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Let $k$ be a field and let $\mathcal{C}=\mathbf{StLin}_k$ be the $\infty$-category of stable infinity categories enriched over the $\infty$-category $\mathbf{Vect}_k$, regarded as a symmetric monoidal $\infty$-category with unit object $\mathbf{Vect}_k$.

The forgetful functor $\operatorname{CAlg}(\mathcal{C}) \to \mathcal{C}$ admits a left adjoint $Sym^*: \mathcal{C} \to \operatorname{CAlg}(\mathcal{C})$. Now let $\mathcal{C}[z]=Sym^*(\mathbf{Vect}_k)$ - i.e. $\mathcal{C}[z]$ is the free stable symmetric monoidal category generated by $\mathbf{Vect}_k$.

My question is: Is the $\infty$-category $\mathcal{C}[z]$ compactly generated in $\mathcal{C}$? (Recall that a category $\mathcal{A}$ is compactly generated if there exists a subcategory $\mathcal{A}_0 \subseteq \mathcal{A}$ and an equivalence $\mathcal{A} \simeq Ind(\mathcal{A}_0)$.)

More generally: If $\mathcal{D} \in \mathcal{C}$ is compactly generated, is $Sym^*(\mathcal{D})$ also compactly generated?

My idea was to show that the map $\mathbf{Vect}_k \to \mathcal{C}[z]$ corresponding to the identity:$ \mathcal{C}[z] \to \mathcal{C}[z]$ under the adjunction above identifies the image of $k$ with a compact generator of $\mathcal{C}[z]$, but I'm not sure if this would work.

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    $\begingroup$ I don't see why you expect $k$ to be a generator ? It seems more reasonable to expect that $(k, k \otimes k, k \otimes k \otimes k, \dot... )$ would be a set of compact generator, isn't it ? More generally I would expect that if $D$ has a sets of compact generator $E$ then $Sym^*(D)$ has a set of compact generator given by all the tensor product of finite families of elements in $E$. $\endgroup$ – Simon Henry Nov 16 '18 at 8:13
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    $\begingroup$ Do you want $\mathcal C$ to consist of cocomplete ∞-categories? If so the answer is yes, because $Ind(Sym^*(\mathcal D^c))$ has the correct universal property. If not, then no as $Sym^*(\mathcal D)$ is not cocomplete. $\endgroup$ – Marc Hoyois Nov 16 '18 at 15:40
  • $\begingroup$ @MarcHoyois Right, if all the categories $X$ are cocomplete then $Map_{CAlg}(Ind(Sym(D^c)), X)=Map_{CAlg}(Sym(D^c),X)=Map_{StLin}(D^c, X)=Map(D,X)$. Is this correct? $\endgroup$ – leibnewtz Nov 16 '18 at 18:56
  • $\begingroup$ Yes. A reference for this symmetric monoidal universal property of Ind is HA 4.8.1.14. $\endgroup$ – Marc Hoyois Nov 16 '18 at 22:15

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