# Free symmetric monoidal category of compactly generated category is compactly generated

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.

• 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 $E$. – Simon Henry Nov 16 '18 at 8:13
• 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. – Marc Hoyois Nov 16 '18 at 15:40
• @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? – leibnewtz Nov 16 '18 at 18:56
• Yes. A reference for this symmetric monoidal universal property of Ind is HA 4.8.1.14. – Marc Hoyois Nov 16 '18 at 22:15