CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)verified 2026-09-09 (gpt-6-astra)

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The Tor boundary is exactly the obstruction to left exactness after tensoring a fixed short exact sequence

Statement

Assume Dependent Choice (DC), with projective resolutions supplied for the Tor long exact sequence. For $0 \to A \to B \to C \to 0$ of left $R$ -modules and a right module $N$ , the sequence $0 \to N \otimes_{R} A \to N \otimes_{R} B \to N \otimes_{R} C \to 0$ is exact exactly when the connecting homomorphism $\partial : {Tor}_{1}^{R} (N, C) \to N \otimes_{R} A$ is zero. This is vanishing of the boundary map, not necessarily of its domain.

Proof

Given: the displayed short exact sequence, $N$ , supplied resolutions, and The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain.

1.1given

Apply the DC-qualified The long exact Tor sequence in the left-module variable. Its exact degree-zero tail is ${Tor}_{1}^{R} (N, C) \overset{\partial}{\to} N \otimes_{R} A \to N \otimes_{R} B \to N \otimes_{R} C \to 0$ .

2.1step 1.1algebra

By exactness of that tail, exactness already holds at $N \otimes_{R} B$ and $N \otimes_{R} C$ , including surjectivity of the final tensor map. The kernel of $N \otimes_{R} A \to N \otimes_{R} B$ equals $im \partial$ . This uses the stated Tor LES tail, not an undeclared tensor theorem for a commutative ring.

3.1step 2.1algebra∎

Thus injectivity of $N \otimes A \to N \otimes B$ , and hence exactness of the whole tensor sequence, is equivalent to $\partial = 0$ .

Remarks

For any explicitly supplied exact tail of the form in step 1.1, the equivalence proved in steps 2.1–3.1 is choice-free. DC is invoked only to obtain the quoted natural resolution-independent Tor boundary; it is not asserted necessary for this elementary kernel-image implication.

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Sources

Charles A. Weibel, An Introduction to Homological Algebra (standard reference, not scraped)