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Let $A=k[x]$ and $B=k[x]/(x^2)$, let $X$ be the complex $\hskip{.1in}\dots\to 0 \to B\stackrel{x}{\to} B\to 0\to \dots$, and let $Y$ be $\hskip{.1in}\dots\to 0\to k\stackrel{0}{\to}k\to 0\to\dots$. Th …

There's a simple counterexample in Section 8 of Rickard, Jeremy, Morita theory for derived categories, J. Lond. Math. Soc., II. Ser. 39, No. 3, 436-456 (1989). ZBL0642.16034. Let kQ be the path alge …

The class of objects with property (P) is a thick subcategory of $\mathbf{D}(R^{op})$ (i.e., a triangulated subcategory closed under taking direct summands), and contains $R$, so it contains all perfe …

This isn't really an answer to the question, but an example to show how badly things can go wrong. Let $M$ be any indecomposable object of $D^b(X)$. The functor $\operatorname{Hom}(M,-)$, from $D^b(X …

No. For example, let $A$ be any object of $\mathcal{A}$, let $X$ be the complex $$\dots\to0\to A\oplus A\stackrel{\begin{pmatrix}1&0\end{pmatrix}}{\longrightarrow}A\to0\to\dots$$ and $\alpha:X\to X$ t …

I think the answer is no in the following very simple example. Let $R$ be the path algebra of the quiver $\bullet\rightarrow\bullet$ over a field $k$. Then since $R$ is hereditary and of finite repre …

As pointed out in nikola karabatic's answer, a decomposition of M as a direct sum of two-term complexes induces decompositions $H^i=H^i_a\oplus H^i_b$ for each $i$. These have the property that $\xi_i …

Up to shifts, every indecomposable object is of one of the forms described in the question. I don't know an explicit reference, but here's a sketch of a proof. By induction on the length, it's not h …

Yes. The perfect objects in $\mathbf{D}(R)$ are the objects isomorphic to bounded complexes of finitely generated projective modules, and $\mathbb{k}$ is isomorphic in $\mathbf{D}(R)$ to its minimal p …

There is such a sequence, but it's not very interesting. Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle $$F\to Z_{i …

Yes. Let $$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$ be the usual truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^ …

If $M$ is a maximal Cohen-Macaulay module for a Noetherian Gorenstein ring $R$, then it follows easily by induction on the projective dimension of $N$ that $\operatorname{Ext}^i_R(M,N)=0$ for all $i>0 …

Possibly the stack example in pbelman's answer is of this form, but an elementary way to construct examples is by taking infinite products. Let $\{\mathcal{C}_i\}_{i\in I}$ be an infinite collection …

Here's a counterexample that appears in nature. Fix a prime $p$ and a field $k$ of characteristic $p$, and let $G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if $p$ is odd, and $n\ge …

No, not in general. For example, let $R=k[x]/(x^2)$ for a field $k$ and let $X$ be the object $$\dots\stackrel{x}{\to}R\stackrel{x}{\to}R\stackrel{x}{\to}R\to0\to0\to\dots$$ of the derived category o …