False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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FALSE: dinatural transformations compose

Statement

False claim: for functors $P, Q, R : C^{o p} \times C \to D$ and dinatural transformations $α : P \to Q$ and $β : Q \to R$ (Dinatural transformation between functors on $C^{o p} \times C$ ), the componentwise composite $(β_{c} \circ α_{c})_{c}$ is a dinatural transformation $P \to R$ ; so the functors on $C^{o p} \times C$ and the dinatural transformations between them form a category.

Facts & Assumptions

Given: The walking arrow $C$ , with objects $0$ and $1$ and one non-identity morphism $u : 0 \to 1$ , and the category $S e t$ as target.

[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 $C^{o p} \times C$ ).

[L1]

There are a category, three set-valued functors on $C^{o p} \times C$ and two dinatural transformations between them whose componentwise composite is not dinatural: dinatural transformations do not compose in general (Dinatural transformations do not compose in general).

[L2]

For natural $σ : P^{'} \Rightarrow P$ and $τ : Q \Rightarrow Q^{'}$ and dinatural $α : P \to Q$ , composing a dinatural transformation with a natural transformation on either side gives a dinatural transformation (Composing a dinatural transformation with a natural transformation on either side gives a dinatural transformation).

Refutation

technique · direct

1.1F1givenconstruct

The witness is restated in full so that this item stands alone. Take $C$ to be the walking arrow, so that a functor $T : C^{o p} \times C \to S e t$ is four sets and four functions subject to one equation. Let $P$ have all four values a one-element set; let $Q$ have $Q (1, 0) = \emptyset$ and $Q (0, 0), Q (1, 1), Q (0, 1)$ one-element sets; and let $R$ have $R (1, 0) = \emptyset$ , $R (0, 0) = {c}$ , $R (1, 1) = {d}$ and $R (0, 1) = {y_{1}, y_{2}}$ with $y_{1} \neq y_{2}$ , with $R (1_{0}, u) (c) = y_{1}$ and $R (u, 1_{1}) (d) = y_{2}$ . Let $α : P \to Q$ and $β : Q \to R$ be the families whose components are the only functions available between one-element sets.

2.1F1L1step 1.1

Both families are dinatural and their composite is not. The hexagon for $α$ at $u$ is an equation between two functions into the one-element set $Q (0, 1)$ , so it holds; the hexagon for $β$ at $u$ is an equation between two functions out of $Q (1, 0) = \emptyset$ , so it holds; and the two legs of the hexagon for the composite send the element of $P (1, 0)$ to $y_{1}$ and to $y_{2}$ respectively, which differ. This is exactly the witness of [L1], so the displayed claim is false and the dinatural transformations are not the morphisms of a category under componentwise composition.

3.1L2step 2.1∎

What is true is the weaker statement [L2]: a dinatural transformation composed with a natural transformation on either side is again dinatural. So dinaturality is not closed under nothing; it is closed under composition with natural transformations, and the false claim is exactly the extension of that to two dinatural factors.

Remarks

The mechanism is the empty slot. Putting $\emptyset$ in the $(1, 0)$ position of $Q$ and of $R$ makes the hexagon for $β$ an equation between functions with empty domain, so $β$ is dinatural for no reason of its own; the one-element set in the $(0, 1)$ position of $Q$ does the same for $α$ at the other end. Nothing then constrains the composite, whose hexagon has a two-element codomain.

The claim is a genuine trap rather than a careless one, because the analogous statement for natural transformations is true and is what makes functor categories exist. What fails here is that a dinatural transformation has components only on the diagonal, so composing two of them loses the off-diagonal information that each one's hexagon was about.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

F. Loregian, (Co)end Calculus (arXiv:1501.02503v7), Chapter 1 introduction and Exercise 1.2 (standard reference, not scraped)