CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-06

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Extension classes form a set whenever derived Ext one does

Statement

Assume the Axiom of Dependent Choice and the hypotheses of The balanced Ext bifunctor. If ${Ext}^{1} (M, N)$ is a set, then extensions of $M$ by $N$ have a set of equivalence classes: explicitly, there is a set $S$ and an assignment $c$ from extensions to $S$ , every element of $S$ occurs, and $c (e) = c (e^{'})$ exactly when $e$ and $e^{'}$ are equivalent. This is a set realization of the classes; it does not assert that each literal class of isomorphic extension diagrams is itself a set.

Facts & Assumptions

Given: Dependent Choice, the balanced-Ext resolution hypotheses, and the set $S = {Ext}^{1} (M, N)$ .

[F1]

Equivalent extensions have identical connecting classes: Equivalent extensions have the same Ext class.

[F2]

Every derived Ext-one element is represented by an extension: Every Ext-one class is represented by an extension.

[F3]

Extensions with the same connecting class are equivalent: Two extensions with the same Ext class are equivalent.

Proof

technique · direct

1.1givenF1F3construct

For an extension $e : 0 \to N \to E \to M \to 0$ , assign its connecting class $c (e) = δ_{e} (1_{M}) \in S$ . By [F1] and [F3], $c (e) = c (e^{'})$ if and only if $e$ and $e^{'}$ are equivalent. These lemmas concern individual extensions and do not assume a set of all extension classes.

2.1givenF2step 1.1algebra∎

The supplied projective resolution gives the presentation $0 \to \ker (P_{0} (M) \to M) \to P_{0} (M) \to M \to 0$ required by [F2]. Hence every $s \in S$ equals $c (e)$ for some extension $e$ . Together with step 1.1 this makes the already given set $S$ a set realization of precisely the extension equivalence classes. No quotient of a proper class or simultaneous choice of representatives is used.

Depends on

Used by

FALSE: extension classes automatically form a set False statement

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Charles A. Weibel, An Introduction to Homological Algebra, Chapters 3–4 (standard reference, not scraped)
The Stacks Project, Extensions and Ext groups (standard reference, not scraped)