Exercise 6, § 7

Let M be a right A-module, N a left A-module. For $n \geqslant 0$, denote by $\mathscr{T}_n(M, N)$ the set of triples $(P, p, q)$, where P is a complex of finitely generated free A-modules, such that $P_r = 0$ for $r < 0$ and $r > n$, and where $p : P \to N$ and $q : P^* \to M(n)$ are morphisms of A-complexes (we denote by $P^*$ the complex $\mathrm{Hom}_{\mathbf{A}}(P, A)$). We say that two elements $(P, p, q)$ and $(P', p', q')$ of $\mathscr{T}_n(M, N)$ are linked if there exists a morphism $u : P \to P'$ such that $p = p' \circ u$ and $q' = q \circ 'u$; we denote by $T_n(M, N)$ the quotient of $\mathscr{T}_n(M, N)$ by the finest equivalence relation for which two linked elements are equivalent (cf. II, p. 52, exercise 9). a) Now prove that $T_n(M, N)$ identifies with $\mathrm{Tor}_n^A(M, N)$. b) Let $(\mathscr{E}) : 0 \to M' \xrightarrow{i} M \xrightarrow{\pi} M'' \to 0$ be an exact sequence of right A-modules. Describe the connecting homomorphism $\partial : T_n(M'', N) \to T_{n-1}(M', N)$ deduced from $\partial(\mathscr{E}, N)$ by means of the preceding identifications. (If $(P, p, q) \in \mathscr{T}_n(M'', N)$, now prove that one has $\partial(P, p, q) = (\overline{P}, \overline{p}, \overline{q})$, where $\overline{P}$ is the complex obtained by replacing $P_n$ by 0 in P, $\overline{p}$ is the morphism deduced from $p$ and $(\overline{q})^{n-1} : P_{n-1}^* \to M'$ is equal to $u^{n-1}$, $u$ being a morphism from $P^*$ into the complex (Con (i)) (n) such that $\pi \circ u^n = q^n$.)