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Are there common examples of a symmetric monoidal product $\otimes$ that preserves both limits and colimits in each variable?

This question is worded incorrectly, I now realize:

(A) I originally meant that, give diagrams $F_1,F_2: D \to \mathcal{C}$, we have $$ (\text{colim}_{X \in D} F_1(X)) \otimes (\text{colim}F_2(X)) \cong \text{colim}_{X \in D} (F_1(X) \otimes F_2(X)) $$ and $$ (\text{lim}_{X \in D} F_1(X)) \otimes (\text{lim}_{X \in D}F_2(X)) \cong \text{lim}_{X \in D} ( F_1(X) \otimes F_2(X)) $$ where $\otimes$ is the symm. monoidal structure on $\mathcal C$.

(B) But as pointed out by Martin, my original wording was incorrect. And of course it's an important concept to consider monoidal structures that do preserve (co)limits in each variable, so that $$ (\text{colim}_{X \in D} F(X)) \otimes A \cong \text{colim}_{X \in D} (F(X) \otimes A) $$ and likewise for limits. And I am interested in such examples as well.

Here are examples:

  • Tensor product of chain complexes preserves a lot of colimits
  • Direct sum of groups/chain complexes preserves all colimits in the sense of (A) (since it is a colimit)
  • Cartesian product preserves all limits in the sense of (A) (since it is a limit)

But I can't think of a product that preserves both limits and colimits.

It's okay if the symmetric monoidal structure doesn't preserve all limits and colimits. For instance, just preserving totalization and geometric realization would be a good example, or just finite limits and finite colimits.

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The cartesian product in a cartesian closed category preserves colimits in each variable separately, but it doesn't preserve products! –  Zhen Lin Sep 23 '13 at 15:19
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"Direct sum of groups/chain complexes preserves all colimits (since it is a colimit)" - No, $A \oplus \mathrm{colim}_i B_i = \mathrm{colim}_i (A \oplus B_i)$ does not hold for example for the empty diagram or more generally any coproducts. –  Martin Brandenburg Sep 23 '13 at 22:05
    
Martin, sorry for the confusion. I think I used "preserve colimit" in an ambiguous way. What I meant was that if I have two diagrams $F,G: D \to C$ (where $C$ is the category I care about), then $\text{colim} F \oplus \text{colim} D \cong \text{colim} (F \oplus G)$. I'll clarify the question! –  Guest Sep 25 '13 at 3:02
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3 Answers 3

If the category $ Mondl $ is monoidal closed, so that we have a natural isomorphism of the form

$$ Mondl(A \otimes B, C) \cong Mondl(A, MONDL(B,C)) $$

then that says that the functor $Mondl(A,-)$ has a right adjoint and so preserves colimits. If $\otimes$ is symmetric, we get preservation of colimits on the other side.

In a category in which the coproducts and products are naturally isomorphic, such as an abelian category, we will also of course get preservation of products under the same conditions.

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I must confess that I don't understand your notation... What does "Mondl" stand for? –  tetrapharmakon Sep 23 '13 at 16:54
    
@tetrapharmakon I guess it meant the internal hom-functor (since he called $Mondl$ for the external hom, $MONDL$ should be the internal hom that give the closedness of the symmetric monoidal category). –  Giorgio Mossa Sep 24 '13 at 9:01
    
Quite right, Giorgio! I quite like this convention, but it means that you have to give a name for the category which has upper and lower case letters, and $Mon$ could be the category of monoids, so I chose $Mondl$ as uncommitted. In this convention, $CAT$ is the internal hom in the category $Cat$ of small categories and functors. –  Ronnie Brown Sep 24 '13 at 11:25
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In every abelian category, for instance $\mathbf {Ab}$ of abelian groups and more generally the category $R$-$\mathbf{Mod}$ of $R$-modules, the binary product $\times$ is the same as the binary coproduct $\amalg$ because they are both the direct sum $\oplus$.

So in such categories these symmetric monoidal products commute with both limits and colimits, since they are limits and colimits.

Anyway as Martin Brandenburg pointed out this doesn't mean that such tensor product preserve limits or colimits in each variable.

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That's not true. $A \oplus 0$ is not $0$. –  Martin Brandenburg Sep 23 '13 at 21:50
    
@MartinBrandenburg Good point, apparently I didn't read carefully the question and thought it was about finding out monoidal product which commute with colimit and limit. I'm gonna edit thanks. –  Giorgio Mossa Sep 24 '13 at 9:03
    
Well the monoidal product also doesn't preserve colimits of limits. –  Martin Brandenburg Sep 24 '13 at 9:31
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Some trivial examples (perhaps I will come up with more interesting ones later): If $R$ is a commutative ring, then the tensor product on the category of $R$-modules preserves colimits in each variable (simply because it is closed). It preserves finite limits in each variable when it is restricted to the category of flat $R$-modules. Note, however, that this only refers to limits taken inside the larger category of $R$-modules. So I don't say here that the monoidal category of flat $R$-modules is an example. However, if we even restrict to the category of finitely presented flat modules, i.e. locally free modules of finite rank, then this monoidal category is "coclosed" in the sense that each $M \otimes -$ has a left adjoint (namely $\check{M} \otimes -$), hence preserves all limits. More generally, if we start with an arbitrary monoidal category, we get a monoidal category of dualizable objects, whose tensor product preserves all limits in each variable.

Here is a sort of converse (quite weak, I have to admit): Assume that $C$ is a presentable monoidal category such that the tensor product commutes with colimits and limits in each variable. Also, assume that the Eilenberg-Watts Theorem holds in $C$. Then, for every $M \in C$, the functor $M \otimes -$ is accessible and continuous, hence by the adjoint functor theorem has a left adjoint $F : C \to C$. By Eilenberg-Watts we have $F(T)=T \otimes \check{M}$ for some $\check{M}$. Then $\check{M}$ is dual to $M$.

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