Let $\mathrm{K}$ be a field.
Denote $ \mathrm{Vect}_{\mathrm{K}} $ the category of K-vector spaces, $\mathrm{Coalg }^{\mathrm{conil } } $ the category of conilpotent, coaugmented, coassociative coalgebras over K and $\mathrm{Cocoalg }^{\mathrm{conil } } $ the category of conilpotent, coaugmented, coassociative and cocommutative coalgebras over K.
Then the forgetful functors $\mathrm{Coalg }^{\mathrm{conil } } \to \mathrm{Vect}_{\mathrm{K}} $ and $\mathrm{Cocoalg }^{\mathrm{conil } } \to \mathrm{Vect}_{\mathrm{K}} $ that take the cokernel of the coaugmentation admit right adjoints $\mathrm{T}: \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Coalg }^{\mathrm{conil } }$ respectively $\mathrm{R}: \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Cocoalg }^{\mathrm{conil } }$ and we have natural isomorphisms $\mathrm{T}(\mathrm{V}) \cong \oplus_{ i \geq 0} \mathrm{V}^{\otimes i}$ and $\mathrm{R}(\mathrm{V}) \cong \oplus_{ i \geq 0} (\mathrm{V}^{\otimes i})^{ \Sigma_i}$ of K-vector spaces.
Denote $ \mathrm{Calg} $ the category of augmented, associative and commutative algebras over K and $\mathrm{S} : \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Calg}, \ \mathrm{V} \mapsto \oplus_{ i \geq 0} (\mathrm{V}^{\otimes i})_{ \Sigma_i} $ the free commutative algebra functor with its canonical augmentation.
If the field K has char. 0, we have a canonical isomorphism $(\mathrm{V}^{\otimes i})_{ \Sigma_i} \cong (\mathrm{V}^{\otimes i})^{ \Sigma_i}$ of K-vector spaces and so a canonical isomorphism $\mathrm{S}(\mathrm{V}) \cong \mathrm{R}(\mathrm{V})$ of K-vector spaces that makes $\mathrm{S}(\mathrm{V}) $ to a commutative and cocommutative bialgebra.
My question is: Does this generalize to $\infty$-categories in the following sense?
Let $\mathcal{C}$ be a stable presentable symmetric monoidal $\infty$-category.
Then one has $\infty$-categories $\mathrm{Calg}(\mathcal{C})^\mathrm{ }$ and $\mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }:= (\mathrm{Calg}(\mathcal{C^\mathrm{op}})^\mathrm{})^\mathrm{op}$ of augmented, associative and commutative algebras respectively of coaugmented, coassociative and cocommutative coalgebras and one can define conilpotent coalgebras in the following way:
The forgetful functor $\psi: \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } \to \mathcal{C}$ that takes the cokernel of the coaugmentation admits a unique section $\mathrm{E} : \mathcal{C} \to \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ that sends an object $\mathrm{X}$ of $\mathcal{C}$ to its co-square-zero extension, whose coaugmentation has cocernel $\mathrm{X}$ with 0-comultiplication $0: \mathrm{X} \to \mathrm{X} \otimes \mathrm{X}.$
Denote $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil }$ the smallest full subcategory of $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ that contains the essential image of $\mathrm{E}$ and closed under small colimits. $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil } \subset \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } $ is a colocalization.
By the adjoint functor theorem $\psi: \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } \to \mathcal{C}$ admits a right adjoint and so the restriction $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil } \to \mathcal{C}$ of $\psi$ admits a right adjoint $\mathrm{R}.$
Similarly one defines $ \mathrm{Coalg}(\mathcal{C})^\mathrm{conil }$ and shows that the forgetful functor $ \mathrm{Coalg}(\mathcal{C})^\mathrm{conil } \to \mathcal{C}$ admits a right adjoint $\mathrm{T}.$
- Question: Given an object $\mathrm{X}$ of $\mathcal{C} $ do we have a canonical equivalence $\mathrm{T}(\mathrm{X}) \simeq \oplus_{ i \geq 0} \mathrm{X}^{\otimes i}$ in $\mathcal{C} $?
The forgetful functor $\mathrm{Calg}(\mathcal{C})^\mathrm{ } \to \mathcal{C} $ that takes the kernel of the augmentation admits a left adjoint $\mathrm{S}$ and we have a canonical equivalence $\mathrm{S}(\mathrm{X}) \simeq \oplus_{ i \geq 0} (\mathrm{X}^{\otimes i})_{ \Sigma_i} $ in $ \mathcal{C} $ for $\mathrm{X} \in \mathcal{C}.$
One can promote $\mathrm{S} $ to an oplax symmetric monoidal functor $\mathcal{C}^\times \to \mathrm{Calg}(\mathcal{C})^\otimes $ that yields a functor $$\bar{\mathrm{S}} : \mathcal{C} \simeq \mathrm{Cocoalg}(\mathcal{C}^\times) \to \mathrm{Cocoalg}(\mathrm{Calg}(\mathcal{C})^\otimes)=: \mathrm{Bialg}(\mathcal{C})$$ lifting $\mathrm{S} $.
