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" ?

  • $\begingroup$ 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 $\endgroup$ Oct 2, 2021 at 13:53
  • $\begingroup$ @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. $\endgroup$ Oct 4, 2021 at 17:02
  • $\begingroup$ @DmitriPavlov : ah - ok, I would not call this "$\otimes^L I$" in this case, but the question is clearer now. $\endgroup$ Oct 4, 2021 at 17:09
  • $\begingroup$ The tensor product with the unit preserves weak equivalences so it is already derived. $\endgroup$ Oct 5, 2021 at 18:44

1 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 $C$.

  • $\begingroup$ Do you mean "the underlying object $Q$ is a cofibrant object of $M$" ? $\endgroup$ Oct 5, 2021 at 14:19
  • $\begingroup$ 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 ? $\endgroup$ Oct 5, 2021 at 14:47
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 5, 2021 at 16:32
  • $\begingroup$ @AdrienMORIN: Yes, the underlying object of Q is cofibrant as an object of M. $\endgroup$ Oct 5, 2021 at 16:32

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