6
$\begingroup$

I'm in a situation where I'd like to prove $Q(E\otimes E) \simeq QE \otimes QE$ for a monoid $E$ in a symmetric monoidal model category. I know it's not true in general that $Q(E\otimes F)\simeq QE \otimes QF$ (with no monoid hypothesis), see e.g. link, where it fails for dgcat. Since dgcat is a pretty nice category, I'm starting to wonder if I have any hope at all of the weak equivalence, but I haven't seen anything in the literature where $E$ is a monoid, so I figured I'd ask. If nothing else, it would be nice to have some word from the experts as to whether this is a reasonable hope or not. I'm also willing to make some assumptions on the model category, e.g. that it satisfies the monoid axiom.

Improve this question
$\endgroup$
6
  • $\begingroup$ 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). $\endgroup$
    – David Roberts ♦
    Commented Aug 26, 2011 at 1:46
  • $\begingroup$ @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. $\endgroup$
    – David White
    Commented Aug 26, 2011 at 2:38
  • 1
    $\begingroup$ 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 $C$. $\endgroup$
    – D.-C. Cisinski
    Commented Aug 26, 2011 at 9:49
  • 1
    $\begingroup$ @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. $\endgroup$
    – David White
    Commented Aug 26, 2011 at 12:34
  • 1
    $\begingroup$ This is true for coconnective dg-algebras, according to arxiv.org/abs/1112.2360. $\endgroup$
    – AAK
    Commented Jan 29, 2015 at 20:57

1 Answer 1

Reset to default
2
$\begingroup$

I asked this question a very long time ago, when I was just starting to do research in abstract homotopy theory. This is a classic case of an xy problem, where I had a proof in mind, and by asking this question, I realized it was a foolish way to proceed. I later came up with a much better proof, and it's part of this paper.

The comments above show that the condition I asked for is too strong to expect it to hold in many examples of interest. Today, thanks to a paper by Maximilien Peroux, I learned of a concrete counterexample. This condition I wanted would imply a functorial lax comonoidal cofibrant replacement $Q$ on the category of monoids in $M$. In his Remark 3.3, Maximilien points out that the existence of such a $Q$ is related to whether or not the $\infty$-category of monoids satisfies cocommutative rigidification (his Definition 3.1). He later provides concrete counterexamples where this fails (Remark 5.6), such as the category of monoids in symmetric spectra, which was exactly the kind of example of a symmetric monoidal model category that I would not want to rule out by assuming my $M$ satisfied the (far too restrictive) condition that $Q(E\otimes F) \simeq QE \otimes QF$ for monoids $E$ and $F$.

Improve this answer
$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .