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May 31, 2022 at 17:34 comment added Tyler Lawson I will think about the finite projective dimension issue (it seems very plausible using something like a Cartan-Eilenberg resolution) but I am not immediately sure.
May 31, 2022 at 17:33 comment added Tyler Lawson @Z.M If M is an R-module of projective dimension 1 and X is an object of D(R), then any map M -> H_n(X) can be realized by a (non-unique) map M[n] -> X in D(R). This allows you to show that any object X of D(ℤ) is quasi-isomorphic to ⊕ (H_n X) [n] without needing boundedness for an inductive argument.
May 31, 2022 at 16:23 comment added Z. M Thanks for reminding the projective dimension of $\mathbb Z$, but I am still a bit confused: this implies that any homologically bounded below object in $D(\mathbb Z)$ is a direct sum of homological groups, but I am not sure whether this is still true for unbounded objects (and even in $\operatorname{Sp}$, your example is not bounded below). Furthermore, I wonder whether you tend to believe that, whenever the projective dimension is finite, there is no example that every truncation of two objects is isomorphic but the two objects are not isomorphic?
May 31, 2022 at 15:09 comment added Tyler Lawson @Z.M I am not 100% sure about what should play the role of $\Omega^\infty$ in your question, but I feel reasonably confident that the answer will be "no". Because $\Bbb Z$ has projective dimension 1, two objects in $D(\Bbb Z)$ are equivalent if and only if they have isomorphic homology groups, and so any functors which can detect these groups can also reflect equivalences.
May 31, 2022 at 7:50 comment added Z. M What about the same question for $D(\mathbb Z)$ in place of $\operatorname{Sp}$?
May 31, 2022 at 1:54 comment added kiran Wow, just the kind of think I was looking for!
May 31, 2022 at 1:49 vote accept kiran
May 30, 2022 at 23:16 history edited Tyler Lawson CC BY-SA 4.0
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May 30, 2022 at 23:08 history answered Tyler Lawson CC BY-SA 4.0