Lemma 2

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

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

The connecting homomorphism

H^n(\mathrm{Homgr}_A(Z, C')) \to H^n(\mathrm{Homgr}_A(B, C'))

associated to (12) is equal to $(-1)^{n+1} H(\mathrm{Homgr}(i, 1))$ .

Proof. Indeed, let $a \in Z^n(\mathrm{Homgr}_A(Z, C'))$ ; this is a morphism of complexes of ascending degree $n$ from $Z$ to $C'$ , whose values are therefore in $Z(C')$ . Since the exact sequence (1) is split ( $B$ being projective), $a$ extends to an element $b$ of $\mathrm{Homgr}_A^n(C, Z(C'))$ . By definition, the image of the class of $a$ by the connecting homomorphism in question is the class in $H^n(\mathrm{Homgr}_A(B, C'))$ of the homomorphism $u$ from $B(C)$ to $C'$ such that for $x \in C$ , we have

u(dx) = Db(x) = d'b(x) - (-1)^n b(dx) = (-1)^{n+1} b(dx) = (-1)^{n+1} a(dx),

hence the assertion.

The homology exact sequence associated to (12) therefore gives the exact sequence → H^n(\mathrm{Homgr}_A(Z, C')) \xrightarrow{\mathrm{H}(\mathrm{Homgr}(i,1))} H^n(\mathrm{Homgr}_A(B, C')) \xrightarrow{\mathrm{H}(\mathrm{Homgr}(\delta,1))} H^{n+1}(\mathrm{Homgr}_A(C, C')) \xrightarrow{\mathrm{H}(\mathrm{Homgr}(j,1))} H^{n+1}(\mathrm{Homgr}_A(Z, C')) \to \ldots .

Moreover, since $Z$ is projective, we obtain from (II_p) exact sequences 0 \to \mathrm{Hom}_A(H_p, H'^q) \to \mathrm{Hom}_A(Z_p, H'^q) \to \mathrm{Hom}_A(B_p, H'^q) \to \mathrm{Ext}_A^1(H_p, H'^q) \to 0 , hence, by passing to products, exact sequences 0 \to \mathrm{Homgr}_A^n(H, H') \to \mathrm{Homgr}_A^n(Z, H') \to \mathrm{Homgr}_A^n(B, H') \to \prod_{p+q=n} \mathrm{Ext}_A^1(H_p, H'^q) \to 0 . Finally, we have the canonical homomorphisms of No. 1:

\begin{align*} \lambda_B &= \lambda(B, C') : \mathrm{H}(\mathrm{Homgr}_A(B, C')) \to \mathrm{Homgr}_A(B, H') , \\ \lambda_Z &= \lambda(Z, C') : \mathrm{H}(\mathrm{Homgr}_A(Z, C')) \to \mathrm{Homgr}_A(Z, H') , \\ \lambda_C &= \lambda(C, C') : \mathrm{H}(\mathrm{Homgr}_A(C, C')) \to \mathrm{Homgr}_A(H, H') , \end{align*}

whence the diagram with exact rows of page X.97.

This diagram is commutative by construction of the homomorphisms $\lambda$ . Moreover, since the complexes B and Z are split and projective, $\lambda_B$ and $\lambda_Z$ are bijective ( $\lambda$ , p. 84, cor. 1). One deduces from this, on the one hand that $\lambda_C^n$ is surjective, with kernel equal to $\mathrm{Im}\ H^n(\mathrm{Homgr}(\delta, 1))$ , and on the other hand that $\varphi^{n-1} \circ \lambda_B^{n-1}$ is surjective, with kernel equal to $\mathrm{Ker}\ H^n(\mathrm{Homgr}(\delta, 1))$ . The theorem follows immediately from this.

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

Cited by

Nothing.

Cites

Nothing.