A computation of $Ext^1(M,N)$ (how to derive the commutative diagram for $Ext^1(M,N)$?). I am trying to understand the proof of Proposition 5.6 in the paper on page 17. How to derive the commutative diagram for $Ext^1(M,N)$? Using the projective resolution
:
\begin{align}
\cdots \to \oplus_{v \in V} P_v \overset{D}{\to} \oplus_{u \in U} P_u \to L_I \to 0,
\end{align}
I obtained the part 
\begin{align}
Hom(\oplus_{u \in U} P_u,L_J) \overset{D^*}{\to} Hom(\oplus_{v \in V} P_v, L_J).
\end{align}
How to derive the other parts of the commutative diagram? Thank you very much.


By laziness I'll oversimplify the notation:
$$
M \xrightarrow{D} N \rightarrow L_I \rightarrow 0
$$
Now break this sequence in two pieces, noting that $\Omega L_I = imD$,
$$
M \xrightarrow{f} \Omega L_I \rightarrow 0, \\ 0 \rightarrow \Omega L_I \xrightarrow{g} N \rightarrow L_I \rightarrow 0
$$
We apply $Hom(-, L_J)$ to the first:
$$
0 \rightarrow Hom(\Omega L_I, L_J) \xrightarrow{f^\ast} Hom(M,L_J)
$$
This gives the vertical part of the diagram. Applying $Hom(-, L_J)$ to the second sequence and taking the long exact sequence of Ext modules leads us to
$$
0\to  Hom(L_I, L_J) \to Hom(N, L_J) \xrightarrow{g^\ast} Hom(\Omega L_I, L_J) \rightarrow\\ \rightarrow Ext^1(L_I, L_J) \to Ext^1(N, L_J) = \{0\}
$$
which givers the horizontal part of the diagram. The last module is trivial since $N$ is projective. 
Also note that $D=g\circ f$ hence $D^\ast = f^\ast \circ g^\ast$ which means that the triangle in the diagram commutes. 
