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

1.1

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

L1L2

1.2

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

L1F1

1.3

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 $F$ .

L1F2

2.1

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

step 1.1step 1.2step 1.3∎

Depends on

Used by

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

Emily Riehl, Category Theory in Context, Theorem 2.2.4 (standard reference, not scraped)
Tom Leinster, Basic Category Theory, Theorem 4.2.1 (standard reference, not scraped)

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