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

whose right vertical arrow is the given map. We can factor this map through the short exact sequence

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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