The De Rham Complex Homotopy and Mayer Vietoris

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De rham cochain complex

Definition

Let $M$ be a finite-dimensional Hausdorff second-countable smooth manifold without boundary. Its real de Rham cochain complex is $(Ω^{∙} (M), d)$ , where $Ω^{k} (M)$ is the space of smooth $k$ -forms for $0 \leq k \leq \dim M$ and is $0$ otherwise; $d$ has degree $+ 1$ .

These are the sections in A smooth differential $k$ -form. The identity $d^{k + 1} d^{k} = 0$ in The exterior derivative squares to zero makes this an instance of Cochain complex in an abelian category. On the empty manifold each section space is the zero vector space. Whenever a product with $[0, 1]$ is used, forms mean smooth forms up to the endpoints, locally extendible across them.

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.

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Closed and exact differential forms

Definition

For the complex De rham cochain complex, put $Z^{k} (M) = \ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))$ and $B^{k} (M) = im (d : Ω^{k - 1} (M) \to Ω^{k} (M))$ . A form is closed if it belongs to $Z^{k}$ and exact if it belongs to $B^{k}$ .

If $ω = d η$ , then $d ω = d^{2} η = 0$ by The exterior derivative squares to zero, so $B^{k} \subseteq Z^{k}$ . In particular $B^{0} = 0$ , since $Ω^{- 1} = 0$ . The zero form is both closed and exact in every degree.

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.

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De rham cohomology

Definition

The real de Rham cohomology is $H_{d R}^{k} (M) = Z^{k} (M) / B^{k} (M)$ , with $Z^{k}, B^{k}$ as in Closed and exact differential forms.

This is 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.

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.

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Zero and out of range de rham cohomology

Statement

$H_{d R}^{k} (M) = 0$ if $k < 0$ or $k > \dim M$ . If $M = \emptyset$ , its cohomology vanishes in every degree.

Facts & Assumptions

Given: The de Rham complex with zero out-of-range terms.

[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.

Proof

technique · direct

1.1

If $k < 0$ or $k > \dim M$ , the cycle space is a subspace of $Ω^{k} (M) = 0$ , so $Z^{k} = B^{k} = 0$ . Its quotient is therefore zero.

F1given

2.1

On the empty manifold there is just one section of each form bundle, namely the zero section. Thus in every degree the quotient is again $0 / 0 = 0$ , proving the empty clause as well.

F1given∎

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.

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Zero th de rham cohomology is locally constant functions

Statement

$H_{d R}^{0} (M)$ is the algebra of locally constant real functions. For nonempty connected $M$ it is canonically $R$ .

Facts & Assumptions

Given: A smooth real function $f$ on $M$ ; $B^{0} = 0$ .

[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]

The local coordinate formula for the exterior derivative: Let $(U, x^{1}, \dots, x^{n})$ be a smooth chart on a smooth manifold and $ω$ a smooth $k$ -form on $U$ , with $k \geq 0$ . Summing over increasing $k$ -tuples $I$ , and writing $d x^{I} = d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , if $ω = \sum_{I} ω_{I} d x^{I}$ , then $d ω = \sum_{I} d ω_{I} \land d x^{I} .$

[F3]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: Let $a < b$ . Suppose $G : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . If $f : [a, b] \to R$ is Riemann integrable and $f (x) = G^{'} (x) (a < x < b),$ then $\int_{a}^{b} f = G (b) - G (a) .$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

Proof

technique · direct

1.1

In a coordinate ball, $d f = \sum_{i} \partial_{i} f d x^{i}$ . If $d f = 0$ , for points $x, y$ in that ball set $h (t) = f (x + t (y - x))$ . The chain rule gives $h' (t) = \sum_{i} (y_{i} - x_{i}) \partial_{i} f = 0$ ; the fundamental theorem gives $f (y) - f (x) = \int_{0}^{1} 0 d t = 0$ . In dimension zero each coordinate ball is a singleton, so the same constancy conclusion holds.

F2F3given

2.1

Conversely, a locally constant function is smooth and has zero coordinate derivatives, hence $d f = 0$ . Since $B^{0} = 0$ , the quotient in degree zero identifies each such function with itself, preserving addition and multiplication.

F1F2step 1.1

3.1

If $M$ is nonempty and connected, fix $p \in M$ . The level set $f^{- 1} (f (p))$ and its complement are open by local constancy; connectedness forces the complement empty. Thus $f$ is the constant $f (p)$ , and $a \mapsto (p \mapsto a)$ is the asserted algebra isomorphism. On the empty manifold the function space is zero.

step 1.1step 2.1given∎

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.

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Pullback is a morphism of de rham complexes

Statement

A smooth map $F : M \to N$ induces a degree-zero real cochain map $F^{*} : Ω^{∙} (N) \to Ω^{∙} (M)$ .

Facts & Assumptions

Given: A smooth map $F : M \to N$ .

[F1]

De rham cochain complex: Let $M$ be a finite-dimensional Hausdorff second-countable smooth manifold without boundary. Its real de Rham cochain complex is $(Ω^{∙} (M), d)$ , where $Ω^{k} (M)$ is the space of smooth $k$ -forms for $0 \leq k \leq \dim M$ and is $0$ otherwise; $d$ has degree $+ 1$ . These are the sections in def-smooth-differential-k-form. The identity $d^{k + 1} d^{k} = 0$ in thm-the-exterior-derivative-squares-to-zero makes this an instance of def-cochain-complex-in-an-abelian-category. On the empty manifold each section space is the zero vector space. Whenever a product with $[0, 1]$ is used, forms mean smooth forms up to the endpoints, locally extendible across them.

