A Proof of Yoneda's Lemma

Yoneda’s Lemma says that an object is determined by how every other object maps into it. Let $\mathcal{C}$ be a locally small category, let $A \in \mathcal{C}$, and let $F : \mathcal{C} \to \mathbf{Set}$ be a functor. This is the covariant version; the contravariant form uses $F : \mathcal{C}^{op} \to \mathbf{Set}$.

The lemma gives a natural isomorphism:

$$\operatorname{Nat}(\operatorname{Hom}_{\mathcal{C}}(A, -), F) \cong F(A).$$

The Map

Given a natural transformation $\eta : \operatorname{Hom}_{\mathcal{C}}(A, -) \Rightarrow F$, send it to Evaluating at $\mathrm{id}_A$ picks out the element of $F(A)$ represented by $\eta$.

$$\Phi(\eta) = \eta_A(\mathrm{id}_A) \in F(A).$$

This defines a function

$$\Phi : \operatorname{Nat}(\operatorname{Hom}_{\mathcal{C}}(A, -), F) \to F(A).$$

Now go the other way. Given $x \in F(A)$, define a natural transformation $\Psi(x) : \operatorname{Hom}_{\mathcal{C}}(A, -) \Rightarrow F$ by specifying its component at each object $B$:

$$\Psi(x)_B(f) = F(f)(x) \qquad \text{for } f : A \to B.$$

Naturality

We need to check that $\Psi(x)$ is natural. For a morphism $g : B \to C$, the naturality square asks for

$$F(g)(\Psi(x)_B(f)) = \Psi(x)_C(g \circ f).$$

Using the definition of $\Psi(x)$ and functoriality of $F$,

$$F(g)(\Psi(x)_B(f)) = F(g)(F(f)(x)) = F(g \circ f)(x) = \Psi(x)_C(g \circ f).$$

So $\Psi(x)$ is indeed a natural transformation.

The Inverses

First, start with $x \in F(A)$. Then

$$\Phi(\Psi(x)) = \Psi(x)_A(\mathrm{id}_A) = F(\mathrm{id}_A)(x) = x.$$

So $\Phi \circ \Psi = \mathrm{id}_{F(A)}$.

Now start with a natural transformation $\eta$. Let

$$x = \Phi(\eta) = \eta_A(\mathrm{id}_A).$$

For any $f : A \to B$, naturality of $\eta$ for the morphism $f$ gives

$$F(f)(\eta_A(\mathrm{id}_A)) = \eta_B(f \circ \mathrm{id}_A) = \eta_B(f).$$

But the left side is exactly $\Psi(x)_B(f)$, so

$$\Psi(\Phi(\eta))_B(f) = \eta_B(f)$$

for every object $B$ and every morphism $f : A \to B$. Hence $\Psi \circ \Phi$ is the identity.

Therefore

$$\operatorname{Nat}(\operatorname{Hom}_{\mathcal{C}}(A, -), F) \cong F(A).$$

This is Yoneda’s Lemma.