The opposite of the covariant hom functor

Question

I need to define the opposite of the covariant hom functor $\mathbb C(C,-):\mathbb C\to \mathbf {Set}$ for some fixed object $C$ of the category $\mathbb C$, that is, I need to define $\mathbb C(C,-)^{op}:\mathbb C^{op}\to \mathbf {Set}^{op}$. After some struggle, it seems to me that I must define $\mathbb C(C,-)^{op}(C')$ to be $\mathbb C^{op}(C,C')$. Which puzzles me since I expected $\mathbb C(C,-)^{op}(C')$ to be $\mathbb C(C,C')$ or equivalently $\mathbb C^{op}(C',C)$. Is my definition correct? If yes, why? If not, what is the correct definition?

Somebody answered that indeed the correct definition is $\mathbb C(C,C')$, but then how would you define $\mathbb C(C,-)^{op}(f^{op}:C'\to C'')$, which is a morphism in $\mathbf{Set}^{op}$? This is where got stuck and was forced to use the seemingly incorrect definition for the hom functor action on objects.

Answer 1

You are correct that it should be the case that $\mathbb{C}(C,{-})^{\mathrm{op}}(C') = \mathbb{C}(C,C')$. More generally, if $F : \mathbb{C} \to \mathbb{D}$ is a functor, then $F^{\mathrm{op}} : \mathbb{C}^{\mathrm{op}} \to \mathbb{D}^{\mathrm{op}}$ is defined on objects by $F^{\mathrm{op}}(C)=F(C)$.

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Edit to answer your updated question:

In general, the functor $F^{\mathrm{op}} : \mathbb{C}^{\mathrm{op}} \to \mathbb{D}^{\mathrm{op}}$ is defined on morphisms by $$F^{\mathrm{op}}(f : A \leftarrow B) = F(f) : F(A) \leftarrow F(B)$$ Here I'm writing $f : A \leftarrow B$ to mean that $f$ is a morphism from $A$ to $B$ in $\mathbb{C}^{\mathrm{op}}$, which is the same thing as a morphism from $B$ to $A$ in $\mathbb{C}$.

This case is no different. A morphism $f : C' \leftarrow C''$ in $\mathbb{C}^{\mathrm{op}}$ (that is, a morphism $f : C'' \to C'$ in $\mathbb{C}$) yields $$\mathbb{C}(C,-)^{\mathrm{op}}(f) = \mathbb{C}(C,-)(f) : \mathbb{C}(C,C') \leftarrow \mathbb{C}(C, C'')$$ This is a morphism from $\mathbb{C}(C,C')$ to $\mathbb{C}(C,C'')$ in $\mathbf{Set}^{\mathrm{op}}$, which is the function from $\mathbb{C}(C,C'')$ to $\mathbb{C}(C, C')$ defined by postcomposition with $f$.