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

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" ?

• Maybe I'm confused but doesn't the unit axiom exactly say that $-\otimes^L I$ is the identity ? In particular it has a canonical strong monoidal structure Oct 2, 2021 at 13:53
• @MaximeRamzi: ⊗^L I is not the identity, it is only weakly equivalent to the identity. The point of the question is whether it is possible to provide a 1-categorical model as a lax/oplax monoidal functor. Oct 4, 2021 at 17:02
• @DmitriPavlov : ah - ok, I would not call this "$\otimes^L I$" in this case, but the question is clearer now. Oct 4, 2021 at 17:09
• The tensor product with the unit preserves weak equivalences so it is already derived. Oct 5, 2021 at 18:44

## 1 Answer

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 $$C$$.

• Do you mean "the underlying object $Q$ is a cofibrant object of $M$" ? Oct 5, 2021 at 14:19
• Could you comment a bit about the last part of my question, i.e. whether the derived structures one obtains through different cofibrant replacements of 1 in Mon(M) are essentially the same ? I suppose it somehow comes from the "homotopical uniqueness" of cofibrant replacements ? Oct 5, 2021 at 14:47
• @AdrienMORIN: If you have two cofibrant replacements Q and Q' of the monoid 1 in Mon(M), they are connected by a zigzag of weak equivalences Q→Q''←Q' in Mon(M). This zigzag induces a zigzag of weak equivalences of lax monoidal functors, as constructed in the answer. Oct 5, 2021 at 16:32
• @AdrienMORIN: Yes, the underlying object of Q is cofibrant as an object of M. Oct 5, 2021 at 16:32