False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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FALSE: a left Kan extension along a fully faithful functor always restricts back to the original functor

Statement refuted

That for every fully faithful $K : C \to D$ and every left Kan extension $(L, η)$ of $F$ along $K$ , the unit components $η_{c} : F (c) \to L (K c)$ are automatically isomorphisms.

The published positive theorem A pointwise Kan extension along a fully faithful functor genuinely extends the original functor shows this is true only under the pointwise hypothesis.

Facts & Assumptions

Given: The discrete category $A$ with objects $ℓ$ and $r$ ; the walking span $D$ with objects $ℓ, m, r$ and non-identity arrows $s : m \to ℓ$ and $t : m \to r$ ; the fully faithful inclusion $K : A ↪ D$ ; the category $E$ with objects $L, L^{'}, M, R^{'}, R$ , identities, and exactly four non-identity arrows $f : L \to L^{'}$ , $g : M \to L^{'}$ , $h : M \to R^{'}$ , and $k : R \to R^{'}$ ; the functor $F : A \to E$ with $F (ℓ) = L$ and $F (r) = R$ ; the functor $G : D \to E$ with $G (ℓ) = L^{'}$ , $G (m) = M$ , $G (r) = R^{'}$ , $G (s) = g$ , and $G (t) = h$ ; and the natural transformation $λ : F \Rightarrow G K$ with components $λ_{ℓ} = f$ and $λ_{r} = k$ .

[F1]

A left Kan extension is initial among pairs $(M, α)$ with $α : F \Rightarrow M K$ (Left and right Kan extensions).

[F2]

A fully faithful functor is bijective on each hom-set (Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors).

[L1]

If such a left Kan extension were pointwise, then the unit component at $r$ would be an isomorphism (A pointwise Kan extension along a fully faithful functor genuinely extends the original functor).

Refutation

technique · direct

1.1

The inclusion $K$ is fully faithful because $A$ is discrete and $D$ has no non-identity arrows between the image objects $ℓ$ and $r$ , so each hom-set map is bijective as required by [F2].

F2given

2.1

The pair $(G, λ)$ is a left Kan extension of $F$ along $K$ . Let $M : D \to E$ and $α : F \Rightarrow M K$ . Because $D$ has arrows $m \to ℓ$ and $m \to r$ , the object $M (m)$ must admit arrows to both $M (ℓ)$ and $M (r)$ . In $E$ the only object with arrows to two targets is $M$ , via $g : M \to L^{'}$ and $h : M \to R^{'}$ . Therefore $M (m) = M$ , $M (ℓ) = L^{'}$ , $M (r) = R^{'}$ , and the two span arrows must be sent to $g$ and $h$ . So $M = G$ . Then $α_{ℓ} : L \to L^{'}$ and $α_{r} : R \to R^{'}$ are forced to be $f$ and $k$ , whence $α = λ$ . Since the only endomorphisms of $L^{'}$ , $M$ , and $R^{'}$ are identities, the only natural transformation $G \Rightarrow M = G$ is the identity. Thus $(G, λ)$ satisfies the universal property [F1].

F1step 1.1

3.1

The components $λ_{ℓ} = f : L \to L^{'}$ and $λ_{r} = k : R \to R^{'}$ are not isomorphisms, since $E$ has no arrows $L^{'} \to L$ or $R^{'} \to R$ . Therefore the unrestricted claim is false. By [L1], this also shows that the witness is not pointwise.

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

E. Riehl, Category Theory in Context, 2nd ed., Example 6.2.17 (standard reference, not scraped)