Yoneda Lemma

by iarnab_kundu, Nov 25, 2018, 5:46 PM

We write some of the proofs of the very useful Yoneda Lemma.

Let $\mathcal{C}$ be a locally small category. That means that the $Hom$ -sets of category $\mathcal{C}$ are sets. Let $\text{Sets}$ denote the category of all sets. Given an object $A$ we might construct two functors $h^A:\mathcal{C}\to\text{Sets}$ and $h_A:\mathcal{C}^{op}\to\text{Sets}$ respectively. We consider the first covariant functor. If there is a morphism $f:A\to B$ in $\mathcal{C}$ then for every object $C$ we have a map of sets $\text{Hom}(B,C)\to\text{Hom}(A,C)$ given by pre-composition by $f$ . This map is natural in the sense that if there is a morphism $g:C\to D$ between two objects $\mathcal{C}$ then there is a commutative digaram

This means that there is a natural transformation $h^B\Rightarrow h^A$ . Therefore given a morphism $A\to B$ we have a "morphism" of functors $h^B\to h^A$ . On the other hand, if we are given a "morphism" of functors $h^B\to h^A$ , can we say that it is induced by a morphism $A\to B$ ? The answer to the question is given by Yoneda Lemma which we state in three different ways.
Let $\text{Set}^{\mathcal{C}}$ denote the category of covariant functors from $\mathcal{C}$ to $\text{Sets}$ . If we have two such functors, that is $F,G:\mathcal{C}\to\text{Sets}$ then $\text{Hom}(F,G)$ is the collection of all natural transformations between them. It is not clear to me at this juncture whether this is a locally small category or not.
We have a contravariant functor $i:\mathcal{C}^{op}\to\text{Sets}^{\mathcal{C}}$ given by $A\mapsto h^A$ . The calculations above show that this is indeed a functor.

Statement 1 - Let $F:\mathcal{C}\to\text{Sets}$ be a covariant functor. Let $A\mathcal{C}$ be an object. Then there is a natural bijection of sets $\text{Hom}(h^A,F)\to F(A)$ .

Remark - We need to explain the naturality above. Let $\mathcal{C}\times\text{Sets}^{\mathcal{C}}$ be the product category, assuming it exists. Then there are two functors $\mathcal{C}\times\text{Sets}^{\mathcal{C}}\to\text{Sets}$ given respectively by $\text{Hom}(h^A,F)$ and by $F(A)$ . Then Yoneda Lemma states that these two functors are naturally isomorphic.

Corollary 1- Given two objects $A,B\in\mathcal{C}$ , if the two functors $h^A$ and $h^B$ are isomorphic then so are $A$ and $B$ .
Proof- We assume that $\eta$ is an isomorphism, that is there exists $\theta$ such that $\theta\circ\eta$ is the identity map of $h^A$ and $\eta\circ\theta$ is the identity map of $h^B$ . We look at the proof of Yoneda Lemma. Putting $F=h^B$ gives us that any natural transformation $\eta:h^A\to h^B$ is given by $f\in h^B(A)=\text{Hom}(B,A)$ where $f=\eta_A(id_A)$ . Similarly we obtain $g:A\to B$ given by $g=\theta_B(id_B)$ . The proof of Yoneda Lemma tells us that $\eta_A$ is given by pre-composition by $g$ , $\theta_A$ is given by pre-composition by $f$ and therefore their composition $\theta_A\circ\eta_A=(theta\circ\eta)_A$ is given by pre-composition by $g\circ f$ . But $\theta\circ\eta$ is the identity map and is therefore given by the pre-composition with the identity map. Therefore $id_A\circ g\circ f = \text{id}_A$ and hence $g\circ f=\text{id}_A$ .

Similarly $f\circ g=id_B$ .
Corollory 2- The functor $i:\mathcal{C}^{op}\to\text{Sets}^{\mathcal{C}}$ given by $A\mapsto h^A$ is fully faithful.
Proof - We first check that $i$ is a functor. It is clear that identity map is taken to the identity natural transformation. Secondly if we have three objects $A,B,C\in\mathcal{C}$ and three morphisms $f:A\to B, g:B\to C,$ and $h:A\to C$ such that $h= g\circ f$ then we need to check that $i(f)\circ i(g)=i(h)$ . We can check at every object $D\in\mathcal{C}$ . Then $i(f)_D:h^B(D)\to h^A(D)$ is given by pre-composition by $f$ , similarly for g and h. Thus it can be checked.
Now we look at the proof of Yoneda Lemma. We put $F=h^B$ and we obtain a map of sets $\text{Hom}(h^A,h^B)\to h^B(A)=\text{Hom}(B,A)$ given by $\eta\mapsto\eta_A(\text{id}_A)$ . But we also have the above map $\text{Hom}(B,A)\to\text{Hom}(h^A,h^B)$ given by $f\mapsto i(f)$ . But $i(f)_A(\text{id}_A)=id_A\circ f$ since $i(f)$ on objects is given by pre-composition by $f$ . For the other direction, we note that any map $\eta:h^A\to h^B$ is given by $\eta_A(id_A)$ .

This post has been edited 4 times. Last edited by iarnab_kundu, Nov 25, 2018, 9:09 PM

category theory

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Cracking the Nut Open

Yoneda Lemma

by iarnab_kundu, Nov 25, 2018, 5:46 PM

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