Is the free algebra functor over an $\infty$-operad symmetric monoidal?

Question

Suppose $F: \mathcal{O}^\otimes \to \mathcal{P}^\otimes$ is a map of $\infty$-operads, and $\mathcal{C}$ is a symmetric monoidal $\infty$-category that admits small colimits, such that the tensor product preserves small colimits in both variables separately. Corollary 3.1.3.5 in Lurie's "Higher Algebra" shows that the forgetful functor $\operatorname{Alg}_{\mathcal{P}}(\mathcal{C}) \to \operatorname{Alg}_{\mathcal{O}}(\mathcal{C})$ induced by precomposition with $F$ admits a left adjoint $$ \operatorname{Free}: \operatorname{Alg}_{\mathcal{O}}(\mathcal{C}) \to \operatorname{Alg}_{\mathcal{P}}(\mathcal{C}) $$ sending a $\mathcal{O}$-algebra to the free $\mathcal{P}$ algebra generated by it. Both $\operatorname{Alg}_{\mathcal{O}}(\mathcal{C})$ and $\operatorname{Alg}_{\mathcal{P}}(\mathcal{C})$ admit canonical symmetric monoidal structures via the tensor product of algebras in HA Example 3.2.4.4. My question is the following: Is the free algebra functor $\operatorname{Free}$ symmetric monoidal?

My issue in showing this is mostly that the definition of the tensor product of algebras above is done via a simplicial construction, and Lurie doesn't seem to give any universal property that characterizes it. Therefore, before I delve too deep into a simplicial calculation that indicates to become pretty hard, I wanted to ask whether anyone has a different idea about how to show this statement, or knows about such a universal property that might help me.

Edit: Let me sketch a first attempt at a partial answer. For $K \to \operatorname{Fin}$ any simplicial set over the category of finite pointed sets, write $K^\flat$ for the associated preoperad where only identity morphisms are marked. Also, write $\operatorname{Op}(L)^\otimes$ for the fibrant replacement of a preoperad, i.e., the associated $\infty$-operad.

To obtain a map of simplicial sets over $\operatorname{Fin}$, by the definition of the symmetric monoidal structure on $\operatorname{Alg}_{\mathcal{O}}(\mathcal{C})$ we can equivalently give a natural map in $K$ $$ \operatorname{Hom}^{\natural}_{\operatorname{sSet}_{/\operatorname{Fin}}}(K^\flat \odot\mathcal{O}^\natural, \mathcal{C}^\natural) \to \operatorname{Hom}^{\natural}_{\operatorname{sSet}_{/\operatorname{Fin}}}(K^\flat \odot\mathcal{P}^\natural, \mathcal{C}^\natural)$$ where $\mathcal{O}^\natural$ denotes the inert marking, and $\odot$ is the tensor product from HA Notation 2.2.5.5. Such a map exists, namely since fibrant replacement is left adjoint to the inclusion of operads into preoperads, one can use the free algebra functor for the map of operads $$\operatorname{Op}(K^\flat \odot\mathcal{O}^\natural)^\otimes = \operatorname{Op}(K^\flat)^\otimes \otimes \mathcal{O}^\otimes \to \operatorname{Op}(K^\flat \odot\mathcal{P}^\natural)^\otimes$$ for this purpose, which on $K = \operatorname{Fin}$ recovers the free algebra functor we are looking for.

It is however not clear that this map sends coCartesian morphisms to coCartesian morphisms. As a reminder, a morphism $\alpha$ in $\operatorname{Alg}_{\mathcal{O}}(\mathcal{C})$ is coCartesian iff for all $O$ in the underlying category of $\mathcal{O}$, the induced morphism $\alpha(O)$ in $\mathcal{C}^\otimes$ is coCartesian, by HA 3.2.4.3.

Answer 1

The answer is no in general, and I can't think of reasonable conditions under which it is yes. David's answer was about the "wrong" monoidal structure but it is still helpful : the free functor is analogous to extension of scalars along a map of bialgebras, and this "should not" preserve the underlying tensor product (defined as soon as you have some kind of bialgebra structure). (In fact this analogy can be made precise by considering free operads, so that counterexamples yield counterexamples)

What we can say in general is that the forgetful functor is (strong) symmetric monoidal and that the left adjoint is therefore oplax monoidal.

Let's give an example to see that it is just not strong monoidal : say $O$ is the initial operad, and $P$ arbitrary (you can pick the commutative operad for concreteness). In this case, our functor is simply the free $P$-algebra functor $C \to Alg_P(C)$, whose "formula" is $X \mapsto \coprod_n (P(n)\otimes X^{\otimes n})_{h\Sigma_n}$.

Now the canonical map $F(X \otimes Y) \to F(X) \otimes F(Y)$ is the one that sends $(P(n) \otimes (X\otimes Y)^{\otimes n})_{h\Sigma_n}$ to $(P(n) \otimes X^{\otimes n})_{h\Sigma_n} \otimes (P(n)\otimes Y^{\otimes n})_{h\Sigma_n}$

It seems very rare for this to be an equivalence. In fact, given how both sides look, it seems very rare for the even exist an abstract equivalence.

Now one could ask for a lax symmetric monoidal structure instead. By some sort of yoga with adjunctions, this could not be compatible with the symmetric monoidal structure on the right adjoint (because of the above), and so it seems a bit unnatural but one might still ask.

It turns out one can also disprove that. Let me give an example : let $P$ be the associative operad, and again $O$ is initial. If there existed a lax symmetric monoidal structure, $Free(-)$ would send algebras to algebras in algebras i.e. $\mathbb E_2$-algebras (or, if we're dealing with $1$-categories as a special case, commutative algebras). I claim that this is not the case : the free associative algebra on any set with more than one element is non-commutative, even if that set admits a monoid structure.

Answer 2

It is instructive to think about the simplest possible case, and do away with all the language of $\infty$-operads. If $R$ and $S$ are rings, and we have a ring homomorphism $f:R\to S$, then the left adjoint you're interested in is "extension of scalars," from $R$-modules to $S$-modules. If the rings are commutative, then this is monoidal. If the rings are not commutative, that can fail, as you can see from the formula in the linked answer (if you want concrete counterexamples, look at change of groups in modular representation theory).

What normally happens is that the right adjoint is lax monoidal. In my work with Donald Yau we get deep into the weeds of these lifted adjoint situations, and we use the comonoidal structure map of the left adjoint, induced from the lax monoidal structure of the right adjoint (see Def 2.3.2 here). My point is that the map actually goes the wrong way.

You might also be interested in this partial positive result