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Setup:

Let $A$ be a self-injective algebra (so projective = injective for modules) and let $D^b(A)$ and $K^b(A)$ be the bounded derived category and the full subcategory consisting of the perfect complexes (complexes whose modules are all projective). The singularity category is defined to be the quotient $D^b_\mathtt{sg}(A) = D^b(A)/K^b(A)$.

Next let $\mathtt{stmod}(A)$ be the stable module category. This is the category whose objects are homomorphisms and whose maps are module homomorphisms modulo those that factor through a projective.

Question:

In his paper "Derived categories and stable equivalence" Jeremy Rickard proves that the stable module category and the singularity category are equivalent. The functor $$F\colon\mathtt{stmod}(A) \to D^b_\mathtt{sg}(A)$$ sends a module $M$ to the complex which is $M$ in degree $0$ and $0$'s in all other degrees. My question is about Rickards proof. He first proves that $F$ is exact and full. That's easy. My problem is when he shows that $F$ is faithful and dense. First faithful, he writes:

Suppose $\alpha\colon X \to Y$ is map for which $F\alpha = 0$, and suppose that $\alpha$ sits in a triangle $X \to Y \to Z \to$; then the identity of $FY$ factors through $FY \to FZ$, so, since $F$ is full, there is a map $\beta\colon Y \to Y$, factoring through $Y \to Z$, such that $F\beta$ is an isomorphism. But then the mapping cone of $\beta$ is sent to zero by $F$, so $\beta$ is an isomorphism, so $Y \to Z$ is a split monomorphism and $\alpha$ is zero.

Maybe I'm misunderstanding what he means by the identity factoring through $FY \to FZ$. Can't we just take $\beta$ to be the identity on $Y$? And the mapping cone is a complex, so how can we apply the functor $F$ to it?

My second question is about how he proves that $F$ is dense. He says that an object $$M^\ast = 0 \to M^r \to \cdots \to M^s \to 0$$ of $D^b_\mathtt{sg}(A)$ can be represented as a complex of projectives $$P^\ast = \cdots \to P^r \to P^{r + 1} \to \cdots \to P^s \to 0$$ such that $P^\ast$ has zero homology in degrees less than $r$. Now I can construct a map of chain complexes $P^\ast \to M^\ast$ which is a quasi-isomorphism, but lower than degree $r$ it's a projective resolution which doesn't necessarily terminate. If it's not a bounded chain complex how can it be an element of $D^b(A)$?

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    $\begingroup$ The triangle $X \to Y \to Z \to$ in the part in italics is constructed in $\operatorname{\mathtt{stmod}}A$, not in $D_{\mathtt{sg}}^b(A)$. That is, $Z$ can be obtained by taking the push-out under $\alpha$ and a monic $X \to I$ into an injective. // Since $F$ is exact, $FX \to FY \to FZ \to$ is a distinguished triangle. In particular $FY \to FZ$ is a weak cokernel of $F(\alpha) \colon FX \to FY$ (see e.g. the first paragraph on page 5 here). Since $1_{FY} \circ F(\alpha) = 0$ the desired factorization follows. $\endgroup$ – Martin May 1 '13 at 7:57
  • $\begingroup$ Concerning the second question: Notice that $D^b(A)$ usually denotes the closure under isomorphisms in $D(A)$ of the image of the category of bounded complexes. That is: $D^b(A)$ really consists of the complexes that are acyclic in all degrees $|n| \gg 0$. $\endgroup$ – Martin May 2 '13 at 7:09
  • $\begingroup$ I was wondering if that might be the case but every reference I've looked at defines $D^b(A)$ as the localization of a quotient of $\mathrm{Ch}^b(A)$. I suppose these two definitions probably yield equivalent categories. I'll have to think about that one. $\endgroup$ – Jim May 2 '13 at 7:29
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    $\begingroup$ Yes, they are equivalent. The equivalence is just a matter of truncation. Keller has a careful presentation of these issues (with only brief indications of the proofs) in section 11 of his Derived categories and their uses. See the remarks after Lemma 11.6 and Lemma 11.7. $\endgroup$ – Martin May 2 '13 at 7:46
  • $\begingroup$ Sweet, thanks for all your help. If you wanted to summarize this in an answer I'll accept it. $\endgroup$ – Jim May 2 '13 at 8:01
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$\require{AMScd}$The factorization you ask about is an instance of the slogan "distinguished triangles are just like exact sequences".

