Proposition 8

Tag 0DX5, permanent. Cite it as [Bourbaki, Tag 0DX5].

Let $f : M_1 \to M$ be a homomorphism of $A$ *-modules* and

\begin{array}{cccccccccc} (\mathscr{E}) & 0 & \longrightarrow & N' & \xrightarrow{u} & N & \xrightarrow{v} & N'' & \longrightarrow & 0 \\ & & & g' \downarrow & & g \downarrow & & g'' \downarrow & & \\ (\mathscr{E}_1) & 0 & \longrightarrow & N'_1 & \xrightarrow{u_1} & N_1 & \xrightarrow{v_1} & N''_1 & \longrightarrow & 0 \end{array}

\begin{array}{ccccc} \operatorname{Ext}(M,N') & \xrightarrow{\operatorname{Ext}(1,u)} & \operatorname{Ext}(M,N) & \xrightarrow{\operatorname{Ext}(1,v)} & \operatorname{Ext}(M,N'') \xrightarrow{\delta(M,\mathscr{G})} \operatorname{Ext}(M,N') \xrightarrow{\operatorname{Ext}(1,u)} \operatorname{Ext}(M,N)\\[6pt] \Big\uparrow\vcenter{\rlap{$\scriptstyle\varphi_M(N')$}} && \Big\uparrow\vcenter{\rlap{$\scriptstyle\varphi_M(N)$}} && \Big\uparrow\vcenter{\rlap{$\scriptstyle\varphi_M(N'')$}} \qquad\qquad\Big\uparrow\vcenter{\rlap{$\scriptstyle\varphi_M(N')$}}\\[6pt] H(\operatorname{Homgr}(L(M),N')) & \xrightarrow{H(\operatorname{Homgr}(1,u))} & H(\operatorname{Homgr}(L(M),N)) & \xrightarrow{H(\operatorname{Homgr}(1,v))} & H(\operatorname{Homgr}(L(M),N'')) \xrightarrow{\partial(M\mathscr{G})} H(\operatorname{Homgr}(L(M),N')) \xrightarrow{H(\operatorname{Homgr}(1,u))} H(\operatorname{Homgr}(L(M),N)) \end{array}

a commutative diagram of $A$ -modules with exact rows. The diagram of $k$ -modules

\begin{array}{ccc} \mathrm{Ext}_A(M, N'') & \xrightarrow{\delta(M, \mathscr{E})} & \mathrm{Ext}_A(M, N') \\ \mathrm{Ext}(f, g'') \downarrow & & \mathrm{Ext}(f, g') \downarrow \\ \mathrm{Ext}_A(M_1, N''_1) & \xrightarrow{\delta(M, \mathscr{E}_1)} & \mathrm{Ext}_A(M_1, N'_1) \end{array}

is commutative. This follows from X, p. 31, prop. 2 applied to the commutative diagram

\begin{array}{ccccccccc} & & \mathrm{Homgr}(1, u) & & & & \mathrm{Homgr}(1, v) & & \\ 0 \to \mathrm{Homgr}_A(L(M), N') & \to & \mathrm{Homgr}_A(L(M), N) & \to & \mathrm{Homgr}_A(L(M), N'') & \to & 0 \\ \mathrm{Homgr}(L(f), g') \downarrow & & \mathrm{Homgr}(L(f), g) \downarrow & & \mathrm{Homgr}(L(f), g'') \downarrow & & \\ 0 \to \mathrm{Homgr}_A(L(M_1), N'_1) & \to & \mathrm{Homgr}_A(L(M_1), N_1) & \to & \mathrm{Homgr}_A(L(M_1), N''_1) & \to & 0 \end{array}

Let $N$ be an $A$ -module, and

(\mathscr{F})

0 \to M' \xrightarrow{r} M \xrightarrow{s} M'' \to 0

an exact sequence of

A

-modules; the sequence of complexes

(\mathscr{F}_N)

0 \longrightarrow \mathrm{Homgr}_A(M'', I(N)) \xrightarrow{\mathrm{Homgr}(s, 1)} \mathrm{Homgr}_A(M, I(N))

\xrightarrow{\mathrm{Homgr}(r, 1)} \mathrm{Homgr}_A(M', I(N)) \longrightarrow 0

is exact (X, p. 83, prop. 2,

a

) ; let

\partial(\mathscr{F}_N) : \mathrm{H}(\mathrm{Homgr}_A(M', I(N))) \to \mathrm{H}(\mathrm{Homgr}_A(M'', I(N)))

be the corresponding connecting homomorphism.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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