# direct proof of the dual statement of the Yoneda lemma

The dual statement of the Yoneda lemma should read:

Given any object $A$ in a locally small category $\mathsf{C}$ and any functor $F: \mathsf{C} \to \mathsf{Sets}$, we have an isomorphism $$\text{Hom}_{\mathsf{Sets^C}}(F, \text{Hom}_{\mathsf{C}}(A, -)) \cong FA$$ which is natural in both $F$ and $A$.

As an exercise, I tried to prove this co-Yoneda lemma directly (without arguing by duality with the presheaf version) but I realized that using the same technique as in the presheaf version doesn't seem to work.

Recall that in the usual proof of the presheaf version, you would chase the identity element of $\text{Hom}_{\mathsf{C}}(A, A)$ around a naturality square and show that a natural transformation from $\text{Hom}_{\mathsf{C}}(-, A)$ to any other presheaf is uniquely determined by where it sends this identity element.

The issue with applying this technique to the co-Yoneda lemma is that when we write out a corresponding naturality square for a transformation $\alpha : F \Rightarrow \text{Hom}_{\mathsf{C}}(A, -)$, we end up chasing an element of $FA$ around the diagram, and this set doesn't have any obvious "distinguished" element.

Am I missing something obvious here? Or is there a different technique that would yield a direct proof (without arguing by duality with the presheaf version)?

• You can't prove it, because it's not true. – Zhen Lin Mar 17 '15 at 19:48
• You dualised too many things at once. The correct statement is $\mathrm{Hom}(\mathrm{Hom}(A, -), F) \cong F A$. – Zhen Lin Mar 17 '15 at 20:03
• There is no error in the text. Note that $\mathcal{C}^\vee$ is defined to be $[\mathcal{C}, \mathbf{Set}]^\mathrm{op}$. So I could equally well have said you have not dualised enough things. – Zhen Lin Mar 17 '15 at 20:48
• Well, the $(-)^\mathrm{op}$ has to go somewhere – you get either $\mathcal{C}^\mathrm{op} \to [\mathcal{C}, \mathbf{Set}]$ or $\mathcal{C} \to [\mathcal{C}, \mathbf{Set}]^\mathrm{op}$. I don't see any compelling reason to prefer one to another. – Zhen Lin Mar 18 '15 at 8:18
• Of course, the nice thing about $[\mathcal C,\mathbf{Set}]^{\mathrm{op}}$ is that it's the free completion of $\mathcal C$ (when $\mathcal C$ is small). More simply, if you want to think of the Yoneda embedding as an embedding of $\mathcal C$ into something rather than an embedding of $\mathcal C ^{\mathrm{op}}$ into something, of course you've got to go this way. – tcamps Mar 19 '15 at 1:11

$$\text{Hom}_{\mathsf{Sets^C}}(\text{Hom}_{\mathsf{C}}(A, -), F) \cong FA$$