# Does equivalence of derived categories preserve boundedness?

Let $$\mathcal{C}$$ be an abelian category and consider the derived category $$D(C)$$. Suppose that $$F: D(\mathcal{C}) \to D(\mathcal{C})$$ is an auto-equivalence.

My question: must $$F$$ preserve the property of being bounded/bounded above/bounded below.

Said differently, if $$A$$ is an object of $$D(\mathcal{C})$$ whose cohomology is bounded/bounded above/bounded below, does $$F(A)$$ also have cohomology which is bounded/bounded above/bounded below?

If the answer to this question is negative, can one impose natural properties on $$F$$ to make the answer positive?

Not in that level of generality. For example, let $$\mathcal{C}$$ be the product of copies of the category of abelian groups, indexed by $$\mathbb{Z}$$, so an object of $$\mathbb{C}$$ is a family $$\{A_i\mid i\in\mathbb{Z}\}$$ of abelian groups, and an object of $$D(\mathcal{C})$$ is a family $$\{X_i\mid i\in\mathbb{Z}\}$$ of complexes of abelian groups.

Then $$\{X_i\}\mapsto\{X_i[i]\}$$ is an auto-equivalence that doesn’t preserve boundedness (or boundedness above or below).

However, it is true if $$\mathcal{C}$$ is the module category of a ring.

This is because it is well-known (see for example, Proposition 15.72.3 here) that the "perfect complexes" (those isomorphic in $$D(\mathcal{C})$$ to bounded complexes of finitely generated projective modules) can be characterized in the derived category as the "compact objects" (objects $$X$$ such that the functor $$\text{Hom}_{D(\mathcal{C})}(X,-)$$ preserves arbitrary coproducts).

But then the bounded (in cohomology) objects $$Y$$ of $$D(\mathcal{C})$$ can be characterized as the objects such that, for each perfect complex $$X$$, $$\text{Hom}_{D(\mathcal{C})}(X,Y[i])=0$$ for all but finitely many $$i\in\mathbb{Z}$$, with similar characterizations of bounded above and bounded below objects.

Since there are inherent characterizations of these various classes of objects, they must be preserved by any auto-equivalence.

• Thanks a lot for this counterexample! Could I ask you to give an explanation or reference for why the statement is true for the module category of a ring - ideally one which generalizes to other "not too large" categories? (For reference, I am particularly interested in "geometric" examples, such as modules over a ring, coherent sheaves over a scheme, ect.) – user142700 Apr 30 at 20:59
• @user142700 I've added an explanation to my answer. – Jeremy Rickard May 1 at 9:54