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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?

  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...

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  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.

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  • $\begingroup$ Thank you. I wait for your edit(s) when you will find counterexamples $\endgroup$
    – fosco
    Mar 16 '16 at 10:30
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    $\begingroup$ Uhm, why does $\hom(-,A)\times B$ preserve limits? You get stuck in $(\varprojlim \hom(A_i,A))\times B$. $\endgroup$
    – fosco
    Mar 16 '16 at 11:03
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    $\begingroup$ @FoscoLoregian That's a good question. You would agree, I think, that it preserves limits in the special case when $\mathcal B$ is the category of sets. But in general, you might need a "coproducts are disjoint" type condition... $\endgroup$ Mar 18 '16 at 2:54
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    $\begingroup$ As Fosco noticed, $\hom(-,A) \times B$ (almost) never preserves limits: for example, $\hom(A_1 \sqcup A_2, A) \times B = \hom(A_1, A) \times \hom(A_2, A) \times B \neq \left( \hom(A_1,A) \times B \right) \times \left( \hom(A_2,A) \times B \right)$. $\endgroup$ Mar 30 '16 at 17:39
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    $\begingroup$ Oh, I forgot to say: notice in particular that it does not preserve products in the case $\cal B$ is the category of sets! $\endgroup$ Mar 30 '16 at 18:01
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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.

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A "toy example" may serve as a source for guiding intuition. For ${\cal A}=\operatorname{Opens}(X)$ and ${\cal B}=\operatorname{Opens}(Y)$, one has ${\cal A}\otimes{\cal B}\cong\operatorname{Opens}(X\times Y)$, with $A\otimes B$ corresponding to the open rectangle $A\times B\subseteq X\times Y$. Then in terms of functors, general $U\subseteq X\times Y$ correspond to Galois connections $\langle U^\land,U^\lor\rangle$ between $\operatorname{Opens}(X)$ and $\operatorname{Opens}(Y)$, with $U^\land(A)=\bigcup\{B\mid A\times B\subseteq U\}$ and $U^\lor(B)=\bigcup\{A\mid A\times B\subseteq U\}$. The projections ${\cal A}\otimes{\cal B}\to{\cal A},{\cal B}$ are given by $U\mapsto U^\lor(Y),U^\land(X)$, with left adjoints $A\mapsto A\times Y$, $B\mapsto X\times B$ respectively. Then $(A\times Y)^\land(A')$ is $Y$ if $A'\subseteq A$ and $\varnothing$ otherwise, while $(A\times Y)^\lor(B')$ is $A$ if $B'\ne\varnothing$ and $X$ if $B'=\varnothing$. Finally, $A\otimes B=A\times B=(A\times Y)\cap(X\times B)$.

These "ifthenelses" then show that it is not quite straightforward to handle the general case.

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  1. I think if you trace through the proof that $\mathrm{Func}(\mathcal{A}^{\mathrm{op}},\mathcal{B})_R$ has the universal property of $\mathcal{A}\otimes\mathcal{B}$, then the adjoint $\mathcal{A}^{\mathrm{op}}\times (\mathcal{A}\otimes\mathcal{B})\to\mathcal{B}$ of the universal bilinear functor corresponds simply to "evaluation".

I can't think right now of any way to extract from this a description of the universal bilinear functor itself, other than writing down the adjoint functor theorem formula.

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  • $\begingroup$ I didn't understand what you wrote, wrote my own answer, then realized you probably meant the same thing! $\endgroup$ Mar 30 '16 at 17:53

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