[F2]

The exterior derivative commutes with pullback: For every smooth map $F : M \to N$ and every form $ω$ on $N$ , $d (F^{*} ω) = F^{*} (d ω) .$

[F3]

Cochain map: Let $C^{∙}$ and $D^{∙}$ be cochain complexes. A cochain map $f : C^{∙} \to D^{∙}$ is a family of morphisms $f^{n} : C^{n} \to D^{n}$ such that $d_{D}^{n} \circ f^{n} = f^{n + 1} \circ d_{C}^{n}$ for every $n \in Z$ . Thus the upper-index square $\begin{array}{ccc} C^{n} & \overset{f^{n}}{\to} & D^{n} \\ d_{C}^{n} ↓ & ↓ d_{D}^{n} \\ C^{n + 1} & \overset{f^{n + 1}}{\to} & D^{n + 1} \end{array}$ commutes in each degree.

Proof

technique · direct

1.1

Pointwise, $(F^{*} ω)_{p} (v_{1}, \dots, v_{k}) = ω_{F (p)} (d F_{p} v_{1}, \dots, d F_{p} v_{k})$ . This formula is real linear in $ω$ and preserves degree; in coordinates its coefficients are finite sums of smooth coefficients times derivatives of $F$ , hence smooth. The unique maps on zero terms supply the other degrees.

F1given

2.1

For every $ω$ , $d_{M} F^{*} ω = F^{*} d_{N} ω$ . This is precisely the equation required for a cochain map in each degree, so the family just constructed is a cochain map.

F2F3step 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.

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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

1.1

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.

F1F2given

2.1

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.

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.

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De rham cohomology is a contravariant functor

Statement

De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

Facts & Assumptions

Given: Composable smooth maps $F : M \to N$ , $G : N \to P$ and an integer $k$ .

[F1]

Pullback induces a well defined map on de rham cohomology: 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 $k$ .

[F2]

Pullback of forms is smooth functorial and preserves wedges: For a smooth map $F : M \to N$ , pullback sends smooth differential forms on $N$ to smooth differential forms on $M$ , is functorial, and satisfies $F^{*} (α \land β) = F^{*} α \land F^{*} β .$

[F3]

Homology respects identities and composition: For every $n \in Z$ : 1. $H_{n} (1_{C}) = 1_{H_{n} (C)}$ for every chain complex $C$ . 2. If $f : C \to D$ and $g : D \to E$ are chain maps, then $H_{n} (g \circ f) = H_{n} (g) \circ H_{n} (f) .$

Proof

technique · direct

1.1

On forms, pullback satisfies $(G \circ F)^{*} = F^{*} G^{*}$ and ${id}^{*} = id$ . These are identities between cochain maps, with arrows from $P$ to $M$ .

F2given

2.1

Using $C_{n} = Ω^{- n}$ , apply homology at degree $- k$ to these identities. Homology preserves identities and composition; the induced maps are those already defined on de Rham classes. This gives both identities in the statement.

F1F3step 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.

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Wedge with a closed form preserves exactness classes

Statement

If $α \in Ω^{p} (M)$ and $β \in Ω^{q} (M)$ are closed, then $d η \land β = d (η \land β)$ for $η \in Ω^{p - 1} (M)$ and $α \land d θ = (- 1)^{p} d (α \land θ)$ for $θ \in Ω^{q - 1} (M)$ .

Facts & Assumptions

Given: Closed forms $α, β$ of degrees $p, q \geq 0$ , and forms $η, θ$ of the indicated degrees.

[F1]

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^{k} (M) = \ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))$ and $B^{k} (M) = im (d : Ω^{k - 1} (M) \to Ω^{k} (M))$ . A form is closed if it belongs to $Z^{k}$ and exact if it belongs to $B^{k}$ . If $ω = d η$ , then $d ω = d^{2} η = 0$ by thm-the-exterior-derivative-squares-to-zero, so $B^{k} \subseteq Z^{k}$ . In particular $B^{0} = 0$ , since $Ω^{- 1} = 0$ . The zero form is both closed and exact in every degree.

[F2]

The exterior derivative is a graded derivation: Let $M$ be a smooth manifold. The exterior derivative is an $R$ -linear map $d : Ω^{*} (M) \to Ω^{*} (M)$ of degree one. For homogeneous smooth forms $α \in Ω^{p} (M)$ and $β \in Ω^{q} (M)$ , $d (α \land β) = d α \land β + (- 1)^{\deg α} α \land d β .$

Proof

technique · direct

1.1

For $p \geq 1$ , the graded Leibniz rule gives $d (η \land β) = d η \land β + (- 1)^{p - 1} η \land d β = d η \land β$ . For $p = 0$ , $η = 0$ in degree $- 1$ and the equality is $0 = 0$ .

