1
$\begingroup$

This is a very basic question about definitions, but I haven't been able to find the answer to it online. If we let $\mathscr{A}$ be an Abelian category, then for any object $A\in\mathscr{A}$, we can define the Hom functor $\mathrm{Hom}_\mathscr{A}(A,-)$.

What category does this take values in? I know that it can take values in $\mathbf{Ab}$, but in the case where $\mathscr{A}=R\hbox{-}\mathbf{Mod}$, the functor seems to take values in $\mathscr{A}$. Which convention holds in general?

$\endgroup$
3
$\begingroup$

By definition, it takes values in $Ab$.

The special thing about $\mathscr{A}=_RMod$ is that it is an enriched category, enriched over itself, see this wonderful page or Kelly's book.

$\endgroup$
  • 3
    $\begingroup$ If $M,N$ are (left) $R$-modules, then in general the abelian group $\hom_R(M,N)$ is not an $R$-module in any way, but rather a $Z(R)$-module. It is thus an $R$-module if $R$ is commutative. It is an $R$-module if $M,N$ have some extra bimodule structure, or if $R$ is a Hopf algebra, but that's another story. $\endgroup$ – Pedro Tamaroff Jan 6 '17 at 4:07
  • $\begingroup$ Yes!! That's true! Good eye (I was assuming he was working with a CRing) $\endgroup$ – AIM_BLB Jan 7 '17 at 20:38
  • $\begingroup$ (Yes, then I saw the [commutative-algebra] tag, though.) $\endgroup$ – Pedro Tamaroff Jan 7 '17 at 21:16
  • $\begingroup$ Stil, fun fact always! I havent thought about NC-geo in a while lol $\endgroup$ – AIM_BLB Jan 7 '17 at 21:47
0
$\begingroup$

In a $V$-enriched category, the natural "hom object" of natural transformations between functors $F$ and $G$ is the end $\int V(F,G)$. As a special limit in $V$, it is an object of $V$.

In the Ab-enriched case, modules are functors into Ab, and so hom-objects between them (that is, module homeomorphisms) also form an abelian group.

They also admit a module structure over the center of the domain.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.