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.

  • 4
    $\begingroup$ 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. $\endgroup$
    – Drew Heard
    Commented Mar 5, 2021 at 18:45
  • $\begingroup$ @DrewHeard does that localizing subcategory even have products? $\endgroup$ Commented Mar 6, 2021 at 23:26
  • 2
    $\begingroup$ @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 $W$. $\endgroup$
    – Drew Heard
    Commented Mar 7, 2021 at 14:30
  • $\begingroup$ @DrewHeard you're right, so it's just that $i$ doesn't preserve products. $\endgroup$ Commented Mar 7, 2021 at 14:52

1 Answer 1


As noted in the comments, there is no reason for the functor to have a left adjoint in general, as the inclusion will not preserve limits.

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.


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