F1F2given

2.1

For $q \geq 1$ , $d (α \land θ) = d α \land θ + (- 1)^{p} α \land d θ = (- 1)^{p} α \land d θ$ . Multiplying by $(- 1)^{p}$ proves the second equality; if $q = 0$ , $θ = 0$ proves it directly. Thus either exact change has an explicit primitive.

F1F2given∎

Source locator

Lee, Chapter 17, p.441 (closed/exact); graded Leibniz rule in the declared exterior-calculus supplier. The two primitive formulas are derived explicitly.

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Wedge product descends to de rham cohomology

Statement

The formula $[α] [β] = [α \land β]$ defines a bilinear, associative, graded-commutative product on $H_{d R}^{∙} (M)$ , with unit $[1]$ .

Facts & Assumptions

Given: Closed forms $α, β$ of degrees $p, q \geq 0$ .

[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]

Wedge with a closed form preserves exactness classes: If $α \in Ω^{p} (M)$ and $β \in Ω^{q} (M)$ are closed, then $d η \land β = d (η \land β)$ for $η \in Ω^{p - 1} (M)$ and $α \land d θ = (- 1)^{p} d (α \land θ)$ for $θ \in Ω^{q - 1} (M)$ .

[F3]

Differential forms form a graded commutative algebra: The graded vector space $Ω^{*} (M) : = ⨁_{k \geq 0} Ω^{k} (M)$ with the wedge product is an associative graded-commutative algebra.

[F4]

The exterior derivative is a graded derivation: Let $M$ be a smooth manifold. The exterior derivative is an $R$ -linear map $d : Ω^{*} (M) \to Ω^{*} (M)$ of degree one. For homogeneous smooth forms $α \in Ω^{p} (M)$ and $β \in Ω^{q} (M)$ , $d (α \land β) = d α \land β + (- 1)^{\deg α} α \land d β .$

Proof

technique · direct

1.1

The identity $d (α \land β) = d α \land β + (- 1)^{p} α \land d β = 0$ shows that the proposed product represents a class. If $α' = α + d η$ and $β' = β + d θ$ , all four forms are closed, and $α' \land β' - α \land β = d (η \land β') + (- 1)^{p} d (α \land θ)$ . Hence the product is independent of both representatives.

F1F2F4given

2.1

Bilinearity and associativity follow by applying the quotient map to the corresponding identities of forms. Similarly $α \land β = (- 1)^{p q} β \land α$ gives the graded sign. The constant function $1$ is closed and satisfies $1 \land α = α$ , giving the unit; on the empty manifold it equals the zero element of the zero algebra.

F1F3step 1.1∎

Source locator

Lee, Chapter 17, p.441 (quotient); the graded-algebra and graded-derivation identities are supplied by the declared local exterior-calculus results.

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De rham cohomology ring

Definition

The de Rham cohomology ring is $H_{d R}^{∙} (M) = ⨁_{k \in Z} H_{d R}^{k} (M)$ with $[α] [β] = [α \land β]$ and unit $[1]$ .

By Wedge product descends to de rham cohomology it is a unital graded-commutative real algebra. On the empty manifold it is the zero algebra, with $1 = 0$ ; this convention allows the zero algebra among unital algebras.

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.

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Pullback is a homomorphism of de rham cohomology algebras

Statement

Smooth pullback induces a unital graded real algebra homomorphism $H_{d R}^{∙} (N) \to H_{d R}^{∙} (M)$ .

Facts & Assumptions

Given: A smooth map $F : M \to N$ and closed homogeneous forms $α, β$ on $N$ .

[F1]

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

[F2]

De rham cohomology ring: The de Rham cohomology ring is $H_{d R}^{∙} (M) = ⨁_{k \in Z} H_{d R}^{k} (M)$ with $[α] [β] = [α \land β]$ and unit $[1]$ . By thm-wedge-product-descends-to-de-rham-cohomology it is a unital graded-commutative real algebra. On the empty manifold it is the zero algebra, with $1 = 0$ ; this convention allows the zero algebra among unital algebras.

[F3]

Pullback of forms is smooth functorial and preserves wedges: For a smooth map $F : M \to N$ , pullback sends smooth differential forms on $N$ to smooth differential forms on $M$ , is functorial, and satisfies $F^{*} (α \land β) = F^{*} α \land F^{*} β .$

[F4]

Pullback induces a well defined map on de rham cohomology: 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 $k$ .

Proof

technique · direct

1.1

Pullback is already a degree-preserving linear map on cohomology. The class formula F4 and the wedge formula give $F^{*} ([α] [β]) = [F^{*} (α \land β)] = [F^{*} α \land F^{*} β] = F^{*} [α] F^{*} [β]$ .

F1F2F3F4given

2.1

For functions, $F^{*} 1 = 1 \circ F = 1$ , so F4 gives $F^{*} [1] = [1]$ . Linearity extends step 1.1 from homogeneous elements to their finite sums in the direct sum algebra. These are precisely the multiplicativity and unit conditions, also for the zero target algebra when $M$ is empty.

F2F4step 1.1given∎

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.

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Integration along the unit interval for a differential form

Definition

Let $ω \in Ω^{k} (M \times [0, 1])$ be smooth up to the endpoints. For $k \geq 1$ , its interval integral is the $(k - 1)$ -form $K ω = \int_{0}^{1} β_{t} d t$ , where $ω = α_{t} + d t \land β_{t}$ and both families are tangential to $M$ . Set $K = 0$ on degree zero and on zero terms.

