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You get an isomorphism of functors $\bar{G}\bar{F}\simeq1_\mathcal{A/C}$ by applying $\pi_\mathcal{A}$ to an isomorphism $GF\simeq1_\mathcal{A}$. Just to check we're thinking of the quotient category …

For a simpler, but arguably more artificial, example than Mark's, take $\mathcal{C}$ to be the category with one object and two morphisms. Then the identity functor $\mathcal{C}\to\mathcal{C}$ is "unn …

Take the category of (at most) countable-dimensional vector spaces over your favourite field. Then take the quotient by the Serre subcategory of finite-dimensional vector spaces. (And take a skeletal …

Yes, it's always zero, assuming $D(A)$ means the unbounded derived category. My complexes will be cochain complexes, and $X[1]$ will be $X$ shifted down in degree. First suppose $X$ is bounded below …

Not necessarily. For example, let $k$ be a field, and $R$ the $k$-algebra $k\oplus V$, where $V$ is an infinite dimensional square-zero ideal. Let $C$ be the category of $R$-modules and $T$ the Serre …

I think the paper "A canonical factorization for graph homomorphisms", Barry Fawcett, Can J. Math. 29 (4), 1977, 738-743, answers the question. Theorem 3 states that in $\textbf{Grph}$, strict epimor …

It's not in general true for the category of modules for a ring. For example, let $R=\mathbb{C}[x]/(x^2)$, let $X=R\oplus R$ be the free module on two generators, and let $Y_1$ and $Y_2$ be the submo …

Every finite length indecomposable object in an abelian category has local endomorphism ring (the proof is exactly the same as for modules), and a uniserial object must be indecomposable, so the answe …

I posted the following example of a Grothendieck category with no simple objects in answer to a question on math.stackexchange about seven and a half years ago. I seem to have said at the time that it …

Here's another example for Question 2 that I encountered in nature. In the derived category of modules for a ring, pick one module $M_i$ for each $i\in\mathbb{Z}$, and let $A_i=M_i[i]$. Then the natur …

I think I have an example. Fix a chain of fields $k_\alpha$ indexed by ordinals $\alpha$, where $k_\alpha\subset k_\beta$ is an infinite field extension for all pairs $\alpha<\beta$ of ordinals. Fir …

I don't see how commutativity matters. Suppose $A$ is the filtered colimit of algebras $A_i$ and $x,y\in A$ with $xy=0$. Then $x$ is represented by $x_j\in A_j$ and $y$ by $y_k\in A_k$ for some $j$ a …

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 …

In Peter LeFanu Lumsdaine's answer, he answers most of the question, leaving only the question of why the composition of counits for a chain of right rejective subcategories should be mono. In fact, …

If you take $R$ to be, say, a finite field, so that all epimorphisms and monomorphisms of $R$-modules are pure and split, and + is just vector space duality, and for simplicity restrict to functors th …