Given a natural transformation $\alpha : h^A \Rightarrow F$, the component at $A$ is a function

$\alpha_A : \mathrm{Hom}(A, A) \to F(A)$

We define

$\Phi(\alpha) = \alpha_A(\mathrm{id}_A) \in F(A)$

This is the "evaluation at the identity" map. The key insight is that, by naturality, the entire natural transformation $\alpha$ is determined by this single element.

Given an element $x \in F(A)$, we must construct a natural transformation $\Psi(x) : h^A \Rightarrow F$. For each object $X \in \mathcal{C}$, we define the component

$\Psi(x)_X : \mathrm{Hom}(X, A) \to F(X), \qquad f \mapsto F(f)(x)$

Here $F(f) : F(A) \to F(X)$ is the image of $f : X \to A$ under the contravariant functor $F$ (recall $F : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}$, so a morphism $f : X \to A$ in $\mathcal{C}$ gives $F(f) : F(A) \to F(X)$).

Claim: $\Psi(x)$ is a natural transformation.

We must verify that for every morphism $g : X \to Y$ in $\mathcal{C}$, the following diagram commutes:

where $g^* = h^A(g)$ is precomposition by $g$, i.e., $g^*(f) = f \circ g$.

Verification: Let $f \in \mathrm{Hom}(Y, A)$.

Going right then down: $F(g)\bigl(\Psi(x)_Y(f)\bigr) = F(g)\bigl(F(f)(x)\bigr) = \bigl(F(g) \circ F(f)\bigr)(x) = F(f \circ g)(x)$

where the last equality uses the functor law $F(g) \circ F(f) = F(f \circ g)$ (note the reversal since $F$ is contravariant).

Going down then right: $\Psi(x)_X\bigl(g^*(f)\bigr) = \Psi(x)_X(f \circ g) = F(f \circ g)(x)$

Both paths yield $F(f \circ g)(x)$, so the diagram commutes. Thus $\Psi(x) : h^A \Rightarrow F$ is indeed a natural transformation.

We verify both compositions are identities.

($\Phi \circ \Psi = \mathrm{id}_{F(A)}$): Let $x \in F(A)$. Then

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

where we used the functor law $F(\mathrm{id}_A) = \mathrm{id}_{F(A)}$.

($\Psi \circ \Phi = \mathrm{id}_{\mathrm{Nat}(h^A, F)}$): Let $\alpha : h^A \Rightarrow F$ be a natural transformation. We must show $\Psi(\Phi(\alpha)) = \alpha$, i.e., for every object $X$ and every $f : X \to A$:

$\Psi(\Phi(\alpha))_X(f) = \alpha_X(f)$

The left side is: $\Psi(\Phi(\alpha))_X(f) = F(f)(\Phi(\alpha)) = F(f)(\alpha_A(\mathrm{id}_A))$

Now we use the naturality of $\alpha$ at the morphism $f : X \to A$. The naturality square for $\alpha$ at $f$ gives:

Commutativity says: $\alpha_X \circ f^* = F(f) \circ \alpha_A$. Evaluating both sides at $\mathrm{id}_A$:

$\alpha_X(f^*(\mathrm{id}_A)) = F(f)(\alpha_A(\mathrm{id}_A))$

Since $f^*(\mathrm{id}_A) = \mathrm{id}_A \circ f = f$, the left side becomes $\alpha_X(f)$. Therefore:

$\Psi(\Phi(\alpha))_X(f) = F(f)(\alpha_A(\mathrm{id}_A)) = \alpha_X(f)$

Since this holds for all $X$ and all $f$, we conclude $\Psi(\Phi(\alpha)) = \alpha$.

Hence $\Phi$ and $\Psi$ are mutually inverse, and $\Phi : \mathrm{Nat}(h^A, F) \xrightarrow{\sim} F(A)$ is a bijection. $\square$

We show $\Phi$ is natural in the variable $F$. Let $\sigma : F \Rightarrow G$ be a natural transformation of presheaves. We must show that the following square commutes:

where $\sigma_* : \mathrm{Nat}(h^A, F) \to \mathrm{Nat}(h^A, G)$ is post-composition by $\sigma$, sending $\alpha \mapsto \sigma \circ \alpha$.

Let $\alpha \in \mathrm{Nat}(h^A, F)$.