Use the product structure of Products of smooth manifolds have a canonical product smooth structure, restricted from $M \times R$ . The families are intrinsically $α_{t} = i_{t}^{*} ω$ and $β_{t} = i_{t}^{*} (ι_{\partial_{t}} ω)$ , using Interior product of a form by a vector field; evaluation on tangential tuples and on $(\partial_{t}, v_{1}, \dots, v_{k - 1})$ proves existence and uniqueness of the decomposition. The integral is in the fixed finite-dimensional fibre $⋀^{k - 1} T_{x}^{*} M$ . Coefficients have smooth local extensions across endpoints. Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral supplies parameter differentiation; coordinate independence and full smoothness are proved in The interval homotopy operator is coordinate independent ↗.

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

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The interval homotopy operator is coordinate independent

Statement

The interval operator $K : Ω^{k} (M \times [0, 1]) \to Ω^{k - 1} (M)$ is coordinate independent and maps smooth forms to smooth forms.

Facts & Assumptions

Given: A smooth form $ω$ up to the endpoints of $M \times [0, 1]$ .

[F1]

Integration along the unit interval for a differential form: Let $ω \in Ω^{k} (M \times [0, 1])$ be smooth up to the endpoints. For $k \geq 1$ , its interval integral is the $(k - 1)$ -form $K ω = \int_{0}^{1} β_{t} d t$ , where $ω = α_{t} + d t \land β_{t}$ and both families are tangential to $M$ . Set $K = 0$ on degree zero and on zero terms. Use the product structure of prop-products-of-smooth-manifolds-have-a-canonical-product-smooth-structure, restricted from $M \times R$ . The families are intrinsically $α_{t} = i_{t}^{*} ω$ and $β_{t} = i_{t}^{*} (ι_{\partial_{t}} ω)$ , using def-interior-product-of-a-form-by-a-vector-field; evaluation on tangential tuples and on $(\partial_{t}, v_{1}, \dots, v_{k - 1})$ proves existence and uniqueness of the decomposition. The integral is in the fixed finite-dimensional fibre $⋀^{k - 1} T_{x}^{*} M$ . Coefficients have smooth local extensions across endpoints. thm-differentiation-under-the-integral-sign-on-a-compact-rectangle supplies parameter differentiation; coordinate independence and full smoothness are proved in lem-the-interval-homotopy-operator-is-coordinate-independent.

[F2]

Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral: Let $a < b$ and $c < d$ . Suppose $g, h : [a, b] \times [c, d] \to R$ are continuous and, for every fixed $t \in [c, d]$ , the function $x \mapsto g (x, t)$ is differentiable on $(a, b)$ with derivative $h (x, t)$ . Define $G (x) : = \int_{c}^{d} g (x, t) d t .$ Then $G$ is differentiable on $[a, b]$ as a function on that interval and $G^{'} (x) = \int_{c}^{d} h (x, t) d t (x \in [a, b]) .$ At $a$ and $b$ the derivative is relative and one-sided. The derivative hypothesis is imposed only for interior parameter values; continuity of $h$ supplies its endpoint values.

Proof

technique · direct

1.1

The coefficient family $β_{t} = i_{t}^{*} (ι_{\partial_{t}} ω)$ is intrinsically a form in the fixed fibre at $x$ . A change of coordinates on $M$ multiplies its coefficient vector by the exterior-power transition matrix $A (x)$ , which does not depend on $t$ . Finite-dimensional integration gives $\int_{0}^{1} A (x) β_{t} (x) d t = A (x) \int_{0}^{1} β_{t} (x) d t$ . Thus the local integral expressions transform as a form.

F1given

2.1

Fix a smaller closed coordinate rectangle about a point of $M$ . Each coefficient $b (x, t)$ and all its derivatives are continuous on that rectangle times $[0, 1]$ , by local smoothness up to endpoints. Applying compact-parameter differentiation with the other coordinates fixed gives $\partial_{x_{j}} \int_{0}^{1} b (x, t) d t = \int_{0}^{1} \partial_{x_{j}} b (x, t) d t$ . The right side is jointly continuous, since uniform continuity on the compact rectangle bounds the difference of integrals by the supremum difference of integrands. Repetition for every multi-index proves all coordinate derivatives exist and are continuous. Hence $K ω$ is smooth. For $k = 0$ , $K = 0$ is smooth directly.

F1F2step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham homotopy formula on a product

Statement

For endpoint inclusions $i_{t} : M \to M \times [0, 1]$ , $i_{1}^{*} - i_{0}^{*} = d K + K d$ on smooth forms of every degree.

Facts & Assumptions

Given: Write $ω = α_{t} + d t \land β_{t}$ with tangential families.

[F1]

The interval homotopy operator is coordinate independent: The interval operator $K : Ω^{k} (M \times [0, 1]) \to Ω^{k - 1} (M)$ is coordinate independent and maps smooth forms to smooth forms.

[F2]

The local coordinate formula for the exterior derivative: Let $(U, x^{1}, \dots, x^{n})$ be a smooth chart on a smooth manifold and $ω$ a smooth $k$ -form on $U$ , with $k \geq 0$ . Summing over increasing $k$ -tuples $I$ , and writing $d x^{I} = d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , if $ω = \sum_{I} ω_{I} d x^{I}$ , then $d ω = \sum_{I} d ω_{I} \land d x^{I} .$

[F3]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: Let $a < b$ . Suppose $G : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . If $f : [a, b] \to R$ is Riemann integrable and $f (x) = G^{'} (x) (a < x < b),$ then $\int_{a}^{b} f = G (b) - G (a) .$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

[F4]

Integration along the unit interval for a differential form: For $ω = α_{t} + d t \land β_{t}$ smooth up to the endpoints, $K ω = \int_{0}^{1} β_{t} d t$ in positive degree, while $K = 0$ in degree zero and on zero terms.

[F5]

Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral: On a compact rectangle, a continuous parameter derivative may be passed through the integral when represented by a continuous function.

Proof

technique · direct

1.1

The coordinate differential gives $d ω = d_{M} α_{t} + d t \land (\partial_{t} α_{t} - d_{M} β_{t})$ : the minus sign follows from moving $d_{M}$ past $d t$ . Consequently the definition F4 gives $K d ω = \int_{0}^{1} \partial_{t} α_{t} d t - \int_{0}^{1} d_{M} β_{t} d t$ .

F2F4given

2.1

The fundamental theorem on each coefficient gives the first integral as $α_{1} - α_{0}$ . By F4, $K ω = \int_{0}^{1} β_{t} d t$ ; coefficientwise F5 permits each $M$ -coordinate derivative through this compact parameter integral, so F2 gives $d_{M} K ω = \int_{0}^{1} d_{M} β_{t} d t$ . Since $i_{t}^{*} ω = α_{t}$ , rearrangement proves the formula. In degree zero, $β = 0$ and this is just the fundamental theorem; in top or out-of-range degrees the vanishing terms satisfy the same identity.

F1F2F3F4F5step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham homotopy formula for a smooth homotopy

Statement

If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

Facts & Assumptions

Given: A smooth homotopy $F$ and a smooth form $ω$ on $N$ .

[F1]

De rham homotopy formula on a product: For endpoint inclusions $i_{t} : M \to M \times [0, 1]$ , $i_{1}^{*} - i_{0}^{*} = d K + K d$ on smooth forms of every degree.

[F2]

The exterior derivative commutes with pullback: For every smooth map $F : M \to N$ and every form $ω$ on $N$ , $d (F^{*} ω) = F^{*} (d ω) .$

Proof

technique · direct

1.1

Apply the product identity to the smooth form $F^{*} ω$ : $i_{1}^{*} F^{*} ω - i_{0}^{*} F^{*} ω = d K F^{*} ω + K d F^{*} ω$ . Evaluation on tangent tuples shows $i_{t}^{*} F^{*} ω = F_{t}^{*} ω$ , including functions.

F1given

2.1

Naturality gives $d F^{*} ω = F^{*} d ω$ . Substitution yields $F_{1}^{*} ω - F_{0}^{*} ω = d (K F^{*} ω) + K F^{*} d ω$ , the required identity for every degree; zero terms require no separate extension.

F2step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

Smoothly homotopic maps induce the same de rham map

Statement

Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.

Facts & Assumptions

Given: Smooth maps $f_{0}, f_{1} : M \to N$ joined by a smooth homotopy $F$ .

[F1]

De rham homotopy formula for a smooth homotopy: If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

[F2]

Pullback induces a well defined map on de rham cohomology: 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 $k$ .

[F3]

Chain-homotopic maps induce the same map on homology: If $f, g : C_{∙} \to D_{∙}$ are chain-homotopic chain maps, then for every $n \in Z$ , $H_{n} (f) = H_{n} (g) : H_{n} (C) \to H_{n} (D) .$

Proof

technique · direct

1.1

For every closed $k$ -form $ω$ , the homotopy formula gives $f_{1}^{*} ω - f_{0}^{*} ω = d (K F^{*} ω)$ , since $d ω = 0$ . For $k = 0$ the primitive term is zero, so the functions themselves agree.

F1given

2.1

The two pullbacks therefore give the same quotient class. Equivalently, under $C_{n} = Ω^{- n}$ the operator $K F^{*}$ has degree $+ 1$ and the formula in [F1] is the chain-homotopy identity; the chain-homotopy theorem gives equality on $H_{- k} = H^{k}$ . The well-defined maps on those classes are exactly the induced de Rham maps.

F1F2F3step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham cohomology is smooth homotopy invariant

Statement

A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

Facts & Assumptions

Given: Smooth maps $f : M \to N$ , $g : N \to M$ with smooth homotopies $g f ≃ {id}_{M}$ and $f g ≃ {id}_{N}$ .

[F1]

Smoothly homotopic maps induce the same de rham map: Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.

[F2]

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

[F3]

Pullback is a homomorphism of de rham cohomology algebras: Smooth pullback induces a unital graded real algebra homomorphism $H_{d R}^{∙} (N) \to H_{d R}^{∙} (M)$ .

Proof

technique · direct

1.1

By smooth-homotopy invariance and functoriality, $f^{*} g^{*} = (g f)^{*} = ({id}_{M})^{*} = {id}_{H^{∙} (M)}$ .

