Timeline for When is the cofibrant replacement of a product the product of the cofibrant replacements?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 29, 2023 at 23:08 | vote | accept | David White | ||
| Nov 29, 2023 at 23:07 | answer | added | David White | timeline score: 2 | |
| Jan 29, 2015 at 23:25 | comment | added | David White | Awesome, thanks for sharing! I will definitely read that paper, and I didn't know about it before. | |
| Jan 29, 2015 at 20:57 | comment | added | AAK | This is true for coconnective dg-algebras, according to arxiv.org/abs/1112.2360. | |
| Aug 26, 2011 at 12:34 | comment | added | David White | @Denis-Charles Cisinski: Thanks. I was aware of the result for when the unit is cofibrant. Sadly, in my case it is not cofibrant. Do you know of any results in that case? I'm sad to hear it's "not more reasonable" but I'm very thankful for the advice of an expert to help me gauge the level of difficulty. | |
| Aug 26, 2011 at 9:49 | comment | added | D.-C. Cisinski | What you ask is not more reasonnable than the formula $Q(E\otimes F)\simeq Q(E)\otimes Q(F)$ (which is a complicated way to say that weak equivalences are stable by tensor product). However, if your model category $C$ satisfies the monoid axiom, then the category $Mon(C)$ of monoids is endowed with a model structure; if the unit is cofibrant, then the forgetful functor $Mon(C)\to C$ preserves cofibrant objects (and weak equivalences). Hence, if $E$ is a monoid, there is a morphism of monoids $E'\to E$ which is a trivial fibration in $C$, such that $E'$ is cofibrant in $Mon(C)$ and thus in $C$. | |
| Aug 26, 2011 at 2:38 | comment | added | David White | @David, yes, I'm trying to prove it's monoidal and don't really know how to proceed. I was trying to get the monoidal structure on $QE$ step by step and this was a part of how to get the multiplication. I have some work for the unit map and I think I see how to proceed there. Perhaps there's some way to prove $Q$ is monoidal without going through all those commutative diagrams? If so, I'd love to hear it. | |
| Aug 26, 2011 at 1:46 | comment | added | David Roberts♦ | Clearly, if you have a cofibrant replacement functor which is monoidal, then what you want is true. I assume that this is not obviously true for your case (or this is the fact you want to prove). | |
| Aug 25, 2011 at 22:02 | history | asked | David White | CC BY-SA 3.0 |