# Tensor-Hom Adjunction In Monoidal Categories?

Is there a generalization of the tensor-hom adjunction to monoidal categories, or is it a special property of $\mathsf{Mod}$-$R$?

• Yes, it's called internal hom (see ncatlab.org/nlab/show/internal+hom) – sjvega Dec 22 '14 at 17:10
• @sjvega almost everything there is way over my head. Is there a book which systematically develops these ideas? – user153312 Dec 22 '14 at 17:15
• Try the Gelfand - Manin textbook on Homological Algebra – Avitus Dec 22 '14 at 17:23
• It's also called a monoidal closed category. The case when the monoidal product is the categorical product is espeically noteworthy: then it's called a cartesian closed category. For instance, $\mathsf{Set}$ is cartesian closed. – tcamps Dec 22 '14 at 19:00
• @tcamps So in a monoidal closed category the tensor product is the left adjoint to the $\mathsf {Hom}$ functor? – user153312 Dec 23 '14 at 20:10

In general , there is no one big theorem about monoidal closed categories that justifies or illuminates them. The internal hom does what it says on the tin: it lets you represent your homsets as objects of your category. This allows you to state questions and theorems that simply wouldn't make sense in a non-closed setting. My advice would simply be to keep an eye out for all the myriad ways one uses the fact that the hom of two $R$-modules is again an $R$-module (when $R$ is commutative!). Basically anything you care to ask or state about the homs of $R$-modules can be said more generally in a monoidal closed category. An even more primordial example is $\mathsf{Set}$, and the fact that the functions from one set to another again form a set, but this is harder to notice because it's such a basic fact -- although note for example that $[X,2]$ is the powerset of $X$.

As for references -- I'll give a few explicitly, and hopefully I'll used good enough terminology that you can Google stuff to find out more. Some of the Wikipedia articles have good lists of examples, too. And even though the nlab can be intimidating, if you keep your eyes open there you can at least pick up something here and there. Since monoidal closed categories are so fundamental, it's surprising to realize that it's kind of hard to find information about them consolidated in one place, though. Let me discuss a few points, at least. There's a very short section in Categories for the Working Mathematician.

First off, there are two dual ways to define a monoidal closed category: we can either ask that every functor $X \otimes (-)$ has a right adjoint, or that every functor $(-)\otimes X$ has a right adjoint. One of these should be called left closed and the other right closed, but I won't attempt to say which is which :). According to Kelly (last paragraph of page), certain endofunctor categories satisfy one definition but not the other. A monoidal category which is both left and right closed is called monoidal biclosed. These sorts of categories are often best thought of as 1-object bicategories in which certain Kan extensions/lifts exist.

There is a whole menagerie of different types of monoidal categories, and many of the distinctions have to do with closure or variations or elaborations on the notion of closure. Selinger is a pretty comprehensive reference.

But all this stuff is more than you need to worry about. Most of the time, we're interested in monoidal closed categories which are symmetric (or at least braided), and in this case left and right closed coincide (exercise!), so we're safe to just talk about symmetric monoidal closed categories. Some things to say about these...

• Since $X \otimes (-)$ is a left adjoint, it preserves colimits. Similarly, $[X,-]$ preserves limits.

• A symmetric monoidal closed category which is complete and cocomplete is sometimes called a cosmos. It is the traditional setting to do enriched category theory in, because almost everything you want carries over from the case of $\mathsf{Set}$. For this reason, Kelly's book on enriched category theory that I linked to above starts off talking about symmetric monoidal closed categories.

• One reflection of this fact is that the internal hom provides an enrichment of a symmetric monoidal closed category in itself.

• Another way to look at this is that having an internal hom allows you to internalize a lot of logic in your category. This is related to linear logic, and to the string diagrams discussed in the Selinger paper I linked to.

• When the product is the cartesian product, we have a cartesian closed category, and we can do even more logic and computer science. Toposes are cartesian closed, which is important to how logic is interpreted in them.

• It is very desirable for a monoidal category to be closed. The category $\mathsf{Top}$ is cartesian monoidal, but fails to be cartesian closed. Finding good cartesian closed categories of spaces attracted a significant amount of research in topology a few decades ago. This is one reason why people use compactly generated spaces or simplicial sets instead of general topological spaces in algebraic topology.

• One natural thing to do with a monoidal category is to study monoids and modules, comonoids and comodules, etc in it. If your category is closed, you can import more ideas from $\mathsf{Set}$ or $\mathsf{Ab}$ - for example, a module over a monoid $M$ is equivalently a map $M \otimes X \to X$ satisfying axioms, or a map $M \to [X, X]$ satisfying axioms.

• It's also possible to define a closed category without reference to a monoidal structure. I don't know any interesting non-monoidal examples, though.

• It's hard to make general statements about how rare or plentiful monoidal closed categories are. Some categories admit no monoidal biclosed structure, or exactly one -- $\mathsf{Cat}$ admits exactly two! Others admit many. This diversity is even to be found among categories of $R$-modules by considering different rings $R$. But in most cases, there is at most one monoidal closed structure on a category of any real interest ($\mathsf{Cat}$ is an interesting exception!).

• Great answer! so in $R$-$\mathsf{Mod}$ we define our tensor product and prove the tensor hom adjunction, but closed monoidal categories are defined as monoidal categories in which tensoring (with one fixed argument) has a right adjoint. However, on wiki it says this right adjoint is the currying functor, which seems related but not identical to the usual hom functor, so where's the tensor hom adjunction from $R$-$\mathsf{Mod}$? – user153312 Dec 24 '14 at 10:36
• I'm not sure what you're asking, but I'm in the mood to guess, I suppose. First note that if you use the isomorphism $\mathrm{Hom}(A \otimes B, C) \cong \mathrm{Bilin}(A,B;C)$, then the exact same "currying" and "uncurrying" formulas that work to describe the cartesian-closure adjunction in $\mathsf{Set}$ also work to describe the tensor-hom adjunction in $R-\mathrm{Mod}$. Second, I think you're asking "Where does the tensor-hom adjunction in $R-\mathrm{Mod}$ 'come from'; why should we expect its existence?". – tcamps Dec 24 '14 at 17:11
• In monoidal closed categories, the tensor product is left adjoint to the currying functor. I did not realize the currying functor is exactly the covariant $\mathsf{Hom}$ functor so I couldn't understand where the actual tensor-hom adjunction was. What you said about $\mathsf{Set}$ makes sense to me now since it is cartesian closed. I was a bit afraid to ask "why should we expect the existence of the tensor-hom adjunction in $R$-$\mathsf{Mod}$?", but I would certainly love answer. – user153312 Dec 24 '14 at 19:24

If $\mathcal{C}$ is closed symmentric monoidal category, and $\mathbb{T}$ is symmetric monoidal monad on $\mathcal{C}$, then the category of $\mathbb{T}$-algebras caries a structure of closed symmetric monoidal category.

this result is well known, and was proved be Anders Kock in the '70.

For the definition of symmetric monoidal monad, you can look at this paper.

In our case the monad is of free R-modules.

(I don't have enough points to elaborate...)

• for the papers of Kock: first and second - old and hard to read. – Yitzhak Z Jan 1 '15 at 18:39
• we define bimorphisms (p. 6 of Seal) and the representing object is a coequalizer (p.7) - which coincide in $R-Mod$ with the known quotient of the free $R$-module $R[A\times B]$. The hom-object is an equalizer (p. 7 in the 2nd paper of Kock) and in $R-Mod$ it is precisely the functions that are $R$-homomorphisms. – Yitzhak Z Jan 1 '15 at 18:55