Yoneda Lemma

Let's check naturality carefully in both components of the family

φ_{F, c} : F (c) ≅ Nat (Y (c), F) : ψ_{F, c}

where

φ_{F, c} (x) = (η_{d}^{x} : Hom (d, c) \to F (d), η_{d}^{x} (f) = F (f) (x))

ψ_{F, c} (η) = η_{c} ({id}_{c})

Notation reminder: For a presheaf

F : C^{op} \to Set

and arrow

u : d \to e

we write

F (u) : F (e) \to F (d)

. For

Y (c) = Hom (-, c)

we have

Y (c) (u) : Hom (e, c) \to Hom (d, c)

given by precomposition with

u

1. Naturality in the first variable (in $F$ ) — postcomposition by $τ$

Fix $c \in C$ . Let $τ : F \Rightarrow G$ be a natural transformation of presheaves. The square to prove commutes is:

\begin{array}{ccc} F (c) & \overset{φ_{F, c}}{\to} & Nat (Y (c), F) \\ ↓ τ_{c} & ↓ τ \circ (-) \\ G (c) & \overset{φ_{G, c}}{\to} & Nat (Y (c), G) \end{array}

We must show for every $x \in F (c)$

φ_{G, c} (τ_{c} (x)) = τ \circ φ_{F, c} (x)

Take arbitrary $d \in C$ and $f : d \to c$ . Evaluate both sides at the component $d$ and argument $f$ .

Left-hand side (LHS):

{(φ_{G, c} (τ_{c} (x)))}_{d} (f) = G (f) (τ_{c} (x))

Right-hand side (RHS):

{(τ \circ φ_{F, c} (x))}_{d} (f) = τ_{d} ({φ_{F, c}}_{d} (x) (f)) = τ_{d} (F (f) (x))

But $τ$ is a natural transformation $F \Rightarrow G$ . Naturality at the arrow $f : d \to c$ says

τ_{d} \circ F (f) = G (f) \circ τ_{c}

Apply both sides to $x \in F (c)$ and get

τ_{d} (F (f) (x)) = G (f) (τ_{c} (x))

So RHS = LHS for the chosen $d$ and $f$ . Since $d, f$ were arbitrary, the natural transformations are equal. Thus the square commutes. QED for naturality in $F$ .

2. Naturality in the second variable (in $c$ ) — precomposition by $Y (k)$

Fix $F \in \hat{C}$ . Let $k : c^{'} \to c$ be a morphism in $C$ . The square to prove commutes is:

\begin{array}{ccc} F (c) & \overset{φ_{F, c}}{\to} & Nat (Y (c), F) \\ ↓ F (k) & ↓ (-) \circ Y (k) \\ F (c^{'}) & \overset{φ_{F, c^{'}}}{\to} & Nat (Y (c^{'}), F) \end{array}

We must show for every $x \in F (c)$

φ_{F, c^{'}} (F (k) (x)) = φ_{F, c} (x) \circ Y (k)

Again take arbitrary $d \in C$ and $f : d \to c^{'}$ . Evaluate both sides at component $d$ and argument $f$ .

Left-hand side (LHS):

{(φ_{F, c^{'}} (F (k) (x)))}_{d} (f) = F (f) (F (k) (x))

By functoriality of $F$ (contravariant), $F (k \circ f) = F (f) \circ F (k)$ , so

F (f) (F (k) (x)) = F (k \circ f) (x)

Right-hand side (RHS):

{(φ_{F, c} (x) \circ Y (k))}_{d} (f) = {φ_{F, c}}_{d} (x) (Y_{d} (k) (f))

But $Y_{d} (k) (f) = k \circ f$ (postcompose $f$ with $k$ to get an arrow $d \to c$ ), and ${φ_{F, c}}_{d} (x) (k \circ f) = F (k \circ f) (x)$ . So

{(φ_{F, c} (x) \circ Y (k))}_{d} (f) = F (k \circ f) (x)

Hence LHS = RHS (= $F (k \circ f) (x)$ ) for arbitrary $d, f$ . Therefore the square commutes. QED for naturality in $c$ .

