I am reading the paper"Dominant dimensions, derived quivalences and tilting modules", the link is here:http://link.springer.com/article/10.1007/s11856-016-1327-4.

On page 22,Lemma 4.2 says that let M and N be A-modules, if $N \in add(_A A)$, then the functor $Hom_A(-,T)$ induces an isomorphism of abelian groups: $Hom_A(M,N) \cong Hom_{B^{op}}(Hom_A(N,T),Hom_A(M,T))$. Here T is a tilting A-module, and $B :=End_A(T)$.

In its proof, it says"We use the fact that $_AA$ has an $add(_AT)$-copresentation, that is, there is an exact sequence $0 \rightarrow A \rightarrow T_0 \rightarrow T_1$ of A-modules with $T_0,T_1 \in add(_AT)$ such that the sequence $Hom_A(T_1,T) \rightarrow Hom_A(T_0,T) \rightarrow Hom_A(A,T) \rightarrow 0$ is still exact."

So I want to ask:

- Why $_AA$ has an $add(_AT)$-copresentation?
- How to use this fact get the isomorphism $Hom_A(M,N) \cong Hom_{B^{op}}(Hom_A(N,T),Hom_A(M,T))$? Thank you.