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.