I am currently reading Higher Algebra by Jacob Lurie and I have a question regarding equivalences of $\mathcal{O}$-monoidal categories.
Let $\mathcal{O}$ be an $\infty$-operad. Suppose that I have two $\mathcal{O}$-monoidal categories $\mathcal{A}^\otimes\xrightarrow[]{p}\mathcal{O}^\otimes$ and $\mathcal{B}^\otimes\xrightarrow[]{p'}\mathcal{O}^\otimes$. It seems to me that there are two natural definitions of equivalence of $\mathcal{O}$-monoidal categories:
- Weak equivalence: if there exists an $\mathcal{O}$-monoidal functor $F:\mathcal{A}^\otimes\to \mathcal{B}^\otimes$ and the functor is a categorical equivalence of the $\infty$-categories, i.e. there exists a homotopic inverse of $F$ but we are not asking for the inverse to be $\mathcal{O}$-monoidal.
- Strong equivalent: if there exists an $\mathcal{O}$-monoidal functor $F:\mathcal{A}^\otimes\to \mathcal{B}^\otimes$ and an $\mathcal{O}$-monoidal functor $G:\mathcal{A}^\otimes\to \mathcal{B}^\otimes$ such that $F$ and $G$ are homotopic inverses.
I didn't find in J. Lurie's book any mention of the possibility of having two distinct definitions, so I started to suspect that they might be equivalent.
QUESTION: Is it well known that those two definitions are equivalent, or there is an example where a weak equivalence that is not a strong equivalence?
I made the following attempt to prove that the two definitions are equivalent, tinkering with CoCartesian morphisms and some results from Higher Topos Theory. However, I am still unfamiliar with the framework of $\infty$-categories, and I am afraid that I might have overlooked some detail.
ATTEMPT: Let $\mathcal{A}^\otimes$, $\mathcal{B}^\otimes$ be two $\mathcal{O}$-monoidal categories, $F: \mathcal{A}^\otimes\to \mathcal{B}^\otimes$ a $\mathcal{O}$-monoidal functor and $G: \mathcal{B}^\otimes\to \mathcal{A}^\otimes$ homotopy inverse to $F$, so $F$ is a weak equivalence. We wish to prove that $G$ is $\mathcal{O}$-monoidal.
Let $B\xrightarrow{f}B'$ be a $p'$-CoCartesian morphisms of $\mathcal{B}^\otimes$ covering the morphism $X\xrightarrow{\alpha}X'$ of $\mathcal{O}^\otimes$. We want to prove that $G(f)$ is $p$-CoCartesian. Let's first consider the case where $B$ is in the image of $F$ and let $A\in \mathcal{A}^\otimes$ such that $F(A)=B$.
Since $\mathcal{A}^\otimes$ is $\mathcal{O}$-monoidal there exists a $p$-CoCartesian morphism $X\xrightarrow[]{h}X''$ covering $\alpha$ with source $A$.
Since $F$ is an $\mathcal{O}$-monoidal map and $h$ is $p$-CoCartesian, then $F(h)$ is $p'$-CoCartesian and it defines the morphism $\phi$ covering the identity, up to a contractible space of choices. Since both $f$ and $F(h)$ are $p'$-CoCartesian morphism with the same source covering $\alpha$ the morphism $\phi$ must be an equivalence.
The image of $\phi$ by the functor $G$ is again an equivalence, therefore a $p$-CoCartesian morphism. Since $G\circ F$ is homotopic to the identity it is $\mathcal{O}$-monoidal, and $h$ was $p$-CoCartesian therefore $G\circ F(h)$ is $p$-CoCartesian covering $\alpha$. Applying the dual version of [HTT Prop. 2.4.1.7] we obtain that $G(f)$ is $p$-CoCartesian.
Let's now consider the case where $B$ is not in the image of $F$. Since $F$ is essentially surjective, there exist an object $\overline{B}\in \mathcal{B}^\otimes$ in the image of $F$ and an equivalence $\overline{B}\xrightarrow[]{g}B$. From the previous case we obtain that $G(g\circ f)$ is $p$-CoCartesian, and since $g$ is an equivalence then $G(g)$ is $p$-CoCartesian. We can apply again [HTT Prop. 2.4.1.7] to conclude that $G(f)$ is $p$-CoCartesian.
Best Answer
They are indeed the same, and in fact you can make the weak version even weaker : if $F$ is strong $O$-monoidal, it suffices to assume that that $F_o$ is an equivalence for every $o\in O^\otimes_{\langle 1\rangle}$.
By 2.4.4.4. or 3.3.1.5. in HTT it will follow that $F$ is an equivalence. Then it is a general exercise to prove that your attempt works for any fibrations: if $X\to Y$ is a map of fibrations over $S$ which is an equivalence, then it must preserve cocartesian edges.