# Tensor products from internal hom?

Monoidal categories come with tensor products, and sometimes, these categories are biclosed, i.e each restriction of the tensor bifunctor has a right adjoint. If the category happens to be symmetric, then restrictions of the $\otimes$ bifunctor are naturally isomorphic and we can talk about symmetric monoidal closed categories.

If I understand correctly, the internal hom is precisely this right adjoint to $-\otimes A$. Hence, the internal hom comes from $\otimes$ and not the other way around.

(When) Is it possible to first define the internal hom, and then have it induce $\otimes$ from the adjunction? Can we somehow go the other way around and get monoidal structure from somehow defining an internal hom?

In general if you don't want to start with the monoidal structure, you start with a closed category. In a closed category $\mathsf C$, you have a bifunctor $$[-,-] : \mathsf C^{op} \times \mathsf C \to \mathsf C,$$ called the internal hom, and various other data that are somewhat "dual" to the axioms of a monoidal category (I put dual in quotes because this isn't the dual notion of "monoidal category", you just dualize one side of the bifunctor $- \otimes -$ to get the axioms for $[-,-]$). More specifically:

• a unit object $I$,
• a natural isomorphism $\operatorname{id}_{\mathsf C} \cong [I, -]$,
• an extranatural transformation $j_X : I \to [X, X]$ that corresponds to getting the "identity" in $[X,X]$,
• post-composition transformation $[Y,Z] \to [[X,Y], [X,Z]$,

subject to various coherence conditions.*

Then just like the internal hom $[-,X]$ in a monoidal category, if it exists, is the right adjoint to $- \otimes X$, the tensor product in a closed category is the left adjoint to the internal hom (if it exists). In both cases you get a closed monoidal category. It's also possible to define categories enriched over closed categories, in a manner similar to categories enriched over a monoidal category, and the two notions coincide when the category is closed monoidal.

* These data are truly dual to the data you want for a monoidal category. Take the point of view that $\hom(X \otimes Y, Z) \cong \hom(X, [Y,Z])$ as if you were truly in a closed monoidal category. Then you can use the Yoneda embedding to make the correspondence clear:

• The isomorphism $[I, X] \cong X$ becomes $\hom(Y, [I,X]) \cong \hom(Y \otimes I, X)$ naturally in $X$ and $Y$, so you get the right unitor $- \otimes I \cong \operatorname{id}_{\mathsf C}$.
• The transformation $I \to [X,X]$ becomes a transformation $$\hom(Y, I) \to \hom(Y, [X,X]) \cong \hom(Y \otimes X, X).$$ If you take $Y = I$ you get $\hom(I,I) \to \hom(I \otimes X, X)$, and the image of the identity of $I$ because the left unitor.
• The data of the associator is equivalent to an (extra in some variables)natural isomorphism $[A, [B,C]] \cong [A \otimes B, C]$. The relation with the post-composition is trickier and I'll let you peruse the reference.

Of course none of the above is an actual proof, just a sketch of the ideas that go into this. Proofs can be found in:

Kelly, G. Max; Mac Lane, Saunders, Coherence in closed categories. J. Pure Appl. Algebra 1 1971 no. 1, 97–140. doi: 10.1016/0022-4049(71)90013-2.

• Great answer! I have two requests: 1) A reference which states all the coherence conditions and proves the basic theorem from this direction, e.g that a left adjoint indeed yields monoidal structure and that the structures coincide starting from either direction. 2) I would like to get the tensor product of chain complexes as the left adjoint of the internal hom there. How should I define the internal hom in $\mathsf{Ch}_\bullet (A)$? – user153312 Jan 27 '15 at 15:28
• I updated my answer, I think it should answer your point 1. For your point 2, well, you know what the tensor product is, if you find out its right adjoint you will get the internal hom... – Najib Idrissi Jan 27 '15 at 15:43
• Thanks for the reference. I tried working out the exactly what the right adjoint is, but failed. I'll give it some time. – user153312 Jan 27 '15 at 15:46
• If it can help, $[A,B]_n$ is the module of maps of homogeneous degree $n$ (or $-n$ I can never remember), and the differential of $f : A \to B$ is $d_Bf \pm fd_A$. But it would be better to open up a new question if it still gives you trouble. – Najib Idrissi Jan 27 '15 at 15:47
• I'll soon take your advice and ask that question separately. Returning to this one, could you explain explicitly how the Yoneda embedding is used? – user153312 Jan 31 '15 at 13:44

This is only a small addition to Najib's very nice answer which is nevertheless too long for a comment.

It is even interesting to look at the structure of a closed category in the following very restricted setting:

1. We consider only posets $({\mathsf P},\leq)$ considered as a categories.
2. We consider only those closed structures in which the unit is a terminal object (intuitively, these are like the internal homs corresponding to the categorical product $\times$ as the monoidal structure)

Denoting ${\mathsf T}$ the terminal object of ${\mathsf P}$, $[-,-]$ is then a function $\to: ({\mathsf P},\leq)^{\text{op}}\times ({\mathsf P},\leq)\to({\mathsf P},\leq)$ such that:

• For all $P\in{\mathsf P}$ we have $({\mathsf T}\to P) = P$ and ${\mathsf T}\leq (P\to P)$, i.e. $(P\to P) = {\mathsf T}$.

• For all $P,Q,R\in{\mathsf P}$ we have $(Q\to R)\leq ((P\to Q)\to (P\to R))$.

This already looks very 'logical', and indeed: Taking ${\mathsf P}$ to be the equivalence classes of formulae in the implicational fragment of intuitionistic propositional logic, with $\to$ being given by $[\psi]\to[\phi] := [\psi\Rightarrow\phi]$, satisfies these axioms.

On the $1$-categorical level, you find an interesting example by looking at the category which again has implicational propositional formulae as objects but equivalence classes of proofs - in the sense of terms in simply typed $\lambda$-calculus - as morphisms. Note that in this setting there is really no categorical product present, but only the internal hom given by implication.