# Yoneda Lemma: definition of Yoneda functors

In stating the Yoneda Lemma, my category theory book (Kashiwara's Categories and Sheaves) makes the following definition:

Let $\mathcal{C}$ be a $\mathcal{U}$-category, where $\mathcal{U}$ is our universe. We define the big categories $$\mathcal{C}^{\wedge}_{\mathcal{U}}: \text{The category of functors from \mathcal{C}^{\text{op}} to }\mathcal{U}\text{-set}$$ $$\mathcal{C}^{\vee}_{\mathcal{U}}: \text{The category of functors from \mathcal{C}^{\text{op}} to }(\mathcal{U}\text{-set})^{\text{op}}.$$ and the functors $$\mathrm{h}_\mathcal{C}: \mathcal{C} \to \mathcal{C}^{\wedge}_{\mathcal{U}}: X \mapsto \operatorname{Hom}_{\mathcal{C}}(\cdotp, X)$$ $$\mathrm{k}_\mathcal{C}: \mathcal{C} \to \mathcal{C}^{\vee}_{\mathcal{U}}: X \mapsto \operatorname{Hom}_{\mathcal{C}}(X, \cdotp).$$

There is a natural isomorphism $$\mathcal{C}^{\vee} \simeq \mathcal{C}^{\text{op}\wedge \text{op}}$$ and $\mathcal{C^{\vee}}$ is the opposite big category to the category of functors from $\mathcal{C}$ to Set. Hence, for $X \in \mathcal{C}$, $\mathrm{k}_\mathcal{C}(X) = (\mathrm{h}_\mathcal{C^{\text{op}}}(X^{\text{op}}))^{\text{op}}$.

I don't understand why we can't (shouldn't) view things like $\text{Hom}_{\mathcal{C}}(X, \cdotp)$ as functors from $\mathcal{C}$ to Sets. They seem to be saying that somehow, the arrows in $\mathcal{C}^{\vee}$ point the wrong way for that to be the case. What does that mean?

• just try to define what the functor does to a function $f:S\to T$ and see what happens. – Ittay Weiss Oct 18 '14 at 3:20
• @IttayWeiss $\text{Hom}(X, \cdotp):(S \xrightarrow{f}T) \mapsto (\text{Hom}(X, S) \xrightarrow{f \circ} \text{Hom}(X, T)$. What's wrong with that? It should be saying something about the arrows between the functors, right (i.e. natural transformations)? – Eric Auld Oct 18 '14 at 3:30

This is actually pretty easy to justify, but even easier to forget (as I did) or to never notice-as happens to readers of approximately every category theory book except Kashiwara-Schapira. Probably many students guess that $[\mathcal{C}^{op},\mathbf{Set}]$ is the opposite of $[\mathcal{C},\mathbf{Set}]$, even though this is badly wrong-I can think of at least one who did, anyway.
Zoom out, and show in general that the functor categories $[\mathcal{C},\mathcal{D}]$ and $[\mathcal{C}^{op},\mathcal{D}^{op}]$ are anti-equivalent. They're actually dual-isomorphic, so the equivalence is given by the identity on objects. But on morphisms, that is, natural transformations, we have to turn things around. If $\alpha:F\to G$ is a morphism in $[\mathcal{C},\mathcal{D}]$ then $\alpha$ is just an $\text{ob}(\mathcal{C})$-indexed collection of morphisms $\alpha_A:FA\to GA$ in $\mathcal{D}$ satisfying the naturality property. So in our equivalence we will associate $\alpha$ with the exact same indexed collection of morphisms, that is, $(\alpha^{op})_{A^{op}}=(\alpha_A)^{op}$. What else? But $\alpha_A^{op}: G^{op} A^{op}=(GA)^{op}\to (FA)^{op}=F^{op} A^{op}$, so $\alpha^{op}:G^{op}\to F^{op}$.
• Should it say "an $\text{ob}(\mathcal{C})$-indexed collection of morphisms $\alpha_A: FA \to GA$"? Not $FA'$? – Eric Auld Oct 18 '14 at 4:18
• OK, I think I understand your point that $[\mathcal{C},\mathcal{D}]$ and $[\mathcal{C}^{\text{op}}, \mathcal{D}^{\text{op}}]$ are not the same. Can you illustrate why $\text{Hom}(X,\cdotp)$ is an object in the latter rather than the former? – Eric Auld Oct 18 '14 at 4:25
• Yes, thanks for the correction. That functor is an object of both categories-the idea of the answer is to show the two functor categories are opposite, but opposite categories have the same (depending on definitions, a canonically isomorphic class of) objects! Sp when I write things like $F^{op}$ it might be obfuscatory: this is an object in the opposite category, but it's exactly the same as a functor. – Kevin Arlin Oct 18 '14 at 16:20
• Oh, so the point is that we want to define $\mathcal{C}^{\vee}$ and $\mathcal{C}^{\wedge}$ to be opposite categories? – Eric Auld Oct 18 '14 at 17:52
• No, it's $\mathcal C^\vee$ and $[\mathcal C,\mathbf{Set}]$ that are opposite: what you said is the common mistake I mention in my first paragraph. If you think of some examples you'll see there's no way to get a contravariant functor from a covariant one in general. – Kevin Arlin Oct 18 '14 at 19:22