TITLE: On the tensor product of presentable categories
QUESTION [5 upvotes]: I am trying to understand how the tensor product of presentable 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?

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?)
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?
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...

REPLY [7 votes]: I think in 2 the correct definition is that $\text{Func}({\cal A}^\text{op}, {\cal B})_R$ denotes the category of functors which are right adjoints or equivalently in this case, functors that preserve limits. Indeed, $\text{Func}({\cal A}^\text{op}, {\cal B})_R \cong \left(\text{Func}({\cal A}, {\cal B}^\text{op})_L\right)^\text{op}$, and because $\cal A$ is presentable a functor ${\cal A} \to {\cal B}^\text{op}$ is a left adjoint if and only if it preserves colimits.
(I think I know why you mentioned filtered colimits though: if $\cal C$ and $\cal D$ are presentable, then a functor $\cal C \to \cal D$ is a right adjoint if and only if it preserves limits and $\kappa$-filtered colimits for some sufficiently large regular cardinals $\kappa$; but notice that if $\cal A$ is presentable then $\cal A^\mathrm{op}$ is never presentable, so this does not apply here!)
OK, so what is $(A,B)$ sent to in $\text{Func}({\cal A}^\text{op}, {\cal B})_R$? Let $\text{ev}_A : \text{Func}({\cal A}^\text{op}, {\cal B})_R \to {\cal B}$ be the functor of evaluation at $A \in \cal A$ and let $L_A : {\cal B} \to \text{Func}({\cal A}^\text{op}, {\cal B})_R$ be its left adjoint, then the pair $(A,B)$ is sent to $L_A(B) \in \text{Func}({\cal A}^\text{op}, {\cal B})_R$.
I tried to make this more explicit but failed: it tempting to ignore the $_R$ and just look at the left adjoint to $ev_A : \text{Func}({\cal A}^\text{op}, {\cal B}) \to {\cal B}$, which is $\hat{L}_A(B) = \hom(-,A) \times B$, but since $\hom(-,A) \times B$ (almost?) never preserves limits, this $\hat{L}_A$ can't be the $L_A$ we're looking for!
You can find a proof of my $L_A(B)$ claim by reading between the lines in the proof of Proposition 4.8.1.16 of Lurie's Higher Algebra.
EDIT: I finally understood what Mike Shulman is saying, and it's the same as what I just wrote! Oh well, I'll leave this here in case anyone finds it easier to read than Mike's version.