F1F2given

2.1

The other homotopy gives $g^{*} f^{*} = (f g)^{*} = {id}_{H^{∙} (N)}$ . Both maps preserve multiplication and units, so these two identities exhibit inverse graded algebra homomorphisms. The conclusion includes empty manifolds, since a homotopy equivalence to an empty manifold forces both to be empty.

F1F2F3step 1.1given∎

Source locator

Lee, Theorem 17.11, pp.445–446; this item assumes smooth homotopies explicitly, so does not use a continuous smoothing theorem.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham cohomology is continuous homotopy invariant on smooth manifolds

Statement

Assume countable choice ${A C}_{ω}$ . Continuously homotopic smooth maps induce equal de Rham maps. Continuous homotopy equivalences between smooth manifolds induce inverse de Rham graded algebra maps via smooth representatives, independently of those representatives.

Facts & Assumptions

Given: Boundaryless smooth manifolds and countable choice. First let $F : P \to N$ be continuous and smooth on an open $V$ containing a closed set $A$ .

[F1]

Smoothly homotopic maps induce the same de rham map: Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.

[F2]

De rham cohomology is smooth homotopy invariant: A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

[F3]

The standard smooth step function: Let $β$ be the standard flat function. The standard smooth step function is $σ (t) : = \frac{β (t)}{β (t) + β (1 - t)} .$ Because $β$ is smooth and positive on $(0, \infty)$ , the denominator is positive on $(0, 1)$ , while $σ (t) = 0$ for $t \leq 0$ and $σ (t) = 1$ for $t \geq 1$ .

[F4]

Precomposition and postcomposition by continuous maps preserve homotopies, including their relative form: Let $f, g : X \to Y$ be continuous and suppose $f ≃_{A} g$ for a subspace $A \subseteq X$ . 1. If $u : W \to X$ is continuous, $B \subseteq W$ , and $u [B] \subseteq A$ , then $f \circ u ≃_{B} g \circ u$ . 2. If $v : Y \to Z$ is continuous, then $v \circ f ≃_{A} v \circ g$ .

[F5]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): The Axiom of Countable Choice, written ${A C}_{ω}$ , is the following statement. > For every family $(X_{n})_{n \in N}$ of nonempty sets indexed by > $N$ there is a function $f$ with domain $N$ such that > $f (n) \in X_{n}$ for every $n \in N$ . Equivalently, in the vocabulary of def-choice-function: every at most countable family of nonempty sets (def-countable) has a choice function.

[F6]

Products of smooth manifolds have a canonical product smooth structure: Let $(M, S)$ and $(N, T)$ be smooth manifolds of dimensions $m$ and $n$ . Then $M \times N$ with the product topology is a topological $(m + n)$ -manifold. If $A$ and $B$ are smooth atlases with $[A] = S$ and $[B] = T$ , then the set of product charts $A \times B : = {(V \times W, φ \times ψ) : (V, φ) \in A, (W, ψ) \in B}$ is a smooth atlas on $M \times N$ , and the maximal atlas it generates is independent of the presenting atlases: it depends only on $S$ and $T$ . This maximal atlas is the product smooth structure of $M \times N$ .

[F7]

Every smooth manifold embeds in some finite-dimensional Euclidean space: Assume countable choice ${A C}_{ω}$ . Every smooth manifold embeds smoothly in some finite-dimensional Euclidean space. More precisely, for a smooth $n$ -manifold $M$ there are a bounded smooth map $G : M \to R^{4 (2 n + 1)}$ and a smooth nonnegative proper exhaustion $ρ : M \to R$ such that $J : M ⟶ R^{4 (2 n + 1)} \times R, J (p) = (G (p), ρ (p))$ is a proper smooth embedding. In particular the bounded-plus-proper form is available when $M$ is noncompact.

[F8]

Whitney approximation for Euclidean-valued maps: Assume countable choice ${A C}_{ω}$ . Let $F : M \to R^{k}$ be continuous, where $M$ is a smooth manifold, and let $ε : M \to (0, \infty)$ be a positive continuous error function. Then there exists a smooth map $\tilde{F} : M \to R^{k}$ such that $∥ \tilde{F} (p) - F (p) ∥ < ε (p) for all p \in M .$

[F9]

Smooth partitions of unity exist on manifolds: Every open cover of a smooth manifold admits a smooth partition of unity subordinate to it.

[F10]

Smooth partitions of unity subordinate to an open cover: Let $M$ be a smooth manifold and let $(U_{i})_{i \in I}$ be an open cover of $M$ . A family of smooth functions $(ϕ_{i})_{i \in I}$ with $ϕ_{i} : M \to [0, 1]$ is a smooth partition of unity subordinate to $(U_{i})_{i \in I}$ when: 1. the family $(supp (ϕ_{i}))_{i \in I}$ is locally finite; 2. $supp (ϕ_{i}) \subseteq U_{i}$ for every $i \in I$ ; and 3. $\sum_{i} ϕ_{i} (p) = 1$ for every $p \in M$ .

[F11]

The normal addition map for a Euclidean submanifold: Let $S \subseteq R^{m}$ be an embedded smooth submanifold. Using the Euclidean inner product, define its orthogonal normal bundle by $N^{⊥} S : = {(p, v) \in S \times R^{m} : v ⊥ T_{p} S} .$ Local slice charts and orthogonal projection onto $T_{p} S$ give this set its standard smooth rank- $(m - \dim S)$ vector-bundle structure. The normal addition map is $E : N^{⊥} S \to R^{m}, E (p, v) : = p + v .$ It restricts on the zero section to the inclusion $S ↪ R^{m}$ and is the basic model map used to build Euclidean tubular neighbourhoods.