The symmetric monoidal forgetful functor $\mathrm{Calg}(\mathcal{C}) \to \mathcal{C}$ (not the functor that takes the augmentation ideal) induces a forgetful functor $\alpha: \mathrm{Bialg}(\mathcal{C}) \to \mathrm{Cocoalg}(\mathcal{C})$.
This way for every $\mathrm{X} \in \mathcal{C}$ the object $\mathrm{S}(\mathrm{X} )$ carries the structure of a coaugmented conilpotent cocommutative coalgebra in $\mathcal{C}.$
Assume that $\mathcal{C} $ is additionally $\mathbb{Q}$-linear (i.e. receives a left adjoint symmetric monoidal functor from $\mathrm{H}(\mathbb{Q})$-module spectra.)
In this case the norm map $(\mathrm{X}^{\otimes i})_{ \Sigma_i} \to (\mathrm{X}^{\otimes i})^{ \Sigma_i}$ is an equivalence for every $\mathrm{X} \in \mathcal{C} $ and $i \geq 0.$
- Question: Is the composition $\mathcal{C} \xrightarrow{\bar{\mathrm{S}}} \mathrm{Bialg}(\mathcal{C}) \xrightarrow{\alpha} \mathrm{Cocoalg}(\mathcal{C})$ equivalent to $\mathrm{R}?$
By the definition of $\mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil }$ and that for every $\mathrm{X} \in \mathcal{C}$ the coalgebra $\mathrm{S}(\mathrm{X} )$ is conilpotent, the statement of 2. is equivalent to the following condition:
For every $\mathrm{X}, \mathrm{Y} \in \mathcal{C} $ the map $$\mathrm{Cocoalg}(\mathcal{C})(\mathrm{E}(\mathrm{X}), \mathrm{S}(\mathrm{Y})) \to \mathcal{C}(\mathrm{X}, \psi(\mathrm{S}(\mathrm{Y}))) \to \mathcal{C}(\mathrm{X}, \mathrm{Y}) $$ induced by $\psi: \mathrm{Cocoalg}(\mathcal{C}) \to \mathcal{C}$ and composition with the projection to the first factor $ \psi(\mathrm{S}(\mathrm{Y})) \simeq \oplus_{ i > 0} (\mathrm{Y}^{\otimes i})_{ \Sigma_i} \to \mathrm{Y}$ is an equivalence.
Being a section of $\psi$ the functor $\mathrm{E}: \mathcal{C} \to \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ admits a right adjoint $\mathcal{P}$ by the adjoint functor theorem with $\mathcal{P} \circ \mathrm{R} \simeq \mathrm{id}.$
One can show that $\bar{\mathrm{S}} : \mathcal{C} \to \mathrm{Bialg}(\mathcal{C})$ is left adjoint to the composition $\bar{\mathcal{P}}: \mathrm{Bialg}(\mathcal{C}) \xrightarrow{\alpha} \mathrm{Cocoalg}(\mathcal{C}) \xrightarrow{\mathcal{P} } \mathcal{C}$.
If 2. would hold, we would have $\mathrm{id} \simeq \mathcal{P} \circ \mathrm{R} \simeq \mathcal{P} \circ \mathrm{\alpha} \circ \bar{\mathrm{S}} \simeq \bar{\mathcal{P}} \circ \bar{\mathrm{S}}$ being the unit of the adjunction $\bar{\mathrm{S}} : \mathcal{C} \rightleftarrows \mathrm{Bialg}(\mathcal{C}): \bar{\mathcal{P}}$.
So $\mathcal{C}$ would embed into $\mathrm{Bialg}(\mathcal{C})$ via $\bar{\mathrm{S}}.$ Can one expect this?
"Being in char 0" one can think of $\bar{\mathcal{P}}$ as an abelian version of primitive elements and of algebras over the monad $\bar{\mathcal{P}} \circ \bar{\mathrm{S}}$ associated to the adjunction $\bar{\mathrm{S}} : \mathcal{C} \rightleftarrows \mathrm{Bialg}(\mathcal{C}): \bar{\mathcal{P}}$ as abelian Lie algebras that are classically uniquely determined by their underlying object.