Showing that $\text{LMod}_{\mathbb{S}}(\text{Sp}) \simeq \text{Sp}$

Question

$\newcommand{LMod}{\text{LMod}} \newcommand{Sp}{\text{Sp}} \newcommand{SSS}{\mathbb{S}} \newcommand{Fun}{\text{Fun}}\newcommand{Op}{\text{Op}} \newcommand{CAlg}{\text{CAlg}} \newcommand{Alg}{\text{Alg}} \newcommand{LM}{\mathcal{LM}^{\otimes}} $For any $\mathbb{E}_\infty$-ring spectrum $A$, one can consider the ($\infty$-)category of left modules over $A$: $$ \text{LMod}_A(\Sp) := \LMod(\Sp^{\otimes}) \times_{\text{CAlg}(\Sp^{\otimes})}\{A\} . $$ Here $\LMod(\Sp^{\otimes}) = \Alg_{\LM}(\Sp^\otimes)$ for a specific $\infty$-operad $\LM$ encoding "module objects" $(A,M)$, with $A \in \CAlg$, and the fiber is taking over a forgetful functor $\LMod(\Sp^{\otimes}) \to \CAlg(\Sp^{\otimes})$ which only retains the "algebra object".

Similarly to classical algebra, if one considers $A = \SSS$, then one gets the expect equivalence $$ \LMod_{\SSS}(\Sp) \simeq \Sp $$ (which I imagine is also strongly monoidal if one equips $\LMod_{\SSS}(\Sp)$ with the tensor product on the category of modules, but this is another question). One way to prove this is to consider the adjunction $$ \text{forgetful}: \LMod_{\SSS}(\Sp) \rightleftarrows \Sp: - \otimes \SSS $$ and use that the unit is an equivalence, and that $\text{forgetful}$ is conservative, to show that also the counit is an equivalence. This also works more on general over any sym. mon. $\infty$-category $(\mathcal{C}, \otimes, \mathbb{1}_{\mathcal{C}})$.

Question. Is there a more direct approach to show the above equivalence, using only the structure of $\LM$ and its "basic" properties?

The approach above is not particularly complicated but it follows somehow from abstract nonsense, and to develop the adjunction one has to work a bit; it seems to me though that $\LM$ should already encode such an equivalence in a more structural way. Thanks in advance for any suggestion!

Answer 1

The proof I have in mind can be read as an argument in quasicategories and "model-independent" $\infty$-categories at the same time, so I will write $\mathbb{F}_*$ for both the $1$-category, quasicategory and $\infty$-category of pointed finite sets, and I will write $\mathbb{F}_*^\mathrm{int}$ for the wide subcategory on inert morphisms. I also write $\mathcal{LM}^\otimes$ for both the $1$-category Lurie defines in Notation 4.2.1.6 in Higher Algebra and its nerve. Finally, I will prove the statement in an arbitrary symmetric monoidal $\infty$-category $\mathcal{C}^\otimes$, rather than just that of spectra. I will write $I$ for the monoidal unit of $\mathcal{C}^\otimes$.

(Note that I will not use Proposition 4.2.2.12 of Higher Algebra, as it turned out the argument I had originally in mind could be adapted to work for $\mathcal{LM}^\otimes$ as well.)

There is an adjunction $i\colon\mathbb{F}_*^\mathrm{int}\rightleftarrows\mathcal{LM}^\otimes\colon c$, where $i(S)=(S,S)$ is fully faithful, $c(J,S)=S$ is a "collapse functor", the unit is the evident isomorphism, and the counit is $\varepsilon: ic\to\mathrm{id}$ given on an object $(J,S)\in\mathcal{LM}^\otimes$ by the unique map $(S,S)\to(J,S)$ that is the identity on $S$.

Next, we need the notion of a semi-inert morphism, which is Definition 3.3.1.1 in Higher Algebra. It is now clear that $\mathrm{LMod}_I(\mathcal{C}^\otimes)$ is the full subcategory of $\mathrm{Fun}_{\mathbb{F}_*}(\mathcal{LM}^\otimes,\mathcal{C}^\otimes)$ on those functors $F\colon\mathcal{LM}^\otimes\to\mathcal{C}^\otimes$ sending inerts to inerts (equivalently, cocartesian morphisms), and semi-inerts of the form $(J,\varnothing)\to(J',\varnothing)$ to cocartesian morphisms in $\mathcal{C}^\otimes$. These functors are precisely the functors that send all semi-inert morphisms in $\mathcal{LM}^\otimes$ to cocartesian morphisms in $\mathcal{C}^\otimes$, as follows from Lemma 1 below (namely, you have to figure out how semi-inerts in $\mathcal{LM}^\otimes$ look like: they are compositions of an inert followed by a map of the form $(J,S)\to (J',S)$ consisting of an inclusion $J\hookrightarrow J'$; now, Lemma 1 below implies that maps of the latter form are sent to cocartesian morphisms). Since $\varepsilon$ is levelwise semi-inert, this implies that, for any such functor $F$, the natural transformation $F\varepsilon\colon Fic\to F$ is levelwise cocartesian in $\mathcal{C}^\otimes$. If we write $p\colon\mathcal{C}^\otimes\to\mathbb{F}_*$ for the structure map of the symmetric monoidal structure, the map $$p_*\colon\mathrm{Fun}(\mathcal{LM}^\otimes,\mathcal{C}^\otimes)\to\mathrm{Fun}(\mathcal{LM}^\otimes,\mathbb{F}_*)$$ is a cocartesian fibration, and $F\varepsilon$ is a $p_*$-cocartesian morphism because it is levelwise a $p$-cocartesian morphism. The defining property of cocartesian morphisms and the fact that the restriction functor $c^*$ is fully faithful (as $i$ is a fully faithful left adjoint of $c$) then imply that the restriction map $$ i^*\colon\mathrm{Fun}_{\mathbb{F}_*}(\mathcal{LM}^\otimes,\mathcal{C}^\otimes)\to\mathrm{Fun}_{\mathbb{F}_*}(\mathbb{F}_*^\mathrm{int},\mathcal{C}^\otimes) $$ is fully faithful when restricted to those functors that send semi-inert morphisms to cocartesian morphisms. In other words, the forgetful functor $\mathrm{LMod}_I(\mathcal{C}^\otimes)\to\mathcal{C}$ is fully faithful.