[F12]

Normal addition is a local diffeomorphism along the zero section: Let $S \subseteq R^{m}$ be an embedded smooth submanifold, and let $E : N^{⊥} S \to R^{m}$ be its normal addition map. For every $p \in S$ the differential $d E_{(p, 0)} : T_{(p, 0)} (N^{⊥} S) \to T_{p} R^{m} ≅ R^{m}$ is an isomorphism. Consequently, $E$ is a local diffeomorphism at every point of the zero section.

Proof

technique · direct

1.1

If $P$ is empty the smoothing assertion is immediate; if $N$ is empty a map forces $P$ empty. Otherwise the repaired embedding theorem gives a proper smooth embedding $j : N \to R^{m}$ under the assumed countable choice. Put $S = j (N)$ and let $E (p, v) = p + v$ on its orthogonal normal bundle. The bundle has the subspace topology in $S \times R^{m}$ , and $E$ is a local diffeomorphism at every $(p, 0)$ .

F5F7F11F12given

1.2

Apply the partition theorem to ${V, P ∖ A}$ . In its refinement-indexed output assign a function to $V$ whenever its support is contained in $V$ , and to $P ∖ A$ otherwise. Let $χ$ be the sum of the first group. Locally finite smooth sums give $0 \leq χ \leq 1$ and smoothness. Each grouped union of closed supports is closed, since locally it is a finite union; the first is contained in $V$ , and the second misses $A$ . Therefore $supp χ \subseteq V$ , and $χ = 1$ on the open complement of the second union, a neighbourhood of $A$ .

F9F10given

2.1

For $p \in S$ put $V_{a} (p) = {(q, v) : ∣ q - p ∣ < a, ∣ v ∣ < a}$ and let $h (p)$ be the supremum of the radii $0 < a \leq 1$ for which $E ∣_{V_{a} (p)}$ is a diffeomorphism onto its image. Local invertibility and the subspace topology provide at least one such radius, so $0 < h (p) \leq 1$ . Eligibility is downward closed, and every $a < h (p)$ is smaller than an eligible radius. Each point or pair of points in $V_{h (p)} (p)$ therefore lies in a smaller eligible set: $E$ is locally invertible and injective on all of $V_{h (p)} (p)$ .

step 1.1construct

2.2

The countable-choice cost in this partition application can be implemented in its coordinate-ball construction as follows. Form all admissible chart tuples, and for each member of a fixed countable basis contained in an eligible chart use countable choice to select a tuple; these charts cover. Take least suitable integer indices in the exhaustion refinement. For each compact annulus, the set of finite ordered lists of nested chart pairs covering it is nonempty by compactness; countable choice selects those lists and then the countable family of bump functions. Their normalized locally finite sum is the partition used above. This uses no point-indexed choice and no dependent choice.

F5F9step 1.2

3.1

If $0 < a < h (p) - ∣ p - q ∣$ , the triangle inequality gives $V_{a} (q) \subseteq V_{h (p)} (p)$ . The restriction of an injective local diffeomorphism to this open set is a diffeomorphism onto its open image, so $h (q) \geq a$ . Taking suprema, and exchanging $p, q$ , proves $∣ h (p) - h (q) ∣ \leq ∣ p - q ∣$ ; when the lower bound is nonpositive, positivity suffices. Thus $h$ is continuous.

step 2.1algebra

4.1

The set $Ω = {(p, v) : ∣ v ∣ < h (p) / 2}$ is open, and $E$ is locally invertible there. If $E (p, v) = E (q, w)$ , relabel so $h (q) \leq h (p)$ . Then $∣ p - q ∣ \leq ∣ v ∣ + ∣ w ∣ < (h (p) + h (q)) / 2 \leq h (p)$ , and both pairs lie in $V_{h (p)} (p)$ . Its injectivity gives $(p, v) = (q, w)$ . Hence $E : Ω \to U = E (Ω)$ is an injective open local diffeomorphism; its local smooth inverses agree. The map $r = π E^{- 1} : U \to S$ is smooth and satisfies $r (p) = p$ .

F11step 2.1step 3.1

5.1

If $U^{c}$ is nonempty, set $e (x) = \min (1, dist (j F (x), U^{c}) / 2)$ , and otherwise set $e = 1$ . For a nonempty set $B \subseteq R^{m}$ , triangle inequalities and infima give $∣ dist (z, B) - dist (z', B) ∣ \leq ∣ z - z' ∣$ ; distances are finite by fixing one point of $B$ . Openness of $U$ makes this distance positive at $j F (x)$ . Thus $e$ is finite, positive and continuous, and $B_{e (x)} (j F (x)) \subseteq U$ . Euclidean Whitney approximation supplies smooth $Q : P \to R^{m}$ with $∣ Q - j F ∣ < e$ .

