# On the definition of the contravariant hom functor

I'm thinking about why we must define the functor $$h_A=\mathrm{Hom}(\_,A)$$ as a contravariant functor, i.e. if there is a way to choose the morphisms to make it a covariant with same image on objects? I tried as following.
Suppose we have a covariant functor $$F$$ from category $$\mathcal{C}$$ to $$\mathrm{Sets}$$, such that $$FX=\mathrm{Hom}(X,A)$$ for any object $$X$$. By Yoneda $$[h^A,F]\cong FA=\mathrm{Hom}(A,A)$$, where [] is the set of natural transforms and $$h^A=\mathrm{Hom}(A,\_)$$. Now the $$1_A$$ in the right corresponds to an isomorphism on the left, hence $$h^A\cong F$$ , a contradiction.
Is my explanation right? What is the exact way to characterize the functor $$h_A$$(or $$h^A)$$, the naturality with respect to A? And do you have other ideas? Thank you!

• Yoneda lemma is applied to functors of the same variance. Feb 21, 2021 at 17:07
• @solgaleo Sorry, there's a typo on the first line... I have fixed it. Feb 21, 2021 at 17:37

Your argument is incomplete. There are two problematic steps, which both contain errors. Firstly, it's not clear that $$1_A$$ should correspond to an isomorphism on the right hand side (and this is not generally true). Secondly, $$h^A\cong F$$ doesn't contradict anything.

Indeed, there are categories $$\mathcal{C}$$ for which $$\mathcal{C}(-,A)$$ can be made a covariant functor (for all $$A$$). In particular, any category with a functor $$\bar{\cdot}:\mathcal{C}\to \mathcal{C}^{\text{op}}$$ which is the identity on objects will do, since then we get a covariant functor from $$X\mapsto \mathcal{C}(\bar{X},A)$$. A classic example of this is given by groups (regarded as single object categories w/ object $$*$$), $$g\mapsto g^{-1}$$ defines a contravariant functor on a group $$G$$ regarded as a category since $$(gh)^{-1}=h^{-1}g^{-1}$$.

A nontrivial example of such a category is the category of inner product spaces with all linear maps between them. The adjoint of a linear map defines a contravariant functor which is the identity on objects.

When we define $$F$$ this way, the natural transformation $$h^A\to F$$ is going to be defined by $$f\in h^V(W) = \mathcal{C}(V,W)\mapsto \bar{f} \in F(W) = \mathcal{C}(W,V).$$

To construct an $$F$$ for which this is not an isomorphism, we need a functor $$\mathcal{C}\to\mathcal{C}^{\text{op}}$$, which is the identity on objects, but is not fully faithful.

For this, we could take $$G$$ a group, and define $$\bar{g} = 1$$ for all $$g$$. This also defines a contravariant functor, though admittedly a rather boring one. Thus we see that the natural map $$h^A\to F$$ isn't necessarily an isomorphism.

What is true

You've asked why $$\mathcal{C}(-,A)$$ must be contravariant. The answer is sort of simple, it's just that it is contravariant. We don't define it to be contravariant, it just naturally has that structure. We can put other structures on the map from objects to sets to make it a covariant functor, as we've just seen, but it would still have the natural contravariant functor structure.

We can also make a more general statement. It is not always possible to put a covariant structure on the sets given by $$\mathcal{C}(-,A)$$.

Proof.

It suffices to give a counterexample. Consider the category with two objects, $$0$$, and $$1$$, and a single nonidentity map, $$f$$, between them. Call this category $$\newcommand\2{\mathbf{2}}\2$$.

Then $$\2(0,0)=\{1_0\}$$, but $$\2(1,0)=\varnothing$$, so if we had a covariant structure on $$\2(-,0)$$, we would get a map $$f_*: \{1_0\}\to \varnothing$$, which is impossible.