5
$\begingroup$

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...

$\endgroup$

4 Answers 4

3
$\begingroup$
  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.

$\endgroup$
5
  • $\begingroup$ Thank you. I wait for your edit(s) when you will find counterexamples $\endgroup$
    – fosco
    Mar 16, 2016 at 10:30
  • 3
    $\begingroup$ Uhm, why does $\hom(-,A)\times B$ preserve limits? You get stuck in $(\varprojlim \hom(A_i,A))\times B$. $\endgroup$
    – fosco
    Mar 16, 2016 at 11:03
  • 1
    $\begingroup$ @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, 2016 at 2:54
  • 1
    $\begingroup$ 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, 2016 at 17:39
  • 2
    $\begingroup$ 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, 2016 at 18:01
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$
  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.

$\endgroup$
1
  • $\begingroup$ I didn't understand what you wrote, wrote my own answer, then realized you probably meant the same thing! $\endgroup$ Mar 30, 2016 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.