# 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?

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If $A$ is semisimple abelian (every short exact sequence splits), then every acyclic complex is nullhomotopic, so $D(A) = K(A)$ and $K(A)$ is abelian if and only if $A$ is semisimple abelian by a result in Verdier's thesis (somewhere in chapter II if I remember correctly). – t.b. Sep 1 '12 at 22:12
could you tell me why every acyclic complex is nullhomotopic and why $D(A)=K(A)$ in this case? I mean, you are claiming that every quasi-isomorphism is an isomorphism, right? How do you prove it? Where I can find a proof that $K(A)$ is abelian if and only if $A$ is semisimple abelian? (in the second chapter of Methods of homological algebra or in the Verdiers' thesis?) – Mec Sep 1 '12 at 22:30
You can write down a contracting homotopy for an acyclic complex using that the image of every differential is a direct summand of its domain and codomain if A is semisimple. Then by definition you have D(A) = K(A)/Ac(A) but I just argued that Ac(A) = 0 in K(A) for A semisimple. A detailed argument is in Verdier's thesis (Des catégories dérivées des catégories abéliennes), but the main part of the argument is somewhere in Gelfand-Manin, too (where they show that in the semisimple abelian case D(A) is equivalent to the category of complexes with zero differential). – t.b. Sep 1 '12 at 22:45
– Juan S Sep 4 '12 at 10:48

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.