I am reading the paper "Derived categories and stable equivalence", the link is here:http://www.sciencedirect.com/science/article/pii/0022404989900819.
At theorem 2.1, there is an equivalent functor $F: \mathbf{mod}-\Lambda \rightarrow D^b(mod-\Lambda)/K^b(P_{\Lambda})$ on page 308. There are some places I can't understand in its proof:
- To show $F$ is full, it says "since F' is clearly full, and that for no non-zero object $X$ of $\mathbf{mod}-\Lambda$ is $FX \cong0$,since no non-projetive $\Lambda$-module is isomorphic in the derived category to an object of $K^b(P_{\Lambda})$". But if $M \in \Lambda-mod$ has a finite projective resolution $P^{\bullet}$, we can get $M \cong P^{\bullet}$ in $D^b(mod- \Lambda)$, so why no non-projetive $\Lambda$-module is isomorphic in the derived category to an object of $K^b(P_{\Lambda})$?
- To show $F$ is faithful, it says "then the mapping cone of $\beta$ is sent to zero by $F$, so $\beta$ is an isomorphism, so $Y \rightarrow Z$ is a split monomorphism and $\alpha$ is zero", I can't understand this, who can explain it more specific?
- On page 309, it say "the natural map from $P^*$ to $\overline{P^*}$ is an isomorphism in $D^b(mod-\Lambda)/K^b(P_{\Lambda})$" I know a map $\alpha$ is an isomorphism in the homotopy category $K(A)$ iff it's mapping cone is 0. Is it still hold in $D^b(mod-\Lambda)/K^b(P_{\Lambda})$?
- Still on page 309, it says "there is a complex $$Q^*= \cdots \rightarrow P^{r-1} \rightarrow P^r \rightarrow Q^{r+1} \rightarrow \cdots \rightarrow 0$$ which is the projective resolution of some module $M$", so I want to ask how to make sure there must be a module whose projective resolution is the form $\cdots \rightarrow P^{r-1} \rightarrow P^{r} \rightarrow \cdots$?
