# How to get $Hom_A(M,N) \cong Hom_{B^{op}}(Hom_A(N,T),Hom_A(M,T))$?

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."

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.

(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.

• @ Jeremy Rickard In (2), if we take $N=T$, I don't think $FT$ and $GT$ are isomorphic to $B$. But $GT= Hom_{B^{op}}(Hom_A(T,T),Hom_A(M,T)) =Hom_{B^{op}}(B,Hom_A(M,T)) \cong Hom_B(B,Hom_A(M,T)) \cong Hom_A(M,T)=FT.$ Is it right? – Xiaosong Peng Sep 14 '16 at 12:48
• Yes, sorry, I meant $\textrm{Hom}_A(M,T)$, not $B$. I've corrected it. – Jeremy Rickard Sep 14 '16 at 14:22
• @ Jeremy Rickard Get it. Thank you! – Xiaosong Peng Sep 15 '16 at 0:16