Lax Gray tensor product and opposite categories:

Question

For any two strict (infinity, infinity)-categories $A,B$ let $A \otimes B $ be the lax Gray tensor product of $A$ and $B$. Let $A^{op}$ be the opposite (infinity,infinity)-category, where morphisms in all dimensions are reversed. Does $$ (A\otimes B)^{op} \cong A^{op} \otimes B^{op}$$ hold or $$ (A\otimes B)^{op} \cong B^{op} \otimes A^{op}$$ hold ?

One definition of the lax Gray tensor product of (infinity, infinity)-categories is due to Steiner: he constructs the lax Gray tensor product for Steiner (infinity, infinity)-categories, which he extends via colimits to all (infinity, infinity)-categories. Steiner proves an equivalence between the full subcategory of Steiner (infinity, infinity)-categories and a category of socalled Steiner complexes, augmented chain complexes $(A, \partial, \epsilon: A \to \mathbb{Z})$ of abelian groups concentrated in non-negative degrees equipped with a graded submonoid $B$ of $A$ such that for every $n \geq 0$ the monoid $B_n$ and the abelian group $A_n$ are freely generated by the same set. The equivalence sends an (infinity,infinity)-category to the complex whose $n$-chains are freely generated by the $n$-cells modulo the relation identifying sum with composition, the differentials are target -source, the $n$-th submonoid is generated by the $n$-cells. Steiner proves that the tensor product of chain complexes lifts to the category of Steiner complexes and constitutes a monoidal structure. Under Steiner's equivalence forming the opposite of a Steiner (infinity,infinity)-category corresponds to multiplying all differentials $\partial$ of a Steiner complex $(A, \partial, \epsilon: A \to \mathbb{Z}) $ with -1 by definition of the differentials. From this perspective it looks like that the first isomorphism holds since for every chain complexes $$(A, \partial), (A', \partial')$$ we have that $$(A \otimes A', -\partial_{A\otimes A'})= (A, -\partial) \otimes (A',-\partial').$$

On the other hand I read in work of Gagna and others (Gray tensor products and lax functors of $(infinity,2)$-categories, Remark 2.4.) who construct the Gray-tensor product for $(infinity,2)$-categories via scaled simplicial sets that $$ (A \otimes B)^{op} \simeq B^{op} \otimes A^{op}.$$ How does this fit together?

Answer 1

Let me depart from the notation of the question, and write

$$A^{op}$$

for the dual which reverses all odd-dimensional cells, and

$$A^{co}$$

for the dual which reverses all even-dimensional cells. So

$$A^{coop} = A^{opco}$$

is the dual which reverse all cells in all dimensions. We have

$$A^{op} \otimes B^{op} = (B \otimes A)^{op}.$$

We also have

$$A^{co} \otimes B^{co} = (B \otimes A)^{co}.$$

Composing, we have

$$A^{coop} \otimes B^{coop} = (A \otimes B)^{coop}.$$

This answers the question: reversing all cells is a monoidal — not anti-monoidal — functor.