Bijection in Yoneda Lemma for Covariant Functors

Theorem
Let $C$ be a locally small category.

Let $\mathbf{Set}$ be the category of sets. Let $F : C \to \mathbf{Set}$ be a covariant functor.

Let $A\in C$ be an object.

Let $\operatorname{id}_A$ be its identity morphism.

Let $h^A = \operatorname{Hom}(A, -)$ be its covariant hom-functor.

The class of natural transformations $\operatorname{Nat}(h^A, F)$ is a small class, and:
 * $\alpha : \operatorname{Nat}(h^A, F) \to F(A) : \eta \mapsto \eta_A(\operatorname{id}_A)$
 * $\beta : F(A) \to \operatorname{Nat}(h^A, F) : u \mapsto (X \mapsto (f \mapsto (F(f))(u)))$

are reverse bijections.

Outline of proof
The crucial point is that the role of morphisms $f \in h^A(B) = \operatorname{Hom}(A, B)$ is close to their role under the hom functor $h^A$: We have $f = f \circ \operatorname{id}_A = (h^A(f))(\operatorname{id_A})$.

Natural transformations $\eta : h^A \to F$ translate what happens with $f$ under $h^A$ to something involving only $F$, so that $\eta$ is determined by $\eta_A(\operatorname{id}_A) = \alpha(\eta)$.

Making the calculation explicit, we know what the reverse bijection $\beta$ should be.

Proof
The fact that $\operatorname{Nat}(h^A, F)$ is a small class follows when we prove that the mappings are bijections.

Injectivity
Let $\eta \in \operatorname{Nat}(h^A, F)$ and $B \in C$ an object.

Let $f \in h^A(B) = \operatorname{Hom}(A, B)$ be a morphism.

Then

which shows that $\eta$ is determined by $\eta_A(\operatorname{id}_A) = \alpha(\eta)$.

Surjectivity
Let $u \in F(A)$.

Inspired by the result above, we define $\beta(u) = \eta$ by $\eta_B (f) = (F(f))(u)$ for $B \in C$ and $f \in \operatorname{Hom}(A, B)$.

Let $B, D \in C$ and $g : B \to D$ be a morphism.

We have to show that the following diagram commutes:
 * $\xymatrix{

\operatorname{Hom}(A, B) \ar[d]^{\eta_B} \ar[r]^{g_*} & \operatorname{Hom}(A, D) \ar[d]^{\eta_D} \\ F(B) \ar[r]^{F(g)}         & F(D) }$ where $g_* = \operatorname{Hom}(A, g)$ is the postcomposition mapping.

Let $f \in \operatorname{Hom}(A, B)$.

We have:

and

Reverse bijections
By construction, $\beta(\alpha(\eta)) = \eta$ for all $\eta \in \operatorname{Nat}(h^A, F)$.

It remains to show that $\alpha(\beta(u)) = u$ for all $u \in F(A)$.

Let $\eta = \beta(u)$.

Then

Also see

 * Bijection in Yoneda Lemma for Contravariant Functors