A fact about triangulated categories is that (exact) localisation functors and so-called colocalisation functors come in pairs, making an exact localisation triangle. I've tried to come up with less traditional examples.


Let $A$ be a commutative ring, and $I \subset A$ an ideal. Then tensoring by $A / I$ is idempotent and so we can look at the localisation functor on $D(A)$, the derived category. This is exact in the triangulated sense, as tensoring is right-exact. Am I correct in thinking the colocalisation should just be the kernel of this tensor map, i.e. multiplying by $I$? If so, I expect for every complex of modules $M^\bullet$ that we should get a quasi-isomorphism $$I^2 M^\bullet \to I M^\bullet$$ from the natural inclusion. Is there an elementary way of seeing this? I'm finding it difficult to follow the machinery that guarantees a natural isomorphism $\Gamma^2 \to \Gamma$ over $D(A)$ for whichever colocalisation functor $\Gamma$ is actually associated to the derived tensor $A / I \otimes \\_ $ defined over $D(A)$.



Is there an elementary reason that this colocalisation map of complexes is a quasi-isomorphism?
The derived tensor product with $A/I$ is typically not idempotent (because ${\rm Tor}\_{\*}^A(A/I, A/I)$ is nontrivial), so this won't give you a localization sequence.


One case where it is idempotent is when $I = eA$ for $e \in A$ an idempotent. In this case, tensoring with $A/I$ is the same as localizing at $1-e$, and the localization sequence does take the form $eM \to M \to M/eM$.