3. Naturality on the product $\hat{C} \times C$ (general morphism $(τ, k)$ )

A morphism in $\hat{C} \times C$ from $(F, c)$ to $(G, c^{'})$ is a pair $(τ, k)$ with $τ : F \Rightarrow G$ and $k : c^{'} \to c$ . On the functors

p (F, c) = F (c), q (F, c) = Nat (Y (c), F)

the induced maps are

p (τ, k) : F (c) \overset{τ_{c}}{\to} G (c) \overset{G (k)}{\to} G (c^{'})

q (τ, k) : Nat (Y (c), F) \overset{(-) \circ Y (k)}{\to} Nat (Y (c^{'}), F) \overset{τ \circ (-)}{\to} Nat (Y (c^{'}), G)

We need to show the square

\begin{array}{ccc} F (c) & \overset{φ_{F, c}}{\to} & Nat (Y (c), F) \\ ↓ p (τ, k) = G (k) \circ τ_{c} & ↓ q (τ, k) = τ \circ (-) \circ Y (k) \\ G (c^{'}) & \overset{φ_{G, c^{'}}}{\to} & Nat (Y (c^{'}), G) \end{array}

commutes, i.e. for all $x \in F (c)$ ,

φ_{G, c^{'}} (G (k) (τ_{c} (x))) = τ \circ φ_{F, c} (x) \circ Y (k)

Evaluate both sides at an arbitrary $d \in C$ and $f : d \to c^{'}$ .

Left-hand side (LHS):

{(φ_{G, c^{'}} (G (k) (τ_{c} (x))))}_{d} (f) = G (f) (G (k) (τ_{c} (x))) = G (k \circ f) (τ_{c} (x))

(used functoriality $G (f) \circ G (k) = G (k \circ f)$ .)

Right-hand side (RHS):

{(τ \circ φ_{F, c} (x) \circ Y (k))}_{d} (f) = τ_{d} ({φ_{F, c}}_{d} (x) (Y_{d} (k) (f)))

Now $Y_{d} (k) (f) = k \circ f$ , and ${φ_{F, c}}_{d} (x) (k \circ f) = F (k \circ f) (x)$ . Thus

{(τ \circ φ_{F, c} (x) \circ Y (k))}_{d} (f) = τ_{d} (F (k \circ f) (x))

Use naturality of $τ : F \Rightarrow G$ at the arrow $k \circ f : d \to c$ :

τ_{d} \circ F (k \circ f) = G (k \circ f) \circ τ_{c}

Apply both sides to $x \in F (c)$ to obtain

τ_{d} (F (k \circ f) (x)) = G (k \circ f) (τ_{c} (x))

Therefore RHS (= $G (k \circ f) (τ_{c} (x))$ ), which equals LHS. Since $d, f$ were arbitrary, the two natural transformations coincide on all components and values. Thus the product square commutes and $φ$ is natural as a transformation $p \Rightarrow q$ on $\hat{C} \times C$ .

4. Naturality of the inverse $ψ$

Because $φ$ is a natural bijection with inverse $ψ_{F, c} (η) = η_{c} ({id}_{c})$ , $ψ$ is also natural. Concretely:

For $τ : F \Rightarrow G$ :

ψ_{G, c} (τ \circ η) = {(τ \circ η)}_{c} ({id}_{c}) = τ_{c} (η_{c} ({id}_{c})) = τ_{c} (ψ_{F, c} (η))

So postcomposition by $τ$ commutes with $ψ$ .

For $k : c^{'} \to c$ :

ψ_{F, c^{'}} (η \circ Y (k)) = {(η \circ Y (k))}_{c^{'}} ({id}_{c^{'}}) = η_{c^{'}} (k) = F (k) (η_{c} ({id}_{c})) = F (k) (ψ_{F, c} (η))

So precomposition by $Y (k)$ commutes with $ψ$ .

Hence $ψ$ is natural in both variables and is the inverse natural transformation to $φ$ .

Conclusion

We have shown in full elementwise detail that for every arrow $τ : F \Rightarrow G$ in $\hat{C}$ and every arrow $k : c^{'} \to c$ in $C$ all required squares commute. Therefore the family ${φ_{F, c}}$ is natural in both components and gives a natural isomorphism of functors

p (F, c) = F (c) \overset{φ}{\to} q (F, c) = Nat (Y (c), F)

This completes the proof of the Yoneda Lemma.

Yoneda Lemma

1. Naturality in the first variable (in F) — postcomposition by τ

2. Naturality in the second variable (in c) — precomposition by Y(k)

3. Naturality on the product C^×C (general morphism (τ,k))

4. Naturality of the inverse ψ

Conclusion

1. Naturality in the first variable (in $F$ ) — postcomposition by $τ$

2. Naturality in the second variable (in $c$ ) — precomposition by $Y (k)$

3. Naturality on the product $\hat{C} \times C$ (general morphism $(τ, k)$ )

4. Naturality of the inverse $ψ$