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Yes. More generally, if $\mathcal{T}$ is a triangulated category and $\mathcal{S}$ is a thick subcategory, then any morphism $\varphi:M\to N$ of $\mathcal{T}$ that becomes zero in $\mathcal{T}/\mathca …

This follows by applying Theorem 3.8 of Beligiannis' 2000 paper to the opposite categories. $\mathcal{I}$ is a full additive subcategory of $\mathcal{A}$, closed under direct summands. It is covariant …

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^ …

For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimen …

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 …

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

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} …

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 …

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 …

This is Proposition 3.1.2 in the thesis of Parra, "Hearts of $t $-structures which are Grothendieck or module categories".

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 …

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 …

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.

$\textrm{AB4}$ (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 …