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CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A componentwise family between functors need not be a natural transformation

Statement refuted

Choosing one morphism between each pair of object values of two functors does not automatically give a natural transformation.

Facts & Assumptions

Given: The walking-arrow category C=(0→a1) and the category Set.

[L1]

Naturality requires G(a)η0=η1F(a) for the arrow a (Natural transformation and its components).

Counterexample

technique · direct
1.1

Let F:C→Set be constant at the singleton {∗} and let G:C→Set be constant at {0,1}, with both functors sending a to the relevant identity function.

L2
2.1

Define components by η0(∗)=0 and η1(∗)=1. Each is a valid function F(i)→G(i).

step 1.1L2
3.1

At a, the left side of the naturality equation sends ∗ to 0, whereas the right side sends ∗ to 1. Thus G(a)η0≠η1F(a).

step 1.1step 2.1L1
4.1

The family (η0,η1) has a component of the correct type at every object but fails naturality.

step 3.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

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.