False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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FALSE: Ext is covariant in both variables

Statement

Facts & Assumptions

Given: A morphism $a : M^{'} \to M$ , a morphism $b : N \to N^{'}$ , and balanced Ext.

Refutation

technique · direct

1.1givenalgebra

Precomposition sends $f : M \to I^{q} (N)$ to $f \circ a : M^{'} \to I^{q} (N)$ , hence gives ${Ext}^{q} (M, N) \to {Ext}^{q} (M^{'}, N)$ ; its direction is opposite to $a$ .

2.1step 1.1algebra∎

Postcomposition by the comparison induced by $b$ gives ${Ext}^{q} (M, N) \to {Ext}^{q} (M, N^{'})$ . Thus only the second variable is covariant, while the first is contravariant, disproving covariance in both.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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, Chapter 2 (standard reference, not scraped)