If M M M N N N Ext A i ( M , N ) = 0 \operatorname{Ext}_A^i(M, N) = 0 Ext A i ( M , N ) = 0 i > 0 i > 0 i > 0 Homgr A ( 1 , e N ) : Hom A ( M , N ) → Homgr A ( M , I ( N ) ) \operatorname{Homgr}_A(1, e_N) : \operatorname{Hom}_A(M, N) \to \operatorname{Homgr}_A(M, I(N)) Homgr A ( 1 , e N ) : Hom A ( M , N ) → Homgr A ( M , I ( N )) Homgr A ( p M , 1 ) : Hom A ( M , N ) → Homgr A ( L ( M ) , N ) \operatorname{Homgr}_A(p_M, 1) : \operatorname{Hom}_A(M, N) \to \operatorname{Homgr}_A(L(M), N) Homgr A ( p M , 1 ) : Hom A ( M , N ) → Homgr A ( L ( M ) , N )
Remarks. — 3) Let g : N → N ′ g : N \to N' g : N → N ′ A A A
Homgr A ( 1 L ( M ) , g ) ∘ Homgr A ( 1 L ( M ) , e N ) = Homgr A ( 1 L ( M ) , e N ′ ) ∘ Homgr A ( 1 L ( M ) , I ( g ) ) ,
\operatorname{Homgr}_A(1_{L(M)}, g) \circ \operatorname{Homgr}_A(1_{L(M)}, e_N) = \operatorname{Homgr}_A(1_{L(M)}, e_{N'}) \circ \operatorname{Homgr}_A(1_{L(M)}, I(g)),
Homgr A ( 1 L ( M ) , g ) ∘ Homgr A ( 1 L ( M ) , e N ) = Homgr A ( 1 L ( M ) , e N ′ ) ∘ Homgr A ( 1 L ( M ) , I ( g )) , and hence the diagram
H ( Homgr A ( L ( M ) , N ) ) → φ M ( N ) Ext A ( M , N ) H ( Homgr A ( 1 L ( M ) , g ) ) ↓ ↓ Ext A ( 1 M , g ) H ( Homgr A ( L ( M ) , N ′ ) ) → φ M ( N ′ ) Ext A ( M , N ′ )
\begin{array}{ccc}
H(\operatorname{Homgr}_A(L(M), N)) & \xrightarrow{\varphi_{M(N)}} & \operatorname{Ext}_A(M, N) \\
\operatorname{H}(\operatorname{Homgr}_A(1_{L(M)}, g)) \downarrow & & \downarrow \operatorname{Ext}_A(1_M, g) \\
H(\operatorname{Homgr}_A(L(M), N')) & \xrightarrow{\varphi_{M(N')}} & \operatorname{Ext}_A(M, N')
\end{array}
H ( Homgr A ( L ( M ) , N )) H ( Homgr A ( 1 L ( M ) , g )) ↓ H ( Homgr A ( L ( M ) , N ′ )) φ M ( N ) φ M ( N ′ ) Ext A ( M , N ) ↓ Ext A ( 1 M , g ) Ext A ( M , N ′ ) is commutative.
4) Analogously, if f : M ′ → M f : M' \to M f : M ′ → M A A A
H ( Homgr A ( M , I ( N ) ) ) → φ ‾ N ( M ) Ext A ( M , N ) H ( Homgr A ( f , 1 I ( N ) ) ) ↓ ↓ Ext A ( f , 1 N ) H ( Homgr A ( M ′ , I ( N ) ) ) → φ ‾ N ( M ′ ) Ext A ( M ′ , N )
\begin{array}{ccc}
H(\operatorname{Homgr}_A(M, I(N))) & \xrightarrow{\overline{\varphi}_{N(M)}} & \operatorname{Ext}_A(M, N) \\
\operatorname{H}(\operatorname{Homgr}_A(f, 1_{I(N)})) \downarrow & & \downarrow \operatorname{Ext}_A(f, 1_N) \\
H(\operatorname{Homgr}_A(M', I(N))) & \xrightarrow{\overline{\varphi}_{N(M')}} & \operatorname{Ext}_A(M', N)
\end{array}
H ( Homgr A ( M , I ( N ))) H ( Homgr A ( f , 1 I ( N ) )) ↓ H ( Homgr A ( M ′ , I ( N ))) φ N ( M ) φ N ( M ′ ) Ext A ( M , N ) ↓ Ext A ( f , 1 N ) Ext A ( M ′ , N ) is commutative.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.