Remark 2

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

Algebra, chapter X, § 5. Modules d’extensions

The morphism of complexes

\operatorname{Homgr}_{A}(p_{M},\epsilon_{N}):\operatorname{Hom}_{A}(M,N)\to\operatorname{Homgr}_{A}(L(M),I(N))

induces on the homology in degree 0 the isomorphism

\lambda^{-1}_{M,N}:\operatorname{Hom}_{A}(M,N)\to\operatorname{Ext}^{0}_{A}(M,N)

reciprocal to $\lambda_{M,N}$ .

We have $L(1_{M})=1_{L(M)}$ , $I(1_{N})=1_{I(N)}$ , hence by passing to homology

(8)

\operatorname{Ext}_{A}(1_{M},1_{N})=1_{\operatorname{Ext}_{A}(M,N)}\,.

If $f' : M''\to M'$ and $g' : N'\to N''$ are $A$ -module homomorphisms, we have $L(f\circ f')=L(f)\circ L(f')$ , $I(g'\circ g)=I(g')\circ I(g)$ , and thus...

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\operatorname{Ext}_{A}(f\circ f',g'\circ g)=\operatorname{Ext}_{A}(f',g')\circ\operatorname{Ext}_{A}(f,g)\,.

Consider the morphisms of $k$ -complexes

\operatorname{Homgr}_A(L(M), N) \xrightarrow{\operatorname{Homgr}_A(1, e_N)} \operatorname{Homgr}_A(L(M), I(N)) \xleftarrow{\operatorname{Homgr}_A(p_M, 1)} \operatorname{Homgr}_A(M, I(N)),

and the homomorphisms they induce in homology:

H(\operatorname{Homgr}_A(L(M), N)) \xrightarrow{\varphi_{M(N)}} \operatorname{Ext}_A(M, N) \xleftarrow{\overline{\varphi}_{N(M)}} H(\operatorname{Homgr}_A(M, I(N))) .

According to Proposition 4 of X, p. 86,

\operatorname{Homgr}_A(1, e_N)

and

\operatorname{Homgr}_A(p_M, 1)

are homologisms, hence:

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

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