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: 


1. Why $\_AA$ has an $add(\_AT)$-copresentation?
2. 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.



How to get $Hom_A(M,N) \cong Hom_{B^{op}}(Hom_A(N,T),Hom_A(M,T))$?
(1) By the definition of a tilting module there is an exact sequence $0\to A\to T\_0\to\dots\to T\_n\to 0$. Applying $\textrm{Hom}\_A(-,T)$ to this gives an exact sequence, and so $0\to A\to T\_0\to T\_1$ is an $\textrm{add}(T)$-copresentation.


To see this, let $X\_k$ be the image of $T\_k\to T\_{k+1}$, so we have short exact sequences $0\to X\_{k-1}\to T\_k\to X\_k\to 0$ for $k<n$. From the long exact sequences obtained by applying $\textrm{Hom}\_A(-,T)$ to these sequences you can see (starting with $k=n-1$ and working backwards) that $\textrm{Ext}^i\_A(X\_k,T)=0$ and that $0\to\textrm{Hom}(X\_k,T)\to\textrm{Hom}(T\_k,T)\to\textrm{Hom}(X\_{k-1},T)\to 0$ is exact, for all $k<n$ and $i>0$.


(2) Let $FN=\textrm{Hom}\_A(M,N)$ and $GN=\textrm{Hom}\_{B^{op}}\left(\textrm{Hom}\_A(N,T),\textrm{Hom}\_A(M,T)\right)$, so $F$ and $G$ are functors from left $A$-modules to abelian groups, and $\textrm{Hom}\_A(-,T)$ induces a natural transformation $F\to G$. Taking $N=T$, $FT$ and $GT$ are both isomorphic to $\textrm{Hom}\_A(M,T)$, and it is easy to check that the natural map $FT\to GT$ is an isomorphism.


Since $F$ and $G$ are additive functors it follows that $FN\to GN$ is an isomorphism for any $N$ in $\textrm{add}(T)$.


Applying $F$ and $G$ to the exact sequence $0\to A\to T\_0\to T\_1$ gives a commutative diagram
$$\begin{array}{ccccccc}
0&\to&FA&\to&FT\_0&\to&FT\_1\\
&&\downarrow&&\downarrow&&\downarrow\\
0&\to&GA&\to&GT\_0&\to&GT\_1
\end{array}$$
where the last two vertical arrows are isomorphisms and both rows are exact (in the case of the second row this uses the fact that $\textrm{Hom}(T\_1,T)\to\textrm{Hom}(T\_0,T)\to\textrm{Hom}(A,T)\to 0$ is exact). Therefore the first vertical arrow is also an isomorphism.


