The Yoneda Embedding Theorem

Let $\mathbf{C}$ be a locally small category. Then for each object $A$ of $\mathbf{C}$ we define the contravariant representable functor $H^A:=hom_{\mathbf{C}}(\_,A):\mathbf{C}^{op}\rightarrow \mathbf{Set}$. Then by the Yoneda Lemma of the previous post we know that for any presheaf $F$ (functor from $\mathbf{C}^{op}$ to $\mathbf{Set}$) we have $Nat(H^A,F)\cong FA$ as sets under the map $Nat(H^A,F)\ni \eta \mapsto \eta_A(1_A)$. Thus, in particular, if we take $F=H^B$ then $Nat(H^A,H^B)\cong H^B(A) = \mathbf{C}(A,B)$.

A fuctor $F$ between categories $\mathbf{C}$ and $\mathbf{D}$ is called an embedding if it is injective on objects and full and faithful; where it is full when it induces a surjective maps from $\mathbf{C}(A,B)$ to $\mathbf{D}(FA,FB)$ for every $A,B$, is faithful when it induces an injective map from $\mathbf{C}(A,B)$ to $\mathbf{D}(FA,FB)$ for every $A,B$, and hence is full and faithful when it induces a bijection between $\mathbf{C}(A,B)$ to $\mathbf{D}(FA,FB)$ for every $\mathbf{C}$ objects $A$ and $B$.

Again letting $\mathbf{C}$ be a small category, define the Yoneda Embedding $H: \mathbf{C} \rightarrow \mathbf{Set}^{\mathbf{C}^{op}}$ by the assignments
\[A\mapsto H^A,\;\;\;\;\;\;(f:A\rightarrow B)\mapsto Hf :: Hf_C = f\_.\]
That is $H(A) = H^A$ on objects and if $f:A\rightarrow B$ is an arrow, then $Hf$ is the natural transformation $H^A\rightarrow H^B$ with components $(Hf)_C = f\_$ for each object $C$.

We have two things that we need to check for this assignment; first, that $H$ is a functor, which includes checking that $Hf$ is a natural transformation, and, second, that $H$ is an embedding.

Let's check that $H$ is a functor. We begin by showing $Hf$ is actually a natural transformation. Let $C,D$ be two $\mathbf{C}$ objects and $g^{op}:C\rightarrow D$ an arrow in $\mathbf{C}^{op}$ (opposite of the arrow $g:D\rightarrow C$ in $\mathbf{C}$). We must then check that the naturality square commutes, i.e. $(H^Bg)(Hf)_C=((Hf)_D)H^Ag)$. So let $h\in H^AC = \mathbf{C}(C,A)$. Then \[(H^Bg)(Hf)_C(h) = (H^Bg)(fh) =fhg = ((Hf)_D)(hg) = ((Hf)_D)H^Ag)(h).\] Hence the square commutes as desired.

Clearly to show that $H$ respects function composition it suffices to show that for any $f:A\rightarrow B, g:B\rightarrow C$ and any $\mathbf{C}$ object $D$ we have $((Hg)_D)(Hf)_D = (H(gf))_D.$ So let $h\in H^AD = \mathbf{C}(D,A)$. Then
\[((Hg)_D)(Hf)_D(h) = gfh = H(gf))_D(h).\] So $H$ respects composition. Lastly, clearly we have $(H1_A)_C = 1_A\_ = \_ = 1_{H^AC} = (1_{H(A)})(C)$. Therefore $H$ is a functor.

Now let's complete the proof of the Yoneda Embedding Theorem by showing that $H$ is an embedding.

Well, clearly $H$ is 1-1 on objects since $H^A = H^B$ implies $H^AA = \mathbf{C}(A,A) = \mathbf{C}(B,A) = H^BA$. So, in particular, $1_A \in \mathbf{C}(B,A)$, which implies (by uniqueness on domain and codomain) that $A=B$.

For the rest, let $A,B$ be $\mathbf{C}$ objects. Then from the conclusion we drew from the Yoneda Lemma above, we have $Nat(H^A,H^B)\cong H^B(A) = \mathbf{C}(A,B)$.
But then $Nat(H^A,H^B) = \mathbf{Set}^{\mathbf{C}^{op}}(H^A,H^B) = \mathbf{Set}^{\mathbf{C}^{op}}(HA,HB)\cong \mathbf{C}(A,B)$ in set, i.e. $H$ is full and faithful. Hence $H$ is an embedding and the proof is complete.