Going right then down: $\sigma_A(\Phi_{A,F}(\alpha)) = \sigma_A(\alpha_A(\mathrm{id}_A))$

Going down then right: $\Phi_{A,G}(\sigma_*(\alpha)) = \Phi_{A,G}(\sigma \circ \alpha) = (\sigma \circ \alpha)_A(\mathrm{id}_A) = \sigma_A(\alpha_A(\mathrm{id}_A))$

These are equal, so $\Phi$ is natural in $F$. $\square$

We show $\Phi$ is natural in $A$. We must be careful about variance. A morphism $u : A \to B$ in $\mathcal{C}$ induces:

• A natural transformation $h^u = u_* : h^A \Rightarrow h^B$ (post-composition by $u$), which gives a pullback map $u^* : \mathrm{Nat}(h^B, F) \to \mathrm{Nat}(h^A, F)$ by pre-composition with $h^u$.
• A map $F(u) : F(B) \to F(A)$ (since $F$ is contravariant).

Both sides are contravariant in $A$, so the naturality is with respect to the functor $\mathcal{C}^{\mathrm{op}} \to \mathbf{Set}$.

The naturality square we must verify is:

where $u^*(\beta) = \beta \circ h^u$ for $\beta : h^B \Rightarrow F$.

Let $\beta \in \mathrm{Nat}(h^B, F)$.

Going right then down: $F(u)(\Phi_{B,F}(\beta)) = F(u)(\beta_B(\mathrm{id}_B))$

Going down then right: $\Phi_{A,F}(u^*(\beta)) = \Phi_{A,F}(\beta \circ h^u) = (\beta \circ h^u)_A(\mathrm{id}_A) = \beta_A(h^u_A(\mathrm{id}_A)) = \beta_A(u \circ \mathrm{id}_A) = \beta_A(u)$

Now, by the naturality of $\beta$ at $u : A \to B$:

we get $\beta_A(u^*(\mathrm{id}_B)) = F(u)(\beta_B(\mathrm{id}_B))$, i.e., $\beta_A(u) = F(u)(\beta_B(\mathrm{id}_B))$.

Both paths give $\beta_A(u) = F(u)(\beta_B(\mathrm{id}_B))$, so the diagram commutes. Hence $\Phi$ is natural in $A$. $\square$

Apply the Yoneda Lemma with $F = h^B = \mathrm{Hom}(-, B)$:

$\mathrm{Nat}(h^A, h^B) \cong h^B(A) = \mathrm{Hom}(A, B)$

The bijection sends a natural transformation $\alpha : h^A \Rightarrow h^B$ to $\alpha_A(\mathrm{id}_A) \in \mathrm{Hom}(A, B)$, and the inverse sends a morphism $f : A \to B$ to the natural transformation $f_* : h^A \Rightarrow h^B$ given by post-composition with $f$.

Faithfulness: If $f_* = g_*$ as natural transformations, then $(f_*)_A(\mathrm{id}_A) = (g_*)_A(\mathrm{id}_A)$, i.e., $f \circ \mathrm{id}_A = g \circ \mathrm{id}_A$, so $f = g$.

Fullness: Every natural transformation $\alpha : h^A \Rightarrow h^B$ equals $f_*$ where $f = \alpha_A(\mathrm{id}_A)$. This is precisely the content of $\Psi \circ \Phi = \mathrm{id}$. $\square$

Since $\mathbf{y}$ is fully faithful, it reflects isomorphisms. Concretely:

($\Rightarrow$): If $f : A \xrightarrow{\sim} B$ is an isomorphism, then $f_* : h^A \Rightarrow h^B$ is a natural isomorphism with inverse $(f^{-1})_*$.

($\Leftarrow$): Suppose $\alpha : h^A \xrightarrow{\sim} h^B$ is a natural isomorphism. By full faithfulness, $\alpha = f_*$ for some morphism $f : A \to B$, and $\alpha^{-1} = g_*$ for some $g : B \to A$. Then:

Note that post-composition reverses the order: $(g \circ f)_* = f_* \circ g_*$ does not hold in general. Instead, $\mathbf{y}$ is a covariant functor, so $(g \circ f)_* = g_* \circ f_*$. We compute:

$(\mathrm{id}_A)_* = \alpha^{-1} \circ \alpha = g_* \circ f_* = (g \circ f)_*$

Since $\mathbf{y}$ is faithful, $g \circ f = \mathrm{id}_A$. Similarly, $\alpha \circ \alpha^{-1} = \mathrm{id}_{h^B}$ gives $(f \circ g)_* = (\mathrm{id}_B)_*$, so $f \circ g = \mathrm{id}_B$. Therefore $f$ is an isomorphism. $\square$