When is the derived category abelian? I read in the book Methods of homological algebra of Gelfand and Manin that the derived category of an abelian category $A$ is never abelian. Now to me this seems to be wrong, because if $A=0$ then $D(A)=0$ and so it is abelian. Do you know what statement is true? (Like every derived category of a non-zero category is not abelian) and do you know how to prove it?
 A: 
Let $\mathscr A$ be an abelian category. The derived category $D(\mathscr{A})$ is abelian if and only if $\mathscr A$ is semisimple.

Recall that an abelian category is called semisimple if all short exact sequences split. Equivalently, $\mathscr A$ is abelian and for every morphism $f\colon A \to B$ there is a pseudoinverse morphism $g\colon B \to A$, that is, a morphism $g$ such that $fgf = f$ and $gfg = g$: If $\mathscr A$ is semisimple, factor $f$ over its image as $f = i j$, choose a left inverse $k$ for $i$ and a right inverse $l$ for $j$ and put $g = lk$ so that $f g f = (ij)(lk)(ij) = i(jl)(ki)j = ij = f$ and similarly $gfg = g$; for the other direction note that $fgf = f$ and $f$ monic imply that $gf = 1$, so every monic splits, and dually every epic splits.
So, the derived category of $R$-modules is abelian if and only if $R$ is a semisimple ring. See also Lam, A first course in non-commutative rings, Theorem and Definition (2.5), page 27 for this point. More explicitly, the derived category of $k$-vector spaces over a field $k$ is abelian while the derived category of abelian groups isn't abelian.

The main ingredient to answer your question is provided by the following:

Lemma (Verdier). A triangulated category $\mathscr T$ is abelian if and only if every morphism $f\colon A \to B$ is isomorphic to $A' \oplus I \xrightarrow{\begin{bmatrix} 0 & 1_I \\ 0 & 0\end{bmatrix}} I \oplus B'$.

In particular, in an abelian triangulated category $\mathscr T$ every morphism $f$ has a pseudoinverse $g$. Since an abelian category $\mathscr{A}$ embeds fully faithfully into its derived category $D(\mathscr A)$ by identifying an object of $\mathscr{A}$ with a complex concentrated in degree zero, this immediately implies that $\mathscr{A}$ must be semisimple if $D(\mathscr A)$ is abelian. 
Conversely, if $\mathscr{A}$ is semisimple abelian then $D(\mathscr{A})$ is equivalent to the abelian category $\mathscr{A}^{\mathbb{Z}}$ via the functor that sends a complex $A$ to its homology complex $H(A^\bullet)$ with $H^k(A)$ in degree $k$ and zero differentials. This is proved in detail in Section III.2.3, page 146f of Gelfand–Manin's Methods of Homological Algebra.

The proof of the lemma is relatively easy: Certainly, if every morphism is of the described form then $\mathscr{T}$ is abelian because $f$ has kernel $A'$, image $I$ and cokernel $B'$ and that's all we need.
On the other hand, if $\mathscr{T}$ is abelian then every morphism $f\colon A \to B$ factors over its image as $f = me$ with an epimorphism $e\colon A \twoheadrightarrow I$ and a monomorphism $m\colon I \rightarrowtail B$ and this reduces the lemma to the statement:

In a triangulated category all monomorphisms and all epimorphisms split.

Recall that the morphism axiom [TR3] shows that two consecutive morphisms in a distinguished triangle $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} A[1]$ compose to zero. If $f$ happens to be monic then $fh[-1] = 0$ shows that $h[-1] =0$, so $h = 0$. Still assuming $f$ to be monic, apply the homological functor $\operatorname{Hom}(C,{-})$ to the distinguished triangle $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{0} A[1]$ to get the exact sequence
$$
0 \to \operatorname{Hom}(C,A) \to \operatorname{Hom}(C,B) \to \operatorname{Hom}(C,C) \to 0
$$
showing that $g$ has a right inverse and applying the cohomological functor $\operatorname{Hom}({-},A)$ to that distinguished triangle shows that $f$ has a left inverse. It follows from this that our distinguished triangle $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{0} A[1]$ with monic $f$ is isomorphic to the triangle $A \to A \oplus C \to C \to A[1]$ obtained by taking the direct sum of the distinguished triangles $A \to A \to 0 \to A[1]$ and $0 \to C \to C \to 0[1]$.
Coming back to our general morphism $f = me$ and applying the above observation to the epimorphism $e$ and the monomorphism $m$ gives rise to a splitting $A \cong A'\oplus I$ and $B \cong I \oplus B'$, and $f$ factors as desired.

See also these two related MO-threads:


*

*How do I know the derived category is not abelian?

*Splitting in triangulated categories.


The above argument is essentially contained in Chapitre II, Proposition 1.2.9, p.101 and Proposition 1.3.6, p.108 in this part of Verdier's thesis Des catégories dérivées des catégories abéliennes, available electronically on Georges Maltsiniotis's home page who prepared the Astérisque edition of the thesis in the mid-90ies.
