I am trying to understand how the [tensor product of presentable categories](https://ncatlab.org/nlab/show/tensor+product+of+presentable+%28infinity%2C1%29-categories) works: let $\otimes\colon {\cal A}\times {\cal B}\to {\cal A}\otimes{\cal B}$ the universal bilinear functor corresponding to $\text{id}\_{{\cal A}\otimes{\cal B}}$ under the correspondence defining ${\cal A}\otimes{\cal B}$,
$$
\text{Bilin}({\cal A}\times{\cal B}, {\cal E})\cong \text{Func}({\cal A}\otimes{\cal B},{\cal E}).
$$
What is the answer to these questions?


1. Does the essential image of $\otimes$ generate ${\cal A}\otimes{\cal B}$ under colimits? This seems the analogue of $V\otimes W$ being made by formal sums of monomials. Does $\otimes$ have some other remarkable properties (fullness, faithfulness, commutes/reflects/creates co-limits?)
2. We have a fairly explicit (albeit constructed by absolute nonsense) model for ${\cal A}\otimes{\cal B}$, that is $\text{Func}({\cal A}^\text{op}, {\cal B})\_R$ (functors ${\cal A}^\text{op}\to \cal B$ which commute with limits (and filtered colimits? I see different definitions here). This means that to $\otimes$ correspond a canonical functor ${\cal A}\times {\cal B} \to \text{Func}({\cal A}^\text{op}, {\cal B})\_R$, sending $(A,B)$ to... who?
3. The adjoint functor theorem gives $A\otimes -$ and $-\otimes B$ right adjoints $A/-$ and $-\backslash B$; how are these functors ${\cal A}\otimes{\cal B}\to {\cal A,B}$ defined? There are Kan extensions one can write, but...



On the tensor product of presentable categories
1. Yes. You can present $\mathcal A \otimes \mathcal B$ as a certain localization of the free cocompletion of $\mathcal A \times \mathcal B$. By "generate under colimits" I of course mean that you are allowed to take transfinitely-iterated colimits: you can take a colimit whose entries are the values of colimits whose entries are the values of ... for any ordinal. I think I can come up with a non-zero example for which the functor $\mathcal A \times \mathcal B \to \mathcal A \otimes \mathcal B$ is neither full nor faithful, but one doesn't come to me right away --- if I come up with one, I'll edit this answer.
2. In your presentation, I think $(A,B) \mapsto \hom(-,A) \times B$, where if $S$ is a set and $B \in \mathcal B$, then $S \times B = \sqcup\_{s\in S} B$ is the disjoint union of $S$ many copies of $B$.
3. The best you can say in general is that the adjoint to $A\otimes$ is a $\mathcal B$-valued version of $\hom(A,-)$. If your categories are very nice, like presheaf categories, then you can give more direct interpretation of this idea.


