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PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Cohomology of derived hom is ext

Statement

In the mixed bounded range of derived Hom, HnRHom(M,N)HomD(A)(M,N[n]) for every integer n. For objects M,N in degree zero and n0 this is classical Extn(M,N) under the supplied one-sided resolution hypothesis.

Facts & Assumptions

Given: In the mixed bounded range of derived Hom, HnRHom(M,N)HomD(A)(M,N[n]) for every integer n. For objects M,N in degree zero and n0 this is classical Extn(M,N) under the supplied one-sided resolution hypothesis.

[F1]

Derived Hom in the mixed bounded range uses a projective source or injective target, with no-roof cohomology comparisons and a mixed comparison zigzag (Derived hom in the bounded setting).

[F2]

Classical Ext identifies with derived Hom from M[0] to N[n] for n0 (Ext is hom in the derived category).

Proof

1.1

In the projective construction, cycles of degree n in Hom(PM,N) are chain maps PMN[n]. Boundaries are their nullhomotopies, since multiplying a homotopy by (1)n converts the shifted homotopy formula into du=dNu(1)n1udP. Hence the cohomology is HomK(PM,N[n]). The no-roof comparison built into the derived Hom construction identifies it with HomD(M,N[n]). The same calculation for Hom(M,IN) uses the injective no-roof comparison. Zero objects and every integer n are allowed.

F1algebra
2.1

For degree-zero inputs, the classical Ext comparison identifies the last Hom group with the supplied resolution Ext for n0. The identifications use the same cocycles and comparison maps, so they are natural in both variables. The mixed Hom zigzag makes the two one-sided descriptions agree when both are available.

F2step 1.1

Depends on

Used by

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