I am learning category theory and trying to prove the contravariant Yoneda lemma:

Let $\mathcal C$ be a locally small category and $F:\mathcal C\to\mathsf{Set}$ be a contravariant functor. Fix an object $c\in\mathcal C_0$. Then the function $$ \Phi:\operatorname{Nat}\left(\operatorname{Hom}_\mathcal C(-,c),F\right)\to F(c),\qquad \Phi(\alpha)=\alpha_c(\mathrm{id}_{c}) $$ is a bijection.

I want to prove it using the covariant Yoneda lemma and duality, but I am a little confused. This stackexchange question partially explains the answer, but I think it is sketchy and it does not explain how the natural transformations between contravariant functors correspond to the usual natural transformations. This is exactly what I am confused by.

Can somebody give a fully rigorous proof of this fact? Thank you!


1 Answer 1


A contravariant functor $C\to D$ is the same thing as a covariant functor $C^{op}\to D.$ So the category of contravariant functors $C\to D$ is the same thing as the category of covariant functors $C^{op}\to D.$ So a natural transformation between two contravariant functors $ C\to D$ is the same thing as a natural transformation between two covariant functors $C^{op}\to D$.

There is no "rigorous proof" to be had here... it is just a matter of definitions.

The covariant Yoneda lemma is $$ \operatorname{Hom}_{\mathrm{Set}^C}(\operatorname{Hom}_C(c,-), F)\cong F(c)$$ for $c\in C$ and $F:C\to \mathrm{Set}$.

Substituting $C$ with $C^{op}$ and using the fact that $\operatorname{Hom_{C^{op}}(c,-) = \operatorname{Hom}_C(-,c)}$ gives $$ \operatorname{Hom}_{\mathrm{Set}^{C^{op}}}(\operatorname{Hom}_C(-,c), F)\cong F(c)$$ for $c\in C$ and $F: C^{op}\to \mathrm{Set}.$ This is the contravariant Yoneda lemma.

  • $\begingroup$ Would it be more precise to say that $\operatorname{Hom}_{\mathcal C}(-,c)$ and $\operatorname{Hom}_{\mathcal C^\mathrm{op}}(c,-)$ are naturally isomorphic functors from $\mathcal C^\mathrm{op}$ to $\mathsf{Set}$, and hence the collection of natural transformations out of them are in bijection? $\endgroup$
    – Sardines
    Commented Aug 20, 2023 at 22:45
  • $\begingroup$ @Zhuo Maybe. I meant literal identity, though. The way I view it, the objects of $C^{op}$ are literally the objects of $C$ and the set of arrows $c\to d$ in $C^{op}$ is literally the set of arrows $d\to c$ in $C.$ But there may be some benefit to viewing them as corresponding but distinct. $\endgroup$ Commented Aug 20, 2023 at 23:33

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