CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-16

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A reflective inclusion need not preserve even the empty colimit

Counterexample

Let $A$ be the full subcategory of $S e t$ whose only object is a fixed singleton $1$ . Its inclusion $I : A ↪ S e t$ is reflective, but it does not preserve the empty colimit.

Facts & Assumptions

Given: The full singleton subcategory $A \subseteq S e t$ .

[L1]

A full subcategory is reflective when its inclusion has a left adjoint (Reflective full subcategory and reflector).

[L2]

Empty colimits are precisely initial objects (Limits of empty diagrams are terminal objects, and colimits of empty diagrams are initial objects).

[L3]

For locally small categories, an adjunction determines hom-set bijections natural in both variables, and conversely every such natural family of bijections determines a unique unit and counit satisfying the triangle identities, hence a unique adjunction structure (Under local smallness, transposition gives the natural hom-set bijection, and conversely).

Verification

technique · counterexample

1.1L1L3

The constant functor $R : S e t \to A$ at $1$ is left adjoint to $I$ : for every set $X$ , both ${Hom}_{A} (1, 1)$ and ${Hom}_{S e t} (X, 1)$ contain one map, and these bijections are natural in $X$ ; by the converse clause of [L3] they determine a unit and counit satisfying the triangle identities, hence an adjunction $R ⊣ I$ , so [L1] makes $A$ reflective.

1.2L2

The sole object $1$ is initial in $A$ , so it is the empty colimit there by [L2]. Its image $I (1) = 1$ is not initial in $S e t$ , whose initial object is $\emptyset$ .

2.1step 1.1step 1.2∎

Therefore the included empty colimit is not an ambient colimit, so $I$ does not preserve even this colimit.

Depends on

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