# Underlying functor of tensor product in a closed and symmetric monoidal category.

I will follow, for terminology and notation, G. M. Kelly, Basic Concepts of Enriched Category Theory. For sake of a self-contained exposition, I will try to write here all the needed concepts.

Let then $\mathcal{V}=(\mathcal{V}_{0},\ \otimes,\ I,\ a,\ r,\ l)$ be a monoidal category and let $\mathcal{A}$ be an enriched $\mathcal{V}-$category. Then one can define the underling category of $\mathcal{A}$, denoted as $\mathcal{A}_{0}$, whose objects are those of $\mathcal{A}$ and whose hom-sets $\mathcal{A}_{0}(A,B)$ are $\mathcal{V}_{0}(I,\ \mathcal{A}(A,B))$, for $A,B$ objects of $\mathcal{A}_{0}$. For $f\in\mathcal{A}_{0}(A,B)$ and $g\in\mathcal{A}_{0}(B,C)$, the composite $gf$ is given by the composite $$I\overset{l^{-1}}{\longrightarrow}I\otimes I\overset{g\otimes f}{\longrightarrow}\mathcal{A}(B,C)\otimes\mathcal{A}(A,B)\overset{M}{\longrightarrow}\mathcal{A}(A,C),$$ where $M$ is the composition law in $\mathcal{A}$. Furthermore, given a $\mathcal{V}-$functor $T\colon\mathcal{A}\to\mathcal{B}$, the underling functor $T_{0}\colon\mathcal{A}_{0}\to\mathcal{B}_{0}$ acts as $T$ on the objects of $\mathcal{A}_{0}$, while it sends $f\in\mathcal{A}_{0}(A,B)$ to $T_{AB}\circ f$, where the composite is taken in $\mathcal{V}_{0}$.

Suppose now that $\mathcal{V}$ is a closed, symmetric monoidal category and call the commutativity isomorphism $c$. Denote also with $$\pi\colon\mathcal{V}_{0}(X\otimes Y, Z)\overset{\simeq}{\longrightarrow} \mathcal{V}_{0}(X,[Y,Z])$$ the adjunction isomorphism. It is then possible to see $\mathcal{V}$ as a category enriched over itself (which we keep on calling $V$): the objects of $\mathcal{V}$ are those of $\mathcal{V}_{0}$, while the hom-object $\mathcal{V}(X,Y)$ is $[X,Y]$. The composition law $M\colon[Y,Z]\times [X,Y]\to [X,Z]$ is the arrow corresponding under $\pi$ to the composite $$([Y,Z]\times [X,Y])\times X\overset{a}{\longrightarrow}[Y,Z]\times ([X,Y]\times X)\overset{1\otimes e}{\longrightarrow} [Y,Z]\times Y\overset{e}{\longrightarrow} Z$$ There is an isomorphism between the underlying category of $\mathcal{V}$ as an enriched category over itself and $\mathcal{V}_{0}$ which, as far as I understand, should send a morphism $f\colon A\rightarrow B$ in the underlying category (so $f\colon I\rightarrow [A,B]$) to $\pi^{-1}(f)\circ l^{-1}$. Therefore, we can identify those two ordinary categories.

We can also define a $\mathcal{V}-$ functor $Ten\colon \mathcal{V\otimes V}\to\mathcal{V}$. Here $\mathcal{V}\otimes\mathcal{V}$ has object-class given by $\mathcal{V}_{0}\times\mathcal{V}_{0}$ and $(\mathcal{V}\otimes\mathcal{V})((X,Y),\ (X',Y')):= [X,X']\otimes [Y,Y']$. The $\mathcal{V}-$functor $Ten$ sends an object $(X,Y)$ to $X\otimes Y$, while $Ten_{(X,Y),(X',Y')}$ corresponds under $\pi$ to the composite $$([X,X']\otimes [Y,Y'])\otimes (X\otimes Y)\overset{m}{\longrightarrow} ([X,X']\otimes X)\otimes ([Y,Y']\times Y)\overset{e\otimes e}{\longrightarrow} X'\otimes X.$$ Here $m:(W\otimes X)\otimes (Y\otimes Z)\simeq (W\otimes Y)\otimes (X\otimes Z)$ is the middle-four interchange isomorphism and $e\colon [Y,Z]\otimes Y\longrightarrow Z$ is the evaluation morphism associated to $\pi$ (the unit of the adjunction). One gets that $e(Ten\otimes 1)=(e\otimes e)m$.

