Yoneda Lemma

Yoneda Embedding

The Yoneda embedding is the functor $y:\mathbf{C}\to {\mathbf{Sets}}^{{\mathbf{C}}^{\text{op}}}$ taking $C\in \text{obj}\mathbf{C}$ to the contravariant representable functor: $y(C)={\text{Hom}}_{\mathbf{C}}(-,C):{\mathbf{C}}^{\mathrm{op}}\to \mathbf{Sets},$and taking $f:C\to D$ to the natural transformation: $y(f)={\text{Hom}}_{\mathbf{C}}(-,f):{\text{Hom}}_{\mathbf{C}}(-,C)\to {\mathrm{Hom}}_{\mathbf{C}}(-,D).$

Yoneda Lemma

Let $\mathbf{C}$ be a locally small category. For any object $C\in \text{obj}\mathbf{C}$ and functor $F\in {\mathbf{Sets}}^{{\mathbf{C}}^{\text{op}}}$, there is an isomorphism ${\text{Hom}}_{{\mathbf{Sets}}^{{\mathbf{C}}^{\text{op}}}}(y(C),F)\cong F(C)$which is natural in both $F$ and $C$.

• Naturality in $F$ means that, given any $\theta :F\to G$, the following diagram commutes:
• Naturality in $C$ means that, given any $h:C\to D$, the following diagram commutes:

Theorem

The Yoneda embedding $y:\mathbf{C}\to {\mathbf{Sets}}^{{\mathbf{C}}^{\text{op}}}$ is full and faithful.

Proof For any objects $C,D\in \text{obj}\mathbf{C}$, we have an isomorphism: ${\text{Hom}}_{\mathbf{C}}(C,D)=yD(C)\cong {\text{Hom}}_{{\mathbf{Sets}}^{{\mathbf{C}}^{\text{op}}}}(y(C),y(D))$And this isomorphism is indeed induced by the functor $y$, since it takes an element $h:C\to D$ of $yD(C)$ to the natural transformation ${\theta }_h:yC\to yD$ given by $\begin{array}{rl}({\vartheta }_h{)}_{C^{\prime }}(f:C^{\prime }\to C)& =yD(f)(h)\\ & ={\mathrm{Hom}}_{\mathbf{C}}(f,D)(h)\\ & =h\circ f\\ & =(yh{)}_{C^{\prime }}(f)\end{array}$where $yh:yC\to yD$ has component at $C^{\prime }$:$(yh{)}_{C^{\prime }}:\mathrm{Hom}(C^{\prime },C)⟶\mathrm{Hom}(C^{\prime },D),f\mapsto h\circ f$So, ${\vartheta }_h=y(h)$.

Corollary Yoneda Principle Given objects $A$ and $B$ in any locally small category $\mathbf{C}$, $y(A)\cong y(B)$ implies $A\cong B$, where $y$ is the Yoneda embedding.

Prop If the cartesian closed category $\mathbf{C}$ has coproducts, then $\mathbf{C}$ is distributive, that is, there is always a canonical isomorphism: $(A\times B)+(A\times C)\cong A\times (B+C)$