For any abelian category $A$ (with enough injectives), consider its derived category $D^b(A)$. Suppose we have a complex of this form $$0\to F\to L^{n+1}\to ... \to L^{1} \to G\to 0$$ which is exact. Furthermore we have $Ext^i(G,F)=0$, for $i>n$.
I want to conclude that $G\oplus F[-(n+1)]\cong [L^{n+1}\to ... \to L^{1}]$. It is said that this is implied by the vanishing of the $Ext$ functor, but I don't see why the vanishing of the $Ext$ functor is necessary. There is a canonical map $G\oplus F[-n+1]\to L^{n+1}\to ... \to L^{1}$, which is a quasi-isomorphism, so the $Ext$ shouldn't be involved here?
EDIT: I was mistaken as pointed out in the comments, so I somehow need the $Ext$ vanishing. Maybe it might be helpful to know that $Ext^i(G,F)=Hom_{D(A)}(G,F[i])$?
Maybe we can also consider the roof $G\leftarrow [ F\to L^{n+1}\to ... \to L^{1}]\to [0\to L^{n+1}\to ... \to L^{1} ] $, the left map being a quasi-isomorphism.
Consider the distinguished triangle $$ [0 \to L^{n+1} \to \dots \to L^1] \to [F \to L^{n+1} \to \dots \to L^1] \to F[n+1]. $$ Its middle term is quasi-isomorphic to $G$, hence the right map is zero by the vanishing of $Ext$'s, hence the first term is quasi-isomorphic to the sum $G \oplus F[n]$.