On the derived functor of the tensor product in a monoidal category

Question

Let $(\mathcal{M},\otimes)$ be a symmetric monoidal model category; I'll assume for simplicity that every object is fibrant. Suppose that the unit $I$ is NOT cofibrant. I'm interested in whether the derived tensor product with the unit is oplax/strong monoidal.

On the one hand, since $-\otimes^L I$ is the left derived functor of $-\otimes Q$ where $Q$ is a cofibrant replacement of $I$, an oplax structure on the derived functor could come from a monoidal Quillen adjunction with left adjoint $-\otimes Q$. This requires to find a "natural" diagonal map $Q\to Q\otimes Q$, and that the map $Q\otimes Q \to I$ be a weak equivalence.

On the other hand, the unit axiom seems to indicate that $-\otimes^L I$ is strong monoidal anyway, since $I\otimes^L I\simeq I$; but this isomorphism can be presented by two maps: $Q\otimes Q \to Q\otimes I$ and $Q\otimes Q \to I\otimes Q$. Does this really induce a strong monoidal structure, and are the two structures "the same" ?

Answer 1

Yes, the tensor product with a cofibrant replacement can be turned into a lax monoidal functor, where the lax structure maps are weak equivalences.

Consider the model category $\def\Mon{{\rm Mon}} \Mon(M)$ of monoids in $M$. This model structure exists if $M$ satisfies the monoid axiom, which is almost always true in practice.

Consider a cofibrant replacement $Q$ of the monoid $1$ in $\Mon(M)$. The underlying object of $Q$ is a cofibrant object in $M$. (See, for example, Theorem 6.7 in arXiv:1410.5675, but earlier references probably exist.)

Now, the monoid structure of $Q$ equips the functor $Q⊗-$ with a structure of a lax monoidal functor whose lax structure maps are weak equivalences because of the unit axiom for $C$.