Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a small simplicial category. The category $Fun(A,s\mathrm{Set})$ of simplicial functors from $A$ to simplicial sets can be given the projective model structure in which fibration and weak equivalences are objectwise.

Now assume further that $A$ is a symmetric monoidal category with respect to some binary operarion $\circ: A\times A\to A$. We can define the Day (or convolution) tensor product on $Fun(A,s\mathrm{Set})$ by the following coend: $$F\otimes G(a)=A(-\circ -,a)\otimes_{A\times A}F(-)\times G(-)$$

My question is:

Is it true that the projective model structure is a monoidal model category in the sense that it satisfies the pushout-product axiom and if so is there a place where this is written down ?

share|improve this question
add comment

3 Answers 3

up vote 5 down vote accepted

[This should probably be a comment, since it is so short. Nevertheless, it is an answer to the question.]

The result you ask for is a consequence of proposition 2.2.15 in Sam Isaacson's Ph.D. thesis (Harvard University, 2009). That proposition is stated for combinatorial symmetric monoidal (closed) model categories, and not just for simplicial sets. In this general case, the statement in Sam Isaacson's thesis requires the base category to have virtually cofibrant morphism objects. This condition is automatic for simplicial sets, as all simplicial sets are cofibrant in the usual model structure.

I have no idea who first proved or published the above result.

share|improve this answer
    
Thanks Ricardo, this answers my question. –  Geoffroy Horel Feb 3 '13 at 17:54
    
@Geoffroy: It was my pleasure. –  Ricardo Andrade Feb 3 '13 at 20:31
add comment

This is perhaps more of a comment but I am not reputable enough yet to leave comments.

Without looking in too much detail, I think Shipley and Schwede assert in their paper "algebras and modules in monoidal model categories" on page 502, that when (with the notation of your question) A is the category of finite simplicial sets and the monoidal product is the smash product, then it does indeed give a monoidal model structure. They talk of it very briefly but give a lot of references. There is an online copy here: http://www.math.uni-bonn.de/people/schwede/AlgebrasModules.pdf

Regards,

Tom

share|improve this answer
add comment

I recently thought about this problem for a general model category rather than $sSet$ and for a different tensor product. I'll tell you what I came up with, because maybe the references can help you, and because I want to write it down somewhere anyway. So let's suppose $A$ is a small category and $M$ is a cofibrantly generated monoidal model category. Define a product on $Fun(A,M)$ by $(X\otimes Y)_a = X_a \otimes Y_a$ for $i\in A$.

Fact: If $A$ has finite coproducts then $Fun(A,M)$ satisfies the pushout product axiom. See Sinan Yalin's 2012 paper for a proof. I tried unsuccessfully to prove the converse, but doing so showed the hypothesis about finite coproducts is there to account for the difference between levelwise cofibrations and projective cofibrations in $Fun(A,M)$. So I believe it's necessary.

Fact: If $A$ is a Reedy category then the Reedy model structure on $Fun(A,M)$ satisfies the pushout product axiom. See Barwick Lemma 4.2

If $M$ is combinatorial, we can define the injective model structure on $Fun(A,M)$. Since the cofibrations are defined to be levelwise cofibrations, we get the pushout product axiom here for free. So this covers all the big model structures on diagram categories, for the objectwise tensor product.

As a side note, if $A = \bullet \to \bullet$ then $Fun(A,M)$ is the arrow category and can be given a different monoidal product, namely the box product $f \Box g$ which is used in the pushout product axiom. In work which is soon to appear, Hovey proves $Fun(A,M)$ inherits the pushout product axiom and monoid axiom from $M$.

Of course, none of this addressed the convolution product, which I've never worked with. But hopefully it'll help you, or at least give you some food for thought. I wonder what happens with the Day product on a Reedy model category, for example.

EDIT: The paper has now been submitted, and a preprint is available on Hovey's new website.

share|improve this answer
1  
The box product on $\textrm{Fun}(A,M)$ you mention is simply the Day convolution product coming from the symmetric monoidal structure on $A=(0\to 1)$ given by $a\otimes b=\min(a,b)=a\cdot b=a\wedge b$. Thus, under some conditions, the result you state that $\textrm{Fun}(A,M)$ inherits the pushout-product and monoid axioms from $M$ follows from propositions 2.2.15 and 2.2.16 in Sam Isaacson's thesis (linked in my answer). Do you know in what generality Mark Hovey proves this result? –  Ricardo Andrade Feb 1 '13 at 4:55
    
I feel uncomfortable going into detail on his unpublished work, and I also don't have the preprint in front of me. I remember it's quite general, like $M$ probably only needs to be a cofibrantly generated monoidal model category. It definitely doesn't need to be combinatorial. I know he later assumes the monoid axiom, but I don't think he needs it for the proof that the pushout product axiom is inherited. He may use the hypothesis that cofibrant objects are flat, since that's one of his favorite hypotheses to place in a situation like this. –  David White Feb 1 '13 at 5:05
    
@David: Thanks. –  Ricardo Andrade Feb 1 '13 at 5:15
1  
Okay, I just dug out the preprint. He uses the projective model structure on $Arr(M)$. You don't need cofibrant objects to be flat. You just need $M$ to be cofibrantly generated and a closed symmetric monoidal model category. The diagram category you get is also cofibrantly generated and a symmetric monoidal model category. If $M$ has the monoid axiom, then so does $Arr(M)$. Hope that helps –  David White Feb 1 '13 at 18:12
    
@David: It does. Thank you very much. –  Ricardo Andrade Feb 1 '13 at 21:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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