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 $F : M \to N$ , the formula $F^{*} [ω] = [F^{*} ω]$ defines a linear map $H_{d R}^{k} (N) \to H_{d R}^{k} (M)$ 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 $H_{d R}^{k} (M) = Z^{k} (M) / B^{k} (M)$ , with $Z^{k}, B^{k}$ 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 $ω' - ω = d η$ for some $(k - 1)$ -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 $F : M \to N$ induces a degree-zero real cochain map $F^{*} : Ω^{∙} (N) \to Ω^{∙} (M)$ .

[F3]

A chain map induces a well-defined map on homology: Let $f : C_{∙} \to D_{∙}$ be a chain map. For every $n \in Z$ there is a unique morphism $H_{n} (f) : H_{n} (C) \to H_{n} (D)$ such that the quotient maps from cycles to homology commute with $Z_{n} (f)$ .

Proof

technique · direct

1.1F1F2given

The cochain identity gives $d F^{*} ω = F^{*} d ω = 0$ . If $ω' = ω + d η$ , then $F^{*} ω' - F^{*} ω = F^{*} d η = d (F^{*} η)$ , so both representatives produce the same class.

2.1F3step 1.1∎

Reindex by $C_{n} = Ω^{- n} (N)$ and $D_{n} = Ω^{- n} (M)$ , with unchanged differentials. The cochain identity is the chain-map identity; its cycles and boundaries at $- k$ are exactly $Z^{k}$ and $B^{k}$ . The induced-homology theorem therefore gives the displayed linear map, agreeing with step 1.1 by its quotient property.

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

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition