Descendable probes modulo null probes

Exact statement

Given cochain maps $i^*:C^\bullet(P)\to C^\bullet(A)$ and $q^*:C^\bullet(Q)\to C^\bullet(A)$ , let $D^k=\{a:i^*a\in\operatorname{im}q^*\}$ and $N^k=\ker i^*$ . Then $D$ and $N$ are subcomplexes, and restriction induces $D^\bullet/N^\bullet\cong\operatorname{im}i^*\cap\operatorname{im}q^*\subseteq C^\bullet(A)$ .

StatusProved and exactly audited

External reviewNo documented external or specialist review of this Quotient Arsenal result is recorded.

Context

The general calculus treats a failed lift as a cocycle, its cohomology class as the obstruction, and a primitive—when one exists—as the receipt that repairs the presentation.

Proof or evidence

Pullback commutes with coboundary and the first isomorphism theorem identifies the quotient.

Self-contained cohomological sewing proof and evidence →

Verification notes

The exact archived proofs were reconstructed, the displayed hypotheses and degree conventions were checked, the original finite controls were replayed, and an independent implementation tested the connecting, graph-transport, staged, and missing-subcomplex boundaries.

Limitations

The finite simplicial model does not automatically prove an arbitrary de Rham analogue.

Open work

Specify admissible probe categories before exporting the construction.