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I'll assume that you at least want your triangulated category to have the property that a (countable) coproduct of exact triangles is an exact triangle? Even then, I think this is probably not true i …

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 …

There are some dull counterexamples. If $A$ is a self-injective algebra such that the square of the radical is zero, then the stable module category is a semi-simple $k$-linear category with one simp …

It is possible. Given $\mathscr{T}$, $\mathscr{A}$ and $i_\mathscr{A}^R$, it is often possible to find a subcategory $\mathscr{B}$ of $\mathscr{T}$ so that $\mathscr{A}\cap\mathscr{B}=0$ (and so certa …

If $\mathcal{T}_{1}=\langle M_{1}\rangle_{d_{1}+1}$ and $\mathcal{T}_{2}=\langle M_{2}\rangle_{d_{2}+1}$, then $\mathcal{T}_{1}\ast\mathcal{T}_{2}\subseteq\langle M_{1}\oplus M_{2}\rangle_{d_{1}+d_{2} …

Let $A$ be a triangulated category, and let $B=A\times A$, with $A$ regarded as a full triangulated subcategory of $B$ via the embedding $u(X)=(X,0)$, and let $f:B\to A$ be the functor $f(X,Y)=X\oplus …

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^ …

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 …

$\textrm{AB4}$ (and a fortiori $\textrm{AB3}$) is not enough, as it's not true for the opposite category of a module category, which is $\textrm{AB4}$. Let $R$ be any ring, and consider the inverse s …

$\operatorname{Hom}_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ means that $\operatorname{Hom}_{\mathcal{T}}\left(X, \Sigma^i(Y)\right) = 0$ for all objects $X,Y$ of $\mathcal{M}$ and all integer …

It means that $K^b(\text{proj }A)$ is the smallest full triangulated subcategory containing $\text{add}(T)$. This is spelled out more explicitly in Section 5 of the earlier paper "Morita theory for de …

I don't think (3) implies (1). For example, the opposite category of the category of abelian groups satisfies (3), but is not AB5.

No. Let $k$ be a field, and let $A$ be the algebra of upper triangular $2\times 2$ matrices over $k$, with augmentation map $\pmatrix{a&b\\0&c}\mapsto a$. $A$ and $A^e$ have finite global dimension, s …

One obstruction is that if all Hom-sets are finite dimensional vector spaces, and for all objects $X$ and $Y$, $\text{Hom}(X,Y[i])=0$ for all but finitely many $i$, then any self-equivalence must pres …

There's an elementary proof that if $\Omega$ is a self-equivalence of $\underline{C}$ then $C$ also has enough injectives, and projectives and injectives coincide. In particular, this shows that if $C …