# Turn a coalgebra to an algebra by passing to right adjoint

Suppose I have a coalgebra $$\mathcal{C}$$ in the $$\infty$$-category of presentable stable $$\infty$$-categories (with continuous functors). If the structure functors for $$\mathcal{C}$$ all have continuous right adjoints, then passing to right adjoints I get an algebra structure on $$\mathcal{C}$$. However, in the general situation, I only have a full stable subcategory $$\mathcal{D}$$ of $$\mathcal{C}$$ such that the right adjoints preserve $$\mathcal{D}$$, i.e. they induce $$\mathcal{D}\otimes \mathcal{D}\rightarrow\mathcal{D}$$, and restricting to $$\mathcal{D}$$ they are continuous. Note that $$\mathcal{D}$$ is not necessarily a sub-coalgebra of $$\mathcal{C}$$.

Is there a clean way to show that in the above situation, $$\mathcal{D}$$ naturally induces an algebra structure? All I can say is that all the homotopical data needed are contained in the coalgebra structure of $$\mathcal{C}$$, but this seems not so clean.

• So if I read correctly, you have a monoidal structure on $\mathcal{C}$ (obtained by the right adjoint of your coalgebra structure) and a full subcategory $\mathcal{D} \subset \mathcal{C}$ stable under the tensor product and you want to say it is a monoidal category ? I assume that either you are working without unit or you also assume that the unit is in $\mathbb{D}$ right ? – Simon Henry May 14 at 18:32
• Let's say the unit is in $\mathcal{D}$. The thing that I'm confused about is that it seems the tensor product on the category of presentable stable $\infty$-categories with continuous functors don't extend to the category with all functors. I'm not sure whether that causes any issue on a monoidal structure on $\mathcal{C}$. – Wonderfield May 14 at 20:34