# 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?

• 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

## 1 Answer

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 $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} A$ 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 $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{0} A$ 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 $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{0} A$ with monic $f$ is isomorphic to the triangle $A \to A \oplus C \to C \to A$ obtained by taking the direct sum of the distinguished triangles $A \to A \to 0 \to A$ and $0 \to C \to C \to 0$.

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:

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.

• wow, very nice answer, thank you – Mec Sep 7 '12 at 4:52
• I was thinking this: if in every triangulated category every monomorphisms and epimorphisms split then every abelian triangulated category is semisimple. It doesn't seem that you need Verdier's Lemma. Am I right? – Mec Sep 7 '12 at 18:35
• Yes, that's right. However, it's only a very small step from this splitting of monics and epics to the formulation of Verdier's lemma characterizing abelian triangulated categories via a decomposition property of morphisms, so I thought it was worth spelling it out. The existence of pseudoinverses for all morphisms does not quite imply abelianness of a category. It does for categories which are idempotent complete, but a triangulated category need not be idempotent complete in general. That's why there's a little extra work to do. – t.b. Sep 7 '12 at 18:47