To wit, the axiom on existence of morphisms yields that in a distinguished triangle \begin{CD} X @>u>> Y @>v>> Z @>w>> X[1] \end{CD} $v$ is a weak cokernel of $u$. For if $f \colon Y \to W$ is a morphism such that $fu = 0$ this axiom yields a commutative diagram \begin{CD} X @>u>> Y @>v>> Z @>w>> X[1] \\ @V{}VV @VVfV @VVgV @VVV \\ 0 @>>> W @>1>> W @>>> 0 \end{CD} so that $f = gv$ for a nonunique morphism $g$.


Rickard briefly explains how to construct a distinguished triangle over any morphism in $\operatorname{\mathtt{stmod}}(A)$. Take a representative $\alpha \colon X \to Y$ in $\operatorname{\mathtt{mod}}A$ and choose an injection $i \colon X \to I$ into an injective (it is not necessary that $I$ be the injective hull of $X$). Take the push-out under $\alpha$ and $i$ (in $\operatorname{\mathtt{mod}}A$) to obtain the diagram \begin{CD} A @>i>> I @>>> \Sigma A \\ @V\alpha VV @V\alpha'VV @| \\ B @>\beta>> C @>\gamma>> \Sigma A \end{CD} where $\Sigma X$ is the cokernel of $i$. By definition of the triangulation on the stable category, the image of $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z \xrightarrow{\gamma} \Sigma X$ is a distinguished triangle in $\operatorname{\mathtt{stmod}}(A)$.

Since $F$ is an exact functor, we obtain a distinguished triangle \begin{CD} FX @>F\alpha>> FY @>F\beta>> FZ @>>> FX[1] \end{CD} in $D^{b}_{\mathtt{sg}}(A)$.

If we assume that $F\alpha = 0$, then applying the weak cokernel property of $F\beta$ to $1_{FY} \colon FY \to FY$ we find a morphism $g \colon FZ \to FY$ such that $g F\beta = 1_{FY}$. Since $F$ is full, $g = F\epsilon$ for some morphism $\epsilon \colon Z \to Y$. Now observe that $F(\epsilon\beta) = 1_{FY}$, and the cone of $1_{FY}$ is zero. This means that the cone $C$ of $\epsilon \beta$ in $\operatorname{\mathtt{stmod}}A$ (constructed as $Z$ above) is sent to zero by $F$. Rickard argued earlier that this an only happen if $C$ is projective, so $\epsilon\beta$ is an isomorphism. In other words, $Z$ is isomorphic to $Y \oplus \Sigma X$ and $\beta$ is isomorphic to the inclusion of $Y$ into the first summand of $Y \oplus \Sigma X$. Reformulating again, the triangle $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z \xrightarrow{\gamma} \Sigma X$ is isomorphic to the sum of the two distinguished triangles $0 \to Y \xrightarrow{1} Y \to 0$ and $0 \to 0 \to \Sigma X \xrightarrow{1} \Sigma X$ so that $\alpha$ must be zero.

This argument establishes that only the zero morphism is sent to zero and this is saying that $F$ is faithful.


The point on essential surjectivity (density) of $F$ is resolved by observing that $D^b(A)$ is equivalent to the full subcategory of complexes which are acyclic on the far left and the far right, i.e., in degrees $|n| \gg 0$.

This is explained carefully, but very concisely, in section 11 of Keller's marvellous Handbook of Algebra article Derived categories and their uses.


If you need more background on stable categories, I think it is hard to beat the clarity of Happel's original exposition in Chapter I of Triangulated Categories in the Representation Theory of Finite Dimensional Algebras.

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The identity of $FY$ factoring through $FY\to FZ$ should mean there's a commutative triangle such that the legs are $FY\to FZ\to FY$ and the base is $FY\stackrel{\operatorname{id}_Y}{\to} FY$. So really it just means $FY\to FZ$ has a left inverse. Given this, isn't it apparent that there's no reason in general for $\operatorname{id}:Y\to Y$ to factor through $Y\to Z$? I'm afraid I don't see where a mapping cone comes in: I'm writing this without understanding all of what you're writing about, so it's quite possible I'm off the mark.

I don't think I can help with your second question, as I haven't studied derived categories much yet.

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  • $\begingroup$ Oops, I forgot to type the part about the mapping cone. My bad, I've edited. $\endgroup$ – Jim May 1 '13 at 16:56

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