Given a locally small category $\mathcal{C}$, Wikipedia defines the Hom functors as

enter image description here

At my lectures (for the more specific case of $\mathcal{C}=\operatorname{R-Mod}$ -- the category of R-modules for some ring R) the functor $Hom_R(-, M)$ was defined as a functor between $\operatorname{R-Mod}^{op}\rightarrow \operatorname{AbGrp}$.

  1. I think I understand why the codomain is $\operatorname{AbGrp}$, but I don't understand why the domain is the opposite category $\operatorname{R-Mod}^{op}$? How is it possible that Wikipedia gives the domain to be $\mathcal{C}$?
  2. Also, my notes claim that it follows from the left-exactness of the Hom-functor $Hom_\mathcal{A}(M, -): \mathcal{A} \rightarrow \operatorname{AbGrp}$ (here $\mathcal{A}$ is an arbitrary Abelian category) that $Hom_R(-, M): \operatorname{R-Mod}^{op} \rightarrow \operatorname{AbGrp}$ is also left exact. Could you explain why this is a corollary of the left-exactness of $Hom_\mathcal{A}(M,-): \mathcal{A}\rightarrow\operatorname{AbGrp}$?
  • $\begingroup$ A contravariant functor $F:A\to B$ is a functor $A^{\operatorname{op}}\to B$. This means it "reverses arrows". $\endgroup$ – Douglas Molin May 17 at 11:17
  • 2
    $\begingroup$ The key to 1. is contravariance. Wikipedia (apparently) is perfectly happy to allow covariant and contravariant functors, but some authors insist on writing contravariant functors as covariant functors from the opposite category. $\endgroup$ – Rylee Lyman May 17 at 11:17
  • $\begingroup$ @RyleeLyman so would it be accurate to say that my confusion regarding the first question is just a confusion regarding syntax? Ie that the insistence on the opposite category as domain is just due to the fact that if we didnt then the functor wouldnt commute with composition as expected? $\endgroup$ – gen May 17 at 11:23
  • $\begingroup$ yes, that's right. $\endgroup$ – Rylee Lyman May 17 at 11:42
  • $\begingroup$ @RyleeLyman In my experience I would say essentially all authors use the convention you describe. $\endgroup$ – Kevin Carlson May 17 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.