# The opposite of the covariant hom functor

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.

• Morphisms in the opposite category are just morphisms in the ordinary category but formally reversed. So if you have an arrow in the opposite category reverse it, then apply the functor, then reverse it again – leibnewtz Apr 2 '18 at 14:48

## 1 Answer

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 $F^{\mathrm{op}}(C)=F(C)$.

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 $f$.

• I edited my question to explain my problem better. – alexanderyaacov Apr 2 '18 at 13:26
• I think the OP is mostly confused about where the arrows go – leibnewtz Apr 2 '18 at 14:49
• @leibnewtz: The OP updated their question since I posted my answer. I'll update my answer momentarily. – Clive Newstead Apr 2 '18 at 15:01
• @alexanderyaacov: I've updated my answer. – Clive Newstead Apr 2 '18 at 15:07