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

Question

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.

Answer 1

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 $$\begin{align} & (ε⊗ε)\ m\ ((f⊗g)⊗(1_A⊗1_B))\ (l^{-1}⊗1_{A⊗B})\ l^{-1} \\ & =\ (ε⊗ε)\ ((f⊗1_A)⊗(g⊗1_B))\ (l^{-1}⊗l^{-1}) \\ & =\ ε(f⊗1)l^{-1}⊗ε(g⊗1)l^{-1} \\ & =\ f^\sharp l^{-1}⊗g^\sharp l^{-1} \end{align}$$