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?
 A: 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 $\alpha^{op}:G^{op}\to F^{op}$.
