# localizing subcategories of a nice triangulated category

Suppose that $$D(A)$$ is the derived category of of a ring A. Let $$b\in D(A)$$ be a compact object and $$B$$ the localizing subcategory generated by b (having arbitrary coproduct).

Does the inclusion functor $$D(A)\leftarrow B: i$$ have a left adjoint ? Let $$W: D(A)\rightarrow B$$ the right adjoint to the inclusion functor. Do we have that $$W\circ i=id$$ ? Does W preserve compact objects?

Remark: the existence of a right adjoint to the inclusion functor is well known.

• Is there any reason to suspect this is true? Take $A = D(R)$ for, say, a Noetherian regular local ring $(R,\mathfrak{m},k)$, and take $b = k$. The localizing subcategory generated by $k$ is then the derived category $\mathfrak{m}$-torsion $R$-modules. The inclusion will not, in general, preserve limits. Commented Mar 5, 2021 at 18:45
• @DrewHeard does that localizing subcategory even have products? Commented Mar 6, 2021 at 23:26
• @FernandoMuro For $Y,\{X_i\} \in \text{Loc}(b)$ we have $[Y,W \prod_i {X_i}] = [i(Y),\prod {X_i}] = \prod_i [i(Y),X_i] = \prod_i [Y,W(X_i)]$. So I think the product is just the product in $D(A)$ followed by $W$. Commented Mar 7, 2021 at 14:30
• @DrewHeard you're right, so it's just that $i$ doesn't preserve products. Commented Mar 7, 2021 at 14:52

For the other two questions, the inclusion functor is fully-faithful, which occurs if and only if the unit map $$\text{id} \to W \circ i$$ is an equivalence. Finally, $$A \in D(A)$$ is compact, but $$W(A) \in \text{Loc}(b)$$ will typically not be.
Remark: In the literature, $$W$$ would be called a colocalization functor. A lot of facts and theorems can be found in the literature - for example, Section 3 of Hovey, Palmieri, Strickland.