Lemma 51.20.2. Let R be a ring. Let \alpha : M \to M' be a map of R-modules. Let P_\bullet \to M and P'_\bullet \to M' be resolutions by projective R-modules. Let e \geq 0 be an integer. Consider the following conditions
We can find a map of complexes a_\bullet : P_\bullet \to P'_\bullet inducing \alpha on cohomology with a_ i = 0 for i > e.
We can find a map of complexes a_\bullet : P_\bullet \to P'_\bullet inducing \alpha on cohomology with a_{e + 1} = 0.
The map \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) is zero for all R-modules N and i > e.
The map \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N) is zero for all R-modules N.
Let N = \mathop{\mathrm{Im}}(P'_{e + 1} \to P'_ e) and denote \xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) the canonical element (see proof). Then \xi maps to zero in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N).
The map \text{Tor}_ i^ R(M, N) \to \text{Tor}_ i^ R(M', N) is zero for all R-modules N and i > e.
The map \text{Tor}_{e + 1}^ R(M, N) \to \text{Tor}_{e + 1}^ R(M', N) is zero for all R-modules N.
Then we always have the implications
(1) \Leftrightarrow (2) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5) \Rightarrow (6) \Leftrightarrow (7)
If M is (-e - 1)-pseudo-coherent (for example if R is Noetherian and M is a finite R-module), then all conditions are equivalent.
Proof.
It is clear that (2) implies (1). If a_\bullet is as in (1), then we can consider the map of complexes a'_\bullet : P_\bullet \to P'_\bullet with a'_ i = a_ i for i \leq e + 1 and a'_ i = 0 for i \geq e + 1 to get a map of complexes as in (2). Thus (1) is equivalent to (2).
By the construction of the \mathop{\mathrm{Ext}}\nolimits and \text{Tor} functors using resolutions (Algebra, Sections 10.71 and 10.75) we see that (1) and (2) imply all of the other conditions.
It is clear that (3) implies (4) implies (5). Let N be as in (5). The canonical map \tilde\xi : P'_{e + 1} \to N precomposed with P'_{e + 2} \to P'_{e + 1} is zero. Hence we may consider the class \xi of \tilde\xi in
\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) = \frac{\mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits (P'_{e + 1}, N \to \mathop{\mathrm{Hom}}\nolimits (P'_{e + 2}, N)}{ \mathop{\mathrm{Im}}(\mathop{\mathrm{Hom}}\nolimits (P'_ e, N \to \mathop{\mathrm{Hom}}\nolimits (P'_{e + 1}, N)}
Choose a map of complexes a_\bullet : P_\bullet \to P'_\bullet lifting \alpha , see Derived Categories, Lemma 13.19.6. If \xi maps to zero in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N), then we find a map \varphi : P_ e \to N such that \tilde\xi \circ a_{e + 1} = \varphi \circ d. Thus we obtain a map of complexes
\xymatrix{ \ldots \ar[r] & P_{e + 1} \ar[r] \ar[d]^0 & P_ e \ar[r] \ar[d]^{a_ e - \varphi } & P_{e - 1} \ar[r] \ar[d]^{a_{e - 1}} & \ldots \\ \ldots \ar[r] & P'_{e + 1} \ar[r] & P'_ e \ar[r] & P'_{e - 1} \ar[r] & \ldots }
as in (2). Hence (1) – (5) are equivalent.
The equivalence of (6) and (7) follows from dimension shifting; we omit the details.
Assume M is (-e - 1)-pseudo-coherent. (The parenthetical statement in the lemma follows from More on Algebra, Lemma 15.64.17.) We will show that (7) implies (4) which finishes the proof. We will use induction on e. The base case is e = 0. Then M is of finite presentation by More on Algebra, Lemma 15.64.4 and we conclude from Lemma 51.20.1 that M \to M' factors through a free module. Of course if M \to M' factors through a free module, then \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) is zero for all i > 0 as desired. Assume e > 0. We may choose a map of short exact sequences
\xymatrix{ 0 \ar[r] & K \ar[r] \ar[d] & R^{\oplus r} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K' \ar[r] & \bigoplus _{i \in I} R \ar[r] & M' \ar[r] & 0 }
whose right vertical arrow is the given map. We obtain \text{Tor}_{i + 1}^ R(M, N) = \text{Tor}^ R_ i(K, N) and \mathop{\mathrm{Ext}}\nolimits ^{i + 1}_ R(M, N) = \mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) for i \geq 1 and all R-modules N and similarly for M', K'. Hence we see that \text{Tor}_ e^ R(K, N) \to \text{Tor}_ e^ R(K', N) is zero for all R-modules N. By More on Algebra, Lemma 15.64.2 we see that K is (-e)-pseudo-coherent. By induction we conclude that \mathop{\mathrm{Ext}}\nolimits ^ e(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^ e(K, N) is zero for all R-modules N, which gives what we want.
\square
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