In the category of modules, the tensor product functor is the left adjoint of the covariant Hom functor. Similarly in the category of sets, the Cartesian product functor is the left adjoint of the covariant Hom functor. I’m wondering whether this can be generalized.

Let $C$ be a category where the covariant Hom functor has a left adjoint. My question is, does this left adjoint make $C$ into a monoidal category? That is, does it induce a tensor product on $C$, in the sense that it is the tensor product functor of some tensor product on $C$?

If not, does anyone know of a counterexample?

  • $\begingroup$ You mean that EVERY covariant Hom functor $\mathrm{Hom}(A,-)$ is assumed to have a left adjoint? That's the only interpretation that makes sense. But then you need to be able to link all these somehow. $\endgroup$ – Randall Mar 9 '19 at 4:17
  • $\begingroup$ @Randall Yes, that’s what I mean. If $F_a$ denotes the left adjoint of Hom(a,_) for each object $a$, I want to define the tensor product of $a$ and $b$ as $F_a(b)$. $\endgroup$ – Keshav Srinivasan Mar 9 '19 at 4:19
  • $\begingroup$ The answer is "no," though I will confess I don't remember why. What do you suppose the unit object will be? $\endgroup$ – Randall Mar 9 '19 at 4:24

Sets and modules over a commutative ring share the property that they are closed monoidal. This means they're not only (symmetric) monoidal, but the monoidal product has a right adjoint, which equips them both with a notion of "internal hom," or equivalently an enrichment over themselves.

If you start with just a category $C$, the Hom functor takes values in $\text{Set}$, so a left adjoint to $\text{Hom}(c, -) : C \to \text{Set}$ is, if it exists, a functor $\text{Set} \to C$. This adjoint, if it exists, turns out to be

$$\text{Set} \ni X \mapsto \bigsqcup_X c \in C;$$

in other words it takes a set $X$ and outputs the coproduct of $X$ copies of $c$; you can think of this as a tensoring $X \otimes c$, but note that $X$ is a set, not another object in $C$. This tensoring equips $C$ with the structure, not of a monoidal category, but of a module over the (cartesian) monoidal category $\text{Set}$.

Edit: There is a notion of closed category, which is a category equipped with just an internal hom functor. If the covariant internal hom has an enriched left adjoint, that reconstructs a monoidal structure; see the nLab.

| cite | improve this answer | |
  • $\begingroup$ It's not completely clear in the question, but my impression is that "Hom" is intended to be the internal hom. I think it's just that that's the terminology often used for module categories and then in $\mathbf{Set}$ it happens that the internal and external hom are the same. $\endgroup$ – Derek Elkins left SE Mar 9 '19 at 6:18
  • $\begingroup$ So, the OP could be asking whether a closed category (ncatlab.org/nlab/show/closed+category) such that the internal hom always has a left adjoint is closed monoidal. I don't know whether this is true, but in any case the OP says "category" so I don't think this is actually the OP's question. $\endgroup$ – Qiaochu Yuan Mar 9 '19 at 6:20
  • 1
    $\begingroup$ Ah, apparently it is true: ncatlab.org/nlab/show/… $\endgroup$ – Qiaochu Yuan Mar 9 '19 at 6:53
  • $\begingroup$ That's for closed categories that have a unit, but that article also discusses prounital closed categories, and those may change the story enough. $\endgroup$ – Derek Elkins left SE Mar 9 '19 at 7:30
  • $\begingroup$ Thanks for clarifying, I guess I hadn’t noticed that what was adjoint to the tensor product functor in the module case was the internal Hom functor and not the external Hom functor. So the question I really should have asked is whether an adjoint of the internal Hom functor, if the latter exists, induces a tensor product. So can you add the result you linked to to your answer? $\endgroup$ – Keshav Srinivasan Mar 9 '19 at 13:45

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.