In order to show essential surjectiveness of this functor, we write $q\colon\mathcal{LM}^\otimes\to\mathbb{F}_*$ for the structure map of the $\infty$-operad, and write $\mathrm{incl}\colon\mathbb{F}^\mathrm{int}_*\to\mathbb{F}_*$ for the inclusion functor. It is clear that $qi\simeq\mathrm{incl}$. This means that we have a natural transformation $q\varepsilon\colon \mathrm{incl}\circ c\to q$.

We write $\mathcal{C}^\otimes_\mathrm{cc}$ for the wide subcategory of $\mathcal{C}^\otimes$ on cocartesian morphisms, which defines a left fibration $p'\colon\mathcal{C}^\otimes_\mathrm{cc}\to\mathbb{F}_*$ and hence a left fibration $$ \mathrm{Fun}(\mathcal{LM}^\otimes,\mathcal{C}^\otimes_\mathrm{cc})\to\mathrm{Fun}(\mathcal{LM}^\otimes,\mathbb{F}_*). $$

Now select an object $M\in\mathcal{C}$, which equivalently is an $\infty$-operad morphism $M\colon\mathbb{F}_*^\mathrm{int}\to\mathcal{C}^\otimes_\mathrm{cc}$. Considered as such, we can consider the functor $Mc\colon\mathcal{LM}^\otimes\to\mathcal{C}^\otimes_\mathrm{cc}$ (which does not live over $\mathbb{F}_*$), and find a $p'_*$-cocartesian lift $Mc\to\widetilde{M}$ in $\mathrm{Fun}(\mathcal{LM}^\otimes,\mathcal{C}^\otimes_\mathrm{cc})$ of the natural transformation $q\varepsilon$. The natural transformation $M\cong Mci\to\widetilde{M}i$ of functors $\mathbb{F}_*^\mathrm{int}\to\mathcal{C}^\otimes_\mathrm{cc}$ is still levelwise cocartesian, and hence a $p_*$-cocartesian lift of the natural transformation $q\varepsilon i\simeq\mathrm{id}_{qi}$. This is a natural equivalence, and hence so must be $M\to\widetilde{M}i$. But $\widetilde{M}\colon\mathcal{LM}^\otimes\to\mathcal{C}^\otimes_\mathrm{cc}\hookrightarrow\mathcal{C}^\otimes$ is a functor living over $\mathbb{F}_*$ sending semi-inerts to cocartesian morphisms, so we have shown that the forgetful functor $\mathrm{LMod}_I(\mathcal{C}^\otimes)\to\mathcal{C}$ is also essentially surjective, and as such an equivalence of $\infty$-categories.

Lemma 1. Suppose $\mathcal{C}^\otimes\to\mathbb{F}_*$ is a cocartesian fibration (i.e. $\mathcal{C}$ is a symmetric monoidal $\infty$-category). Consider a morphism $f\colon X\simeq(X_1,\ldots,X_n)\to Y\simeq(Y_1,\ldots,Y_m)$ in $\mathcal{C}^\otimes$ living over $\alpha\colon\langle n\rangle\to\langle m\rangle$ in $\mathbb{F}_*$, and write, given $j=1,\ldots,m$, $\rho_j\colon\langle m\rangle\to\langle1\rangle$ for the inert map sending $j$ to $1$ and the rest to the base point. Now write $\alpha_j=\rho_j\alpha$ and $f_j\colon X\to Y\to Y_j$ for the $j$-th component of $f$ (where $Y\to Y_j$ is the inert lift of $\rho_j$ starting at $Y$). Then $f$ is a cocartesian lift of $\alpha$ iff each $f_j$ is a cocartesian lift of $\alpha_j$.

Proof. The forward direction is clear, as each inert $Y\to Y_j$ is cocartesian and cocartesian morphisms are stable under composition. Conversely, if each $f_j$ is cocartesian, then we use that we can find a cocartesian morphism $X\to\alpha_!X$ and a factorization of $f$ through this map followed by $\alpha_!X\to Y$, which lives over $\mathrm{id}_{\langle m\rangle}$. Proving that $f$ is cocartesian is equivalent to proving that this map $\alpha_!X\to Y$ is cocartesian, so we can at this point without loss of generality assume that $\alpha=\mathrm{id}_{\langle m\rangle}$. Our standing assumption is then that each map $f_j$ is inert, i.e. that each map $X_j\to Y_j$ (of which $f$ consists because of $\mathcal{C}^\otimes_{\langle m\rangle}\simeq\prod_{j=1}^m\mathcal{C}$) is an equivalence. This clearly forces $f$ to be an equivalence, and hence to be cocartesian.