Questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$

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 $D^b(A)$?

Solution 1:

$\require{AMScd}$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.

Solution 2:

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.