[proofplan] Fix objects $A,B \in \mathcal C$. We prove directly that the map $\Psi_{A,B}$ is bijective by constructing an inverse: a natural transformation $\alpha: y(A) \to y(B)$ is sent to the morphism $\alpha_A(\operatorname{id}_A): A \to B$. Naturality forces every component of $\alpha$ to be postcomposition by this single morphism, and evaluating $y(s)$ at $\operatorname{id}_A$ recovers $s$. Since this works for every pair $A,B$, the Yoneda embedding is fully faithful. [/proofplan] [step:Define the candidate inverse by evaluating at the identity of $A$] Fix objects $A,B \in \mathcal C$. Define a function \begin{align*} \Phi_{A,B}: \operatorname{Nat}(y(A),y(B)) &\to \operatorname{Hom}_{\mathcal C}(A,B), \\ \alpha &\mapsto \alpha_A(\operatorname{id}_A). \end{align*} This is well-defined because the component \begin{align*} \alpha_A: y(A)(A) &\to y(B)(A) \end{align*} is a function \begin{align*} \alpha_A: \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(A,B), \end{align*} so $\alpha_A(\operatorname{id}_A)$ is a morphism $A \to B$ in $\mathcal C$. [guided] Fix objects $A,B \in \mathcal C$. To prove that \begin{align*} \Psi_{A,B}: \operatorname{Hom}_{\mathcal C}(A,B) &\to \operatorname{Nat}(y(A),y(B)) \end{align*} is bijective, we need to recover a morphism $A \to B$ from a natural transformation between the represented presheaves. Let $\alpha: y(A) \to y(B)$ be a natural transformation. Its component at the object $A$ is a function \begin{align*} \alpha_A: y(A)(A) &\to y(B)(A). \end{align*} By the definition of the represented presheaves, this is \begin{align*} \alpha_A: \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(A,B). \end{align*} Since $\operatorname{id}_A \in \operatorname{Hom}_{\mathcal C}(A,A)$, the value $\alpha_A(\operatorname{id}_A)$ is a morphism $A \to B$. Therefore define \begin{align*} \Phi_{A,B}: \operatorname{Nat}(y(A),y(B)) &\to \operatorname{Hom}_{\mathcal C}(A,B), \\ \alpha &\mapsto \alpha_A(\operatorname{id}_A). \end{align*} The guiding idea is that the whole natural transformation is determined by what it does to the universal element $\operatorname{id}_A$ of the represented presheaf $y(A)$ at $A$. [/guided] [/step] [step:Show that every natural transformation is postcomposition by its value on $\operatorname{id}_A$] Let $\alpha: y(A) \to y(B)$ be a natural transformation, and define \begin{align*} s := \Phi_{A,B}(\alpha) = \alpha_A(\operatorname{id}_A) \in \operatorname{Hom}_{\mathcal C}(A,B). \end{align*} We prove that $\alpha = y(s)$. Let $X \in \mathcal C$ be an object and let $f \in \operatorname{Hom}_{\mathcal C}(X,A)$. Regard $f: X \to A$ as a morphism of $\mathcal C$. Since $y(A)$ and $y(B)$ are contravariant functors, the morphism $f$ induces functions \begin{align*} y(A)(f): \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(X,A), \\ g &\mapsto g \circ f, \end{align*} and \begin{align*} y(B)(f): \operatorname{Hom}_{\mathcal C}(A,B) &\to \operatorname{Hom}_{\mathcal C}(X,B), \\ h &\mapsto h \circ f. \end{align*} Naturality of $\alpha$ with respect to $f: X \to A$ gives \begin{align*} y(B)(f) \circ \alpha_A = \alpha_X \circ y(A)(f). \end{align*} Evaluating both sides at $\operatorname{id}_A$ gives \begin{align*} y(B)(f)(\alpha_A(\operatorname{id}_A)) &= \alpha_X(y(A)(f)(\operatorname{id}_A)). \end{align*} By the definitions of $y(A)(f)$, $y(B)(f)$, and $s$, this becomes \begin{align*} s \circ f &= \alpha_X(f). \end{align*} Thus, for every object $X \in \mathcal C$ and every morphism $f: X \to A$, \begin{align*} \alpha_X(f) = s \circ f = y(s)_X(f). \end{align*} Therefore $\alpha_X = y(s)_X$ for every object $X$, so $\alpha = y(s)$. [guided] Let $\alpha: y(A) \to y(B)$ be a natural transformation. Define the morphism \begin{align*} s := \Phi_{A,B}(\alpha) = \alpha_A(\operatorname{id}_A) \in \operatorname{Hom}_{\mathcal C}(A,B). \end{align*} We must prove that $\alpha$ is exactly the natural transformation $y(s)$, meaning that each component $\alpha_X$ sends a morphism $f: X \to A$ to $s \circ f$. Fix an object $X \in \mathcal C$ and a morphism $f \in \operatorname{Hom}_{\mathcal C}(X,A)$. The morphism $f: X \to A$ gives, by contravariance of represented presheaves, a function \begin{align*} y(A)(f): \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(X,A), \\ g &\mapsto g \circ f. \end{align*} It also gives a function \begin{align*} y(B)(f): \operatorname{Hom}_{\mathcal C}(A,B) &\to \operatorname{Hom}_{\mathcal C}(X,B), \\ h &\mapsto h \circ f. \end{align*} Naturality of $\alpha$ says that the square determined by the morphism $f: X \to A$ commutes. Written as an equality of functions, this is \begin{align*} y(B)(f) \circ \alpha_A = \alpha_X \circ y(A)(f). \end{align*} Now evaluate this equality at the element $\operatorname{id}_A \in \operatorname{Hom}_{\mathcal C}(A,A)$. We obtain \begin{align*} y(B)(f)(\alpha_A(\operatorname{id}_A)) &= \alpha_X(y(A)(f)(\operatorname{id}_A)). \end{align*} The left-hand side is $s \circ f$, because $s = \alpha_A(\operatorname{id}_A)$ and $y(B)(f)$ acts by precomposition with $f$. The right-hand side is $\alpha_X(f)$, because \begin{align*} y(A)(f)(\operatorname{id}_A) = \operatorname{id}_A \circ f = f. \end{align*} Hence \begin{align*} \alpha_X(f) = s \circ f. \end{align*} Since $X$ and $f: X \to A$ were arbitrary, every component of $\alpha$ agrees with the corresponding component of $y(s)$. Therefore $\alpha = y(s)$. [/guided] [/step] [step:Verify that the two constructions are inverse functions] First let $s \in \operatorname{Hom}_{\mathcal C}(A,B)$. Then \begin{align*} (\Phi_{A,B} \circ \Psi_{A,B})(s) &= \Phi_{A,B}(y(s)) \\ &= y(s)_A(\operatorname{id}_A) \\ &= s \circ \operatorname{id}_A \\ &= s. \end{align*} Thus $\Phi_{A,B} \circ \Psi_{A,B} = \operatorname{id}_{\operatorname{Hom}_{\mathcal C}(A,B)}$. Conversely, let $\alpha \in \operatorname{Nat}(y(A),y(B))$. The previous step proves that if $s = \Phi_{A,B}(\alpha)$, then $\alpha = y(s)$. Therefore \begin{align*} (\Psi_{A,B} \circ \Phi_{A,B})(\alpha) &= \Psi_{A,B}(s) \\ &= y(s) \\ &= \alpha. \end{align*} Hence $\Psi_{A,B} \circ \Phi_{A,B} = \operatorname{id}_{\operatorname{Nat}(y(A),y(B))}$. [/step] [step:Conclude that the Yoneda embedding is fully faithful] The function $\Psi_{A,B}$ has inverse $\Phi_{A,B}$, so $\Psi_{A,B}$ is a bijection for the fixed pair of objects $A,B \in \mathcal C$. Since $A$ and $B$ were arbitrary, for every pair of objects $A,B \in \mathcal C$ the induced function \begin{align*} \operatorname{Hom}_{\mathcal C}(A,B) &\to \operatorname{Nat}(y(A),y(B)), \\ s &\mapsto y(s) \end{align*} is a bijection. Therefore the Yoneda embedding \begin{align*} y: \mathcal C &\to \widehat{\mathcal C} \end{align*} is fully faithful. [/step]