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)?

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    $\begingroup$ You can't prove it, because it's not true. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2015 at 19:48
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    $\begingroup$ You dualised too many things at once. The correct statement is $\mathrm{Hom}(\mathrm{Hom}(A, -), F) \cong F A$. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2015 at 20:03
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    $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2015 at 20:48
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    $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Commented Mar 18, 2015 at 8:18
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    $\begingroup$ 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. $\endgroup$
    – tcamps
    Commented Mar 19, 2015 at 1:11

1 Answer 1


To "answer" this question so it no longer shows up as unanswered:

Zhen Lin pointed out that I had dualized too many things at once (or not enough, from a different point of view) so the correct statement should be

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

so the proof proceeds in the same manner as the usual presheaf version of the Yoneda lemma, if you want to do it directly.


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