Finally, here comes the question. It should be true that the ordinary functor $S$ given as the composite $$\mathcal{V}_{0}\times\mathcal{V}_{0}\longrightarrow (\mathcal{V}\otimes\mathcal{V})_{0}\overset{Ten_{0}}{\longrightarrow}\mathcal{V}_{0}$$ is the tensor product $\otimes\colon\mathcal{V}_{0}\times\mathcal{V}_{0}\longrightarrow\mathcal{V}_{0}$. I can not prove this fact.

My attempt, up to now, has been the following. I have found that, on arrows $(f\colon I\rightarrow [A,A'],\ g\colon I\rightarrow [B,B'])$, $S(f,g)$ corresponds, under $\pi$, to $(e\otimes e)\circ m\circ (((f\otimes g)\circ l^{-1})\otimes 1)$. Under the identification between the underlying category of the $\mathcal{V-}$ category $\mathcal{V}$ and the ordinary category $\mathcal{V}_{0}$, $S(f,g)$ should then be $(e\otimes e)\circ m\circ (((f\otimes g)\circ l^{-1})\otimes 1)\circ l^{-1}$. Theoretically speaking, it seems to me that I should then prove this last arrow to be the same as $\otimes (f,g)$, which, under the above identification, should be $(\pi^{-1}(f)\circ l^{-1})\otimes (\pi^{-1}(g)\circ l^{-1})$, but I can not do this.

Any suggestion, as well as complete solutions to the problem, would be greatly appreciated.

• I wonder how many people have actually read this question completely. Can you shorten it somehow? (If necessary, restrict the audience to more experienced people by shortening) Commented Apr 21, 2014 at 11:49
• The question is clear and well written. Don't shorten it. Commented Apr 27, 2014 at 7:51
• @Turion I admit length might be a problem, but I have tried to mark distinctly the part where the question comes, so that the more expert reader can immediately jump to conclusions. Moreover, it seemed to me that here in MSE there is a sort of implicit rule forcing posters to write down as self-contained questions as possible, especially because, when this is not done, clarifications are usually required by other users. Commented Apr 27, 2014 at 9:50
• @MartinBrandenburg Thanks, that was my hope when I wrote down all those lines. Still, the absence of answers made me think a little bit about what Turion asked... Commented Apr 27, 2014 at 9:52
• I made some corrections, in particular, I changed the $×$ into a $\otimes$ at one point, I think this is what you meant. And don't worry about the length. I'm thinking about this question now and maybe I'll write an answer. Commented May 22, 2015 at 18:13

So you have arrows $f:I\to[A,A']$ and $g:I\to[B,B']$. The functor we are interested in and which you called $S$ is the composite $$\mathcal V_0\times\mathcal V_0\to (\mathcal{V\otimes V})_0\to \mathcal V_0 \\ \mathcal V_0(I,[A,A'])×\mathcal V_0(I.[B,B'])\to \mathcal V_0(I,[A,A']⊗[B,B'])\to \mathcal V_0(I,[A⊗B,A'⊗B']) \\ (f,g)\mapsto (f⊗g)l^{-1}\mapsto \text{Ten}(f⊗g)l^{-1}$$ We want to show that, under the bijection $\pi$, this arrow corresponds to $f^\sharp l^{-1}⊗g^\sharp l^{-1}:A⊗B\to A'⊗ B'$. But $(\text{Ten}(f⊗g)l^{-1})^\sharp l^{-1}$ is, as you correctly computed, equal to $$(ε⊗ε)\ m\ (((f⊗g)l^{-1})⊗(1_A⊗1_B))\ l^{-1}$$ which is \begin{align} & (ε⊗ε)\ m\ ((f⊗g)⊗(1_A⊗1_B))\ (l^{-1}⊗1_{A⊗B})\ l^{-1} \\ & =\ (ε⊗ε)\ ((f⊗1_A)⊗(g⊗1_B))\ (l^{-1}⊗l^{-1}) \\ & =\ ε(f⊗1)l^{-1}⊗ε(g⊗1)l^{-1} \\ & =\ f^\sharp l^{-1}⊗g^\sharp l^{-1} \end{align}