TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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The Yoneda bijection $Nat (C (a, -), F) ≅ F (a)$ is natural in both $a$ and $F$

Statement

Under the hypotheses of Evaluation at the identity gives $Nat (C (a, -), F) ≅ F (a)$ and proves that the natural-transformation collection is a set, the bijections

$E_{a, F} : Nat (C (a, -), F) \overset{≅}{\to} F (a)$

are natural in both variables. Explicitly:

if $h : a \to a^{'}$ and $α : C (a, -) \Rightarrow F$ , then $E_{a^{'}, F} (α \circ C (h, -)) = F (h) (E_{a, F} (α));$
if $η : F \Rightarrow G$ , then $E_{a, G} (η \circ α) = η_{a} (E_{a, F} (α)) .$

Here $C (h, -) : C (a^{'}, -) \Rightarrow C (a, -)$ is precomposition by $h$ . These equations make sense for every locally small $C$ and do not require a functor category with source $C$ .

Facts & Assumptions

Given: A locally small category $C$ , objects $a, a^{'}$ , a morphism $h : a \to a^{'}$ , functors $F, G : C \to S e t$ , natural transformations $α : C (a, -) \Rightarrow F$ and $η : F \Rightarrow G$ , and the category and functor laws.

[L1]

Evaluation is the bijection $E_{a, F} (α) = α_{a} (1_{a})$ , with inverse $x \mapsto (f \mapsto F (f) (x))$ (Evaluation at the identity gives $Nat (C (a, -), F) ≅ F (a)$ and proves that the natural-transformation collection is a set).

[L2]

The hom-assignment is a bifunctor, so $h : a \to a^{'}$ induces the natural transformation $C (h, -)$ with component $k \mapsto k \circ h$ (The hom-assignment $C (-, -) : C^{o p} \times C \to S e t$ is a bifunctor).

[F1]

Naturality of $α$ says $F (h) \circ α_{a} = α_{a^{'}} \circ C (a, h)$ (Natural transformation and its components).

[F2]

Vertical composition is componentwise: $(η \circ α)_{a} = η_{a} \circ α_{a}$ (Identity natural transformation and vertical composition).

Proof

technique · direct

1.1L1L2

By [L2], the component of $C (h, -)$ at $a^{'}$ sends $1_{a^{'}}$ to $1_{a^{'}} \circ h = h$ ; hence $E_{a^{'}, F} (α \circ C (h, -)) = α_{a^{'}} (h)$ .

1.2L1F1

Applying [F1] to $1_{a}$ gives $α_{a^{'}} (h) = F (h) (α_{a} (1_{a})) = F (h) (E_{a, F} (α))$ , proving naturality in $a$ .

1.3L1F2

By componentwise composition, $E_{a, G} (η \circ α) = (η \circ α)_{a} (1_{a}) = η_{a} (α_{a} (1_{a})) = η_{a} (E_{a, F} (α))$ , proving naturality in $F$ .

2.1step 1.1step 1.2step 1.3∎

Steps 1.1--1.3 are the two required naturality squares, and their pointwise formulas use no functor category on a large source.

The Yoneda bijection $Nat (C (a, -), F) ≅ F (a)$ is natural in both $a$ and $F$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The Yoneda bijection Nat⁡(C(a,−),F)≅F(a)Nat(C(a,−),F)≅F(a) is natural in both aa and FF

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The Yoneda bijection $Nat (C (a, -), F) ≅ F (a)$ is natural in both $a$ and $F$