When is the cofibrant replacement of a product the product of the cofibrant replacements?

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.

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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). –  David Roberts Aug 26 '11 at 1:46
@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. –  David White Aug 26 '11 at 2:38
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 $C$. –  Denis-Charles Cisinski Aug 26 '11 at 9:49
@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. –  David White Aug 26 '11 at 12:34