Factoring $(f, g)$ in the other order gives the same result, because each factorisation is a composite in $C^{o p} \times D^{o p} \times C \times D$ of the same pair of morphisms in different slots. At $f = 1_{c}$ the first condition is the identity equation $ω_{(c, d)} = ω_{(c, d)}$ and the chain reduces to the second condition alone; at $g = 1_{d}$ it reduces to the first, so the reduction is not circular.

F2F3F4F5step 1.1step 2.1

4.1

The two directions together give the asserted equivalence. What the converse direction spends is that the source of $ω$ is the constant functor at $X$ : each rewriting in step 3.1 composed a one-variable equation on the target side only, with no source-side action to carry along, and for a dinatural transformation between two varying functors there is such an action in every slot, so the corresponding statement does not follow from this argument and is not asserted.

F2step 3.1∎

Remarks

That dinaturality is fragile under exactly this kind of extension is not a suspicion: Dinatural transformations do not compose in general exhibits two dinatural transformations on this same page whose composite is not dinatural, and the mechanism there is also a source-side action that the constant case does not have.

Both factorisations of $(f, g)$ are checked because they are the two ways the joint equation can be reduced, and an argument that used only one of them would leave open whether the two one-variable conditions had to be imposed in a fixed order. They do not: the four slots act independently.

Depends on

Used by

Fubini: an end over a product index category and the two iterated ends exist together and agree Theorem

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

G. M. Kelly, Basic Concepts of Enriched Category Theory (TAC Reprints 10), (2.7)-(2.9) (standard reference, not scraped)
F. Loregian, (Co)end Calculus (arXiv:1501.02503v7), Theorem 1.3.1 (standard reference, not scraped)