Lemma 51.20.1. Let $R$ be a ring. Let $M \to M'$ be a map of $R$-modules with $M$ of finite presentation such that $\text{Tor}_1^ R(M, N) \to \text{Tor}_1^ R(M', N)$ is zero for all $R$-modules $N$. Then $M \to M'$ factors through a free $R$-module.
Proof. 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 can factor this map through the short exact sequence
51.20.1.1
\begin{equation} \label{local-cohomology-equation-pushout} 0 \to K' \to E \to M \to 0 \end{equation}
which is the pushout of the first short exact sequence by $K \to K'$. By a diagram chase we see that the assumption in the lemma implies that the boundary map $\text{Tor}_1^ R(M, N) \to K' \otimes _ R N$ induced by (51.20.1.1) is zero, i.e., the sequence (51.20.1.1) is universally exact. This implies by Algebra, Lemma 10.82.4 that (51.20.1.1) is split (this is where we use that $M$ is of finite presentation). Hence the map $M \to M'$ factors through $\bigoplus _{i \in I} R$ and we win. $\square$
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