Yoneda Lemma

Yoneda lemma : Let $\mathcal{C}$ be a (locally) small category. Then

naturally in $A\in \mathcal{C}$ and $X\in [\mathcal{C}^{\rm{op}},\rm{Set}]$. Terminology :

$\mathcal{C}$ is a category mentioned in the lemma, $\mathcal{C}^{\rm{op}}$ is the opposite category associated to $\mathcal{C}$.
$\rm{Set}$ is the category with elements as sets and morphisms as functions.
$X:\mathcal{C}^{op}\rightarrow \rm{Set}$ is a functor.
Given $A\in \mathcal{C}$, $H_A$ is the functor $H_A:\mathcal{C}^{\rm{op}}\rightarrow \rm{Set}$ given by $B\mapsto \mathcal{C}(B,A)$.
$[\mathcal{C}^{\rm{op}},\rm{Set}]$ is the category with functors from $\mathcal{C}^{\rm{op}}$ to $\rm{Set}$ as elements and natural transformations between these functors as morphisms.
$[\mathcal{C},\rm{Set}]$ is the category with functors from $\mathcal{C}$ to $rm{Set}$ as elements and natural transformations between these functors as morphisms.

Given $X,A$ as above, $X(A)$ is a set and $[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)$ is a set. Yoneda lemma says that there is a bijection between these sets, natural bijection in both $A$ and $X$. Natural transformation : Let $\mathcal{A},\mathcal{B}$ be two categories and $F,G:\mathcal{A}\rightarrow \mathcal{B}$ be both contravariant or both covariant functors. A natural transformation $\eta:F\rightarrow G$ is a family of arrows (morphisms) $F(A)\xrightarrow{\eta(A)}G(A)$ such that for each $A\xrightarrow{f}A'$ in $\mathcal{A}$ the following appropriate diagram commutes. ql_f5f1b45cf5b9428b42e7bcfd77134d7e_l3

Now that we have defined terminology used in the statement, we will now prove the statement. We will first give a bijection

and then prove that it is natural in $A$ and $X$. Construction of Bijective map : Let $\eta\in [\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)$ i.e., $\eta:H_A\rightarrow X$ is a natural transformation. We want to assign an element in $X(A)$ with this $\eta$. It is only natural to consider the map $\eta(A):H_A(A)\rightarrow X(A)$. The set $H_A(A)$ has a special element namely $1_A\in H_A(A)$, its image $\eta(A)(1_A)\in X(A)$. Define $\Phi(\eta)=\eta(A)(1_A)$. This give a map

$\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$.

We will now prove that this map is injective and surjective. Surjectivity of $\Phi$ : Let $a\in X(A)$. We want to assign a natural trasformation $\eta:H_A\rightarrow X$ with this $a$ such that $\Phi(\eta)=a$. We need to define $\eta(B):H_A(B)\rightarrow X(B)$ for each $B\in \rm{Ob}(\mathcal{C})$. Let $B\in \rm{Ob}(\mathcal{C})$ be fixed and $f\in H_A(B)$ i.e., $f:B\rightarrow A$. The functor $X:\mathcal{C}^{\rm{op}}\rightarrow \text{Set}$ induces $X(f):X(A)\rightarrow X(B)$. We have $X(f)(a)\in X(B)$. Define $\eta(B)(f)=X(f)(a)$. This gives a natural transformation $\eta:H_A\rightarrow X$ and

$\eta(A)(1_A)=X(1_A)(a)=1_{X(A)}(a)=a$.

So, $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$ is surjective.

Injectivity of $\Phi$ : Suppose $\eta_1,\eta_2\in [\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)$ be such that $\Phi(\eta_1)=\Phi(\eta_2)$ i.e., $\eta_1(A)(1_A)=\eta_2(A)(1_A)$. We prove that $\eta_1=\eta_2$ i.e., $\eta_1(B)=\eta_2(B)$ for every $B\in \mathcal{C}$ i.e., $\eta_1(B)(f)=\eta_2(B)(f)$ for every $f\in H_A(B)$.

Fix $B\in \mathcal{C}$ and $f\in H_A(B)$ i.e., $f:B\rightarrow A$ in $\mathcal{C}$. We then have following commutative diagrams ql_9ae4d01e872f982000796aef89e2f7d6_l3

We have

. So,

Thus, $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$ is injective. So, we have bijective correspondence

Bijection is natural in $A$ : We prove that the above bijection is natural in $A$ i.e., given $A,B\in \mathcal{C}$ with $f:B\rightarrow A$ the following diagram is commutative ql_a7dce34a06e3d903e27c7dd5e3e86312_l3

where, for $\eta\in [\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)$, $\varphi(f)$ sends $\eta$ to the composition $\eta\circ f^*: H_B\xrightarrow{f^*} H_A\xrightarrow{\eta} X$. For $\eta\in [\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)$ we have $\varphi(f)(\eta) :H_B\rightarrow X$. Let $C\in \mathcal{C}$ then, we have

We now prove that

We have

The following commutative diagram ql_ec375e8f1d8b7efa8bca8e04d5295373_l3

says that $X(f)(\eta(A)(1_A))=\eta(B)(H_A(f)(1_A))$. As $H_A(f)(1_A)=f$, we have

Now,

We have defined $\varphi(f)$ as $\varphi(f)(\eta)(C)(g)=\eta(C)(f\circ g)$. So,

So, we see that

So, the bijection $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$ is natural in $A$. Bijection is natural in $X$ : We prove that the above bijection is natural in $X$ i.e., for functors $X,Y:\mathcal{C}^{\rm{op}}\rightarrow \rm{Set}$ and a natural transformation $\tilde{\eta}:X\rightarrow Y$ we prove that the following diagram is commutative ql_bcc85b77d21e8dcdb72eac4665076b29_l3

Let $\eta:H_A\rightarrow X$ be a natural transformation. Then, $\varphi(\eta)=\tilde{\eta}\circ \eta:H_A\rightarrow Y$. We have ql_bed391786bb6afe9101e5a5d37298891_l3

So,

So, the bijection $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$ is natural in $X$. Special case: We have seen that $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$ is natural in $A$ and $X$. Let $X=H_B$ for some $B\in \mathcal{C}$ then, we have $\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,H_B)\rightarrow H_B(A)=\mathcal{C}(A,B)$. So, let $\mathcal{D}$ denote the category $[\mathcal{C}^{\rm{op}},\rm{Set}]$. Then,

Here $\mathcal{C}(A,B)$ denotes the collection of morphisms and $\mathcal{D}(H_A,H_B)$ denotes the collection of all natural transformations. We have

Covariant version of Yoneda lemma : Let $\mathcal{C}$ be a (locally) small category. Then

naturally in $A\in \mathcal{C}$ and $X\in [\mathcal{C},\rm{Set}]$, where $H^A:\mathcal{C}\rightarrow \rm{Set}$ is given by $B\mapsto \mathcal{C}(A,B)$. As a special case when $X=H^B$ for some $B\in \mathcal{C}$ we have

Here $\mathcal{C}(B,A)$ denotes the collection of morphisms and $\mathcal{D}(H^A,H^B)$ denotes the collection of all natural transformations. We have