Theorem 1

Let $M$ be a finitely generated $A$-module. Then $\alpha_M$ is a homologism: we have $H_i(D(D(M))) = 0$ for $i \neq 0$ and the homomorphism $\alpha_M$ is bijective.

Let us first take $M = A$. The mapping $e : \Omega \to D(A)$ is a homomorphism (Example 1), hence so also is the mapping $D(e) : D(D(A)) \to D(\Omega)$. The mapping $\tilde{e} : A \to D(\Omega)$ is a homomorphism (Example 2), and we have $D(e) \circ \alpha_A = \tilde{e}$; thus $\alpha_A$ is a homomorphism, which proves the theorem in this case. It follows that $\alpha_M$ is a homomorphism when the $A$-module $M$ is a finitely generated free module.

Let us pass to the general case; we shall prove by induction on the integer $n$ the following assertion:

$(A_n)$ for every finitely generated $A$-module $M$, the homomorphism $H_i(\alpha_M)$ is bijective for $i \leq n$.

This also means that $H_i(D(D(M)))$ is zero for $i \neq 0$ and $i \leq n$, and that $\alpha_M$ is bijective if $n \geq 0$. Observe that $(A_n)$ holds for $n < -d$, where $d$ is the length of the complex $I$; in fact the $A$-module $D(D(M))_i$ is equal to $\bigoplus_p \mathrm{Hom}_A(\mathrm{Hom}_A(M, I^p), I^{p-i})$, and is therefore zero for $i < -d$ and $i > d$.

Let us prove the implication $(A_n) \Rightarrow (A_{n+1})$. Let $M$ be a finitely generated $A$-module. There exists a finitely generated free $A$-module $L$ and an exact sequence $0 \to N \xrightarrow{u} L \xrightarrow{v} M \to 0$.

The sequence $0 \to D(M) \xrightarrow{D(v)} D(L) \xrightarrow{D(u)} D(N) \to 0$ is exact; analogously, if we put $u' = D(D(u))$ and $v' = D(D(v))$, the sequence $0 \to D(D(N)) \xrightarrow{u'} D(D(L)) \xrightarrow{v'} D(D(M)) \to 0$ is exact.

Since $H_i(D(D(L)))$ is zero for $i \neq 0$, we have isomorphisms

this yields the implication $(A_n) \Rightarrow (A_{n+1})$ for $n \neq -1$ and $n \neq 0$. Consider the commutative diagram with exact rows

where $\alpha_L$ is bijective. If $(A_0)$ holds, the homomorphism $\alpha_N$ is likewise bijective, hence $H_0(u')$ is injective and one obtains $H_1(D(D(M))) = 0$, whence $(A_1)$. If $(A_{-1})$

Thus $(A_n)$ is true for every $n$, which proves the theorem.

Let $M$ be a finitely generated $A$-module; put $c = \dim(A) - \dim_A(M)$. Let $D'(M)$ denote the subcomplex of $D(M)$ equal to $\bigoplus_{i < c} D(M)^i \bigoplus Z^c(D(M))$, and

the canonical injection. From the canonical surjection $Z^c(D(M)) \to H^c(D(M))$ and the isomorphism $\varphi(M, I)$ one deduces a morphism of complexes

since $H^i(D(M))$ is zero for $i < c$ (No. 1, prop. 3 a)), $(D'(M)(-c), p)$ is a left resolution of $\mathrm{Ext}_A^c(M, \Omega)$. By A, X, p. 100, th. 1, one has a canonical isomorphism

by composing it with the homomorphism $H_0(D(j)) : H_0(D(D(M))) \to H_0(D(D'(M)))$

one therefore obtains a homomorphism

whence finally, by composition with $\alpha_M$, a canonical homomorphism