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It suffices to require that through any epimorphism in $A'$ from an object of $A'$ onto an object of $A$ some epimorphism in $A$ (onto the same object) would factorize; or the dual condition for monom …

It is hard to think of a general way to obtain an isomorphism of vector bundles from several isomorphisms of cohomology spaces. In particular, there can be moduli of vector bundles, I presume, even o …

It appears that this result may go back to a 1984 letter of Joyal to Grothendieck. The reference to this letter, as well as some other early references, can be found in Example 3.2 in the paper Cotor …

Yes, of course. The coderived and contraderived categories (or more specifically, the absolute derived categories) are defined precisely in such a way that short exact sequences of CDG-modules induce …

In any triangulated category, the necessary and sufficient condition for a distinguished triangle $A\to B\to C\to A[1]$ to split is that the morphism $C\to A[1]$ in this distinguished triangle vanishe …

Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are: For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\bu …

One reference is H. Krause, "Derived categories, resolutions, and Brown representability", Contemporary Math. vol.436, AMS, 2007, p.101-139 or https://arxiv.org/abs/math/0511047 , Section 1.6. Another …

Some construction of derived Hom complexes in an arbitrary $k$-linear Quillen exact category (for any commutative ring $k$) is worked out in the appendix to my paper "Artin-Tate motivic sheaves with f …

Firstly, for any Quillen exact category $\mathcal E$, one can define the derived category $D(\mathcal E)$, as well as its bounded versions $D^+(\mathcal E)$, $D^-(\mathcal E)$, and $D^b(\mathcal E)$. …

I'd guess that if $X$ is a reasonable scheme or stack, then $\operatorname{QCoh}(X)$ is, at least, a Grothendieck abelian category. In particular, it has a generator $G$. If $X$ has resolution proper …

Another and more "modern" approach would be using the construction of the complete cotorsion pair ("a half of an abelian model structure") generated by a generating set of objects in a Grothendieck ab …

This is a good question the answer to which I unfortunately do not know, so let me give answers to three different questions instead. The derived category of comodules over any coalgebra (over a fie …

My favorite (counter)example is this: let $A$, $B$, $C$ be the categories of left modules over some rings $R$, $S$, $T$ (respectively), and let $F$ and $G$ be the functors of tensor product with some …

I have not heard the slogan and perhaps do not understand the context, but it seems to me that this has nothing to do with the derived categories. For any graded quiver (with or without relations) th …