More questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$ This question is a follow-up to an older one about a proof of Rickard that the stable module category can be described as a triangulated Verdier quotient. In fact it's a complemented in that this question asks for clarification of every point that the previous question didn't touch upon.
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 1 (Resolved in comments):
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.
Rickard proves in two sentences that this functor is an exact functor.

A distinguished triangle in $\mathtt{stmod}(A)$ is a triangle
$$X \to Y \to Z \to X[1]$$
coming from a pushout diagram of modules
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{} & X & \ra{} & I & \ra{} & \Omega^{-1} X & \ra{} & 0 \\
 & & \da{} & & \da{} & & \da{} & &  & \\
0 & \ras{} & Y & \ras{} & Z & \ras{} & \Omega^{-1} X & \ras{} & 0 \\
\end{array}$$
where $X \to I$ is the embedding of $X$ into its injective hull. Since short exact sequences of modules give distinguished triangles in the derived category, and since $FI = 0$,
$$FX \to FY \to FZ \to FX[1]$$
is a distinguished triangle.

Could someone clarify what is going on here? I can understand that $X \to I \to \Omega^{-1} X \to X[1]$ and $Y \to Z \to \Omega^{-1} X \to Y[1]$ are distinguished triangles in $D^b(A)$ and that $FI = 0$, but I entirely fail to see how this gets to the conclusion.
Update: This question is now resolved. One uses the defining property of $D^b(A)$ to infer that we have distinguished triangles $X \to I \to \Omega^{-1} X \to X[1]$ and $Y \to Z \to \Omega^{-1} X \to Y[1]$. Passing to the singularity category, $I$ gets killed, along us to deduce an isomorphism between $\Omega^{-1} X$ and $X[1]$, and hence an isomorphism of triangles between $Y \to Z \to \Omega^{-1} X \to Y[1]$ and $Y \to Z \to X[1] \to Y[1]$. The latter is therefore distinguished, and so by shifting, $X \to Y \to Z \to X[1]$ is distinguished too.
Question 2 (Resolved in comments):
Next, it's proved that the functor is full, where he claims the following.

... no non-projective $A$-module is isomorphic in the derived category to an object of $K^b(A)$.

But this seems unlikely to me. If $M$ is a non-projective $A$-module then $M$ admits a projective resolution $P^* \to M$ and if $P^*$ can be made finite then it's an object of $K^b(A)$ quasi-isomorphic to $M$.
Question 3 (Resolved in answer):
Later on, Rickard proves that the functor is essentially surjective. Every object $X$ in $D^b(A) / K^b(A)$ is isomorphic to a complex of projectives $P^* = \cdots \to P^r \to P^{r+1} \to \cdots P^s \to 0$, where $r < 0$ and $P^*$ has zero homology in degrees less than $r$. The previous question already touched upon the fact that $P^*$ is, in fact, not in $D^b(A)$, so I'll ignore this. The natural map from $P^*$ to $\widetilde{P}^* = \cdots \to P^{r-1} \to P^r \to 0 \to \cdots$ is an isomorphism in $D^b(A) / K^b(A)$ since its mapping cone is a bounded complex of projectives. At this point Rickard writes

... and there is a complex
$$Q^* = \cdots \to P^{r-1} P^r \to Q^{r+1} \to \cdots \to Q^0 \to 0$$
which is the projective resolution of some module $M$ and whose natural map to $\widetilde{P}^*$ is an isomorphism in $D^b(A) / K^b(A)$. Thus $P^* \cong FM$.

This strikes me as simply rephrasing what has to be proved, in terms of projective resolutions rather than in terms of modules. Could someone clarify what is being done here?
 A: The first two questions have already been settled in the comments. For the last one:
First, let's pretend we're working in $\textbf{D}(A)/\textbf{K}^b(A)$ instead of  $\textbf{D}^b(A)/\textbf{K}^b(A)$. Then, the argument shows that within  $\textbf{D}(A)/\textbf{K}^b(A)$, any object $X$ of  $\textbf{D}^b(A)/\textbf{K}^b(A)$ is isomorphic to a complex $\ldots\to P^{r-1}\to P^r\to 0$ with no cohomology below $r$. In other words, it's isomorphic to a stalk complex $M[-r]$ concentrated in degree $r$, for some module $M$. This stalk complex is isomorphic to the degre-$0$ talk complex $(\Omega^{-r} M) [0]$ in $\textbf{D}^b(A)/\textbf{K}^b(A)$ which you see by taking a projective resolution $0\to M\to Q^{r+1}\to \ldots\to Q^0\to \Omega^{-r} M\to 0$ to the right. This is, I think, the argument Rickard presents.
You can eliminate the transit through  $\textbf{D}(A)/\textbf{K}^b(A)$ as follows: Given $X\in \textbf{D}^b(A)$, consider a bicomplex $T$ consisting of projective resolutions of the components of $X$. The totalization of this (unbounded!) bicomplex $T$ is an unbounded complex of projectives quasi-isomorphic to $X$, which you then go on to truncate as described. You can avoid expand+truncate step by already truncating the vertical projective resolutions in $T$ in such a way that the totalization of $T$ is projective except for the lowest non-zero entry (this will require truncating the projective resolutions at different places). This will give you a bounded complex $0\to M\to Q^r\to\ldots\to Q^s\to 0$ quasi-isomorphic to $X$, and hence $X\cong M[-r]$ in $\textbf{D}^b(A)/\textbf{K}^b(A)$ as before.
This is a bit sketchy - let me know if you need mor details.
