How to get the following complex acyclic?

Question

I have asked a question here:https://mathoverflow.net/questions/251808/how-to-get-the-following-complex-acyclic. But there is no response. So I repost it here.

Let $\mathcal{C}$ be a triangulated category and $\mathcal{U} \subset \mathcal{C}$ be a full triangulated subcategory of $\mathcal{C}$. We define

$\mathcal{U}^{\bot}=\{X \in \mathcal{C}| Hom(U,X)=0$ for all $U \in \mathcal{U} \}$.

Let $A$ be a finite dimensional k-algebra. $\mathcal{I}$ is the full subcategory of mod $A$ formed by injective modules. $K^b(_A \mathcal{I})$ is the bounded homotopy category of injetive modules. There is a proposition and its proof:

Propostion: If fin.dim(A) < $\infty$, then $K^b(_A\mathcal{I})^{\bot}$=0.

Proof: Assume that $K^b(_AI) \not=0.$ Let $I^{\bullet} \in K^b(_AI)^{\bot}$. We assume that $I^{\bullet}=(I^i,d^i_I)\not=0$. Applying the translation functor if necessary we may assume that $I^i=0$ for $i <0$. Next conside the complex $$ \cdots 0\rightarrow Hom(D(A_A),I^0)\rightarrow Hom(D(A_A),I^1)\rightarrow \cdots$$ Since $I^{\bullet} \in K^b(_A\mathcal{I})^{\bot}$, we infer that this complex is acyclic. Note that proj.dim cok $Hom(D(A_A),d^i)<\infty$. But this contradicts fin.dim$(A)<\infty$.

My question in the proof is that

1. how to use $I^{\bullet} \in K^b(_A\mathcal{I})^{\bot}$ get this complex is acyclic?
2. It says this contradicts fin.dim$(A)<\infty$, because for any n, there is an i suct that proj.dim $cok Hom(D(A_A),d^i)=n$. But wether there is a j such that $I^{k}=0$, for all $k>j$? Then $cok Hom(D(A_A),d^k)=0$, so it doesn't contradict with fin.dim$(A)<\infty$.

Answer 1

1. The degree $k$ homology of $\textrm{Hom}\left(D(A_A),I^\bullet\right)$ is the space of homotopy classes of maps from $D(A_A)$, considered as a complex concentrated in degree $k$, to $I^\bullet$. Since $D(A_A)$ is in $K^b(_A\mathcal{I})$ and $I^\bullet$ is in $K^b(_A\mathcal{I})^\perp$, this must be zero for all $k$.

2. Since $D(A_A)$ contains every indecomposable injective as a direct summand, $\textrm{Hom}\left(D(A_A),I^k\right)=0$ if and only if $I^k=0$. But if $I^k=0$ for all $k>j$ then $I^\bullet$ is in $K^b(_A\mathcal{I})$. Since it's also in $K^b(_A\mathcal{I})^\perp$, it must be isomorphic to zero (since the identity map $I^\bullet\to I^\bullet$ is zero).