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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Pullback induces a well defined map on de rham cohomology

Statement

For a smooth F:MN, the formula F[ω]=[Fω] defines a linear map HdRk(N)HdRk(M) for every integer k.

Facts & Assumptions

Given: A smooth F and a closed k-form ω on N.

[F1]

De rham cohomology: The real de Rham cohomology is HdRk(M)=Zk(M)/Bk(M), with Zk,Bk as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form ω represents a class [ω]. For closed forms ω,ω, equality [ω]=[ω] means precisely ωω=dη for some (k1)-form η. Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

[F2]

Pullback is a morphism of de rham complexes: A smooth map F:MN induces a degree-zero real cochain map F:Ω(N)Ω(M).

[F3]

A chain map induces a well-defined map on homology: Let f:CD be a chain map. For every nZ there is a unique morphism Hn(f):Hn(C)Hn(D) such that the quotient maps from cycles to homology commute with Zn(f).

Proof

technique · direct
1.1

The cochain identity gives dFω=Fdω=0. If ω=ω+dη, then FωFω=Fdη=d(Fη), so both representatives produce the same class.

F1F2given
2.1

Reindex by Cn=Ωn(N) and Dn=Ωn(M), with unchanged differentials. The cochain identity is the chain-map identity; its cycles and boundaries at k are exactly Zk and Bk. The induced-homology theorem therefore gives the displayed linear map, agreeing with step 1.1 by its quotient property.

F3step 1.1

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 17, pp.441–443, Proposition 17.2 and Corollary 17.3; local quotient calculations below.

Depends on

Used by

Dependency tree · two levels

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Sources