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$\newcommand{\V}{\mathscr{V}}\newcommand{\A}{\mathcal{A}}\newcommand{\B}{\mathcal{B}}\newcommand{\C}{\mathcal{C}}$Fix a closed symmetric monoidal category $\V$, writing the product as $\otimes$, the hom-objects as $\underline{\V}(-,-)$ and the monoidal unit as $\ast$.

In my "research" I encountered and solved the following problem:

If $\underline{\A},\underline{\B}$ are $\V$-categories then find a suitable concept of a $\V$-category $\underline{\A\times\B}$ generalising certain phenomena from ordinary category theory

For me, the phenomena were related to adjunctions with parameter and lifting intertwined families of functors $\A\to\C,\B\to\C$ to some kind of combined bifunctor object $\A\times\B\to\C$. This is probably extremely standard, but I wouldn't know.

I define $\underline{\A\times\B}$ to have objects pairs $(\alpha,\beta)$ for $\alpha\in\A,\beta\in\B$ and hom-objects shall be: $$\underline{\A\times\B}((\alpha,\beta),(\alpha',\beta')):=\underline{\A}(\alpha,\alpha')\otimes\underline{\B}(\beta,\beta')$$ Together with obvious definitions of identity arrow and composition. This really does form a $\V$-category, and it served the purposes I needed it to.

My question is, are there situations where we can make sense of this category as a genuine product object? A priori I think there need not be any enriched projection functor $\underline{\A\times\B}\to\underline{\A}$, because that demands a way of defining arrows: $$\underline{\A}(\alpha,\alpha')\otimes\underline{\B}(\beta,\beta')\overset{??}{\longrightarrow}\underline{\A}(\alpha,\alpha')$$ Which I don't see how to do. Moreover, it's not clear to me what the underlying category $\underline{\A\times\B}_0$ should be: it should in some sense "contain" the product of underlying categories $(\underline{\A})_0\times(\underline{\B})_0$ since for a pair of arrows $f:\alpha\to\alpha',g:\beta\to\beta'$ in the underlying categories I have an arrow: $$\ast\cong\ast\otimes\ast\overset{\overline{f}\otimes\overline{g}}{\longrightarrow}\underline{\A}(\alpha,\alpha')\otimes\underline{\B}(\beta,\beta')$$Which defines an arrow in $\underline{\A\times\B}_0$. But to the converse, I don't see how we can characterise generic arrows: $$\ast\overset{?}{\longrightarrow}\underline{\A\times\B}(-,-)$$So there could conceivably be more arrows in $\underline{\A\times\B}_0$, and these additional arrows are currently mysterious to me.

I am motivated to ask this question by the fact that if $\V=\mathsf{Set}$ - or more generally, if $\V$ is a Cartesian closed category with $\times$ as its monoidal product - then it is clear that $\underline{\A\times\B}$ is the same thing as the ordinary product of ordinary categories. So, what conditions can one impose on $\V$ to understand $\underline{\A\times\B}$ as a product? Or at least, to understand what the underlying category of $\underline{\A\times\B}$ looks like? As it stands, I can note only one interesting property of $\underline{\A\times\B}$:

If $\underline{\C}$ is a $\V$-category and for each $(\alpha,\beta)\in\A\times\B$ there are $\V$-functors $F^\alpha:\underline{\B}\to\underline{\C},F^\beta:\underline{\A}\to\underline{\C}$ satisfying suitable a 'intertwining' relationship, we can uniquely lift these data to a $\V$-functor $F:\underline{\A\times\B}\to\underline{\C}$.

Are there any more?

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    $\begingroup$ Have you looked into Basic concepts of enriched category theory by Kelly? $\endgroup$ Commented Jul 26, 2023 at 16:08
  • $\begingroup$ @მამუკაჯიბლაძე No, thanks. I had not encountered this book $\endgroup$
    – FShrike
    Commented Jul 26, 2023 at 16:27
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    $\begingroup$ As you say right at the end, your construction (which I'd rather call $A\otimes B$) represents a notion of bilinear morphism from $(A,B)$ to $C.$ This is a universal characterization that's in principle all you really need to know. It gives a monoidal structure on $V$-Cat which is closed under reasonable conditions, but certainly isn't going to be a product in general any more than $x\otimes y$ is a product in $V.$ The underlying category's morphisms are just points of $A(a,a')\otimes B(b,b'),$ but again there's nothing we can say in general about $x\otimes y$'s points. $\endgroup$ Commented Jul 26, 2023 at 22:16
  • $\begingroup$ @KevinArlin Ok, thanks. It's good to know I'm not missing something $\endgroup$
    – FShrike
    Commented Jul 27, 2023 at 10:08
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    $\begingroup$ @FShrike This is the content of chapter 2 of the book that was suggested by მამუკა ჯიბლაძე $\endgroup$
    – AT0
    Commented Jul 27, 2023 at 12:43

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