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 $D^b(A)$?