F5F8step 4.1

6.1

Put $R = χ j F + (1 - χ) Q$ . It is smooth on $V$ , and outside $supp χ$ it agrees locally with $Q$ , hence is globally smooth. It agrees with $j F$ near $A$ , and $∣ R - j F ∣ = (1 - χ) ∣ Q - j F ∣ < e$ . Consequently $G = j^{- 1} r R$ is smooth, agrees with $F$ near $A$ , and $j^{- 1} r ((1 - s) j F + s R)$ is a continuous homotopy from $F$ to $G$ fixed on $A$ : its whole segment stays in $B_{e} (j F)$ .

step 4.1step 5.1step 1.2

7.1

Now let $H : M \times [0, 1] \to N$ continuously join smooth maps $f_{0}, f_{1}$ . On the smooth product $P = M \times R$ , put $λ (t) = σ (3 t - 1)$ and $F (x, t) = H (x, λ (t))$ . It is smooth on $V = M \times ((- \infty, 1 / 3) \cup (2 / 3, \infty))$ , where it equals an endpoint map. The closed set $A = M \times ((- \infty, 1 / 4] \cup [3 / 4, \infty))$ lies in $V$ . Step 6.1 gives a smooth map fixed near these collars. Its restriction to $M \times [0, 1]$ is a smooth homotopy with exact endpoints $f_{0}, f_{1}$ . Thus their induced maps agree.

F1F3F6step 6.1

8.1

Taking $A = V = \emptyset$ and $χ = 0$ in step 6.1 gives a smooth representative of every continuous map, with an explicit homotopy to it. If continuous $f, g$ are homotopy inverses, choose smooth representatives $f', g'$ . Composition of homotopies and their concatenation give $g' f' ≃ id$ and $f' g' ≃ id$ . Step 7.1 smooths each of these endpoint homotopies; smooth homotopy invariance now makes $(f')^{*}, (g')^{*}$ inverse graded algebra maps. Two smooth representatives of one continuous class are continuously homotopic by concatenation, so step 7.1 also proves independence.

F2F4step 6.1step 7.1∎

Source locator

Lee, Theorem 6.21, pp.136–137; normal addition and tubular retraction, pp.137–141; Theorem 6.26, p.141; Theorem 17.11, pp.445–446. The bounded supremum tube proof and smaller closed collars are the explicit local construction here, supported by the declared normal-addition local inverse. The current published embedding and absolute approximation proofs, repaired 2026-09-09, were read in full; no historical relative-Whitney assertion is imported.

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Radial contraction of a star shaped domain

Definition

For an open $U \subseteq R^{n}$ star-shaped about a specified $c \in U$ , the radial contraction is $F : U \times [0, 1] \to U$ , $F (x, t) = c + t (x - c)$ .

Star-shaped open subsets of Euclidean space says exactly that each displayed value lies in $U$ . The coordinate expression is polynomial, so its restriction is smooth up to both endpoints; $F (x, 0) = c$ and $F (x, 1) = x$ . The centre is part of the data, so $U$ is nonempty. For $n = 0$ the unique nonempty domain is a point and the formula is constant.

Source locator

Lee, Theorem 17.14, p.447.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

Poincare lemma for differential forms on star shaped domains

Statement

Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

Facts & Assumptions

Given: A domain $U$ star-shaped about $c$ , and a closed $k$ -form $ω$ with $k \geq 1$ .

[F1]

Radial contraction of a star shaped domain: For an open $U \subseteq R^{n}$ star-shaped about a specified $c \in U$ , the radial contraction is $F : U \times [0, 1] \to U$ , $F (x, t) = c + t (x - c)$ . def-star-shaped-open-subset-of-rn says exactly that each displayed value lies in $U$ . The coordinate expression is polynomial, so its restriction is smooth up to both endpoints; $F (x, 0) = c$ and $F (x, 1) = x$ . The centre is part of the data, so $U$ is nonempty. For $n = 0$ the unique nonempty domain is a point and the formula is constant.

[F2]

De rham homotopy formula for a smooth homotopy: If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

Proof

technique · direct

1.1

Take the radial homotopy from the constant map to the identity. Its time-zero pullback on positive-degree forms vanishes because the differential of the constant map is zero. The homotopy formula and $d ω = 0$ give $ω = d (K F^{*} ω)$ , so $η = K F^{*} ω$ is a smooth primitive.

F1F2given

2.1

For $c = 0$ , $d F_{(x, t)} \partial_{t} = x$ and $d F_{(x, t)} (v, 0) = t v$ . The contraction coefficient therefore equals $t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1})$ . Integrating gives the displayed formula. When $k = 1$ the factor is $1$ , including $t = 0$ ; for $k > 1$ the integrand is smooth and vanishes there. Degrees exceeding the dimension have zero form and zero primitive.

F2step 1.1∎

Source locator

Lee, Theorem 17.14, p.447; the explicit primitive follows by evaluating the interval operator.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

Closed differential forms are locally exact

Statement

Every closed smooth differential form of positive degree is locally exact.

Facts & Assumptions

Given: A closed $k$ -form $ω$ on a smooth manifold, $k \geq 1$ , and a point $p$ .

[F1]

Poincare lemma for differential forms on star shaped domains: Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

[F2]

Pullback of forms is smooth functorial and preserves wedges: For a smooth map $F : M \to N$ , pullback sends smooth differential forms on $N$ to smooth differential forms on $M$ , is functorial, and satisfies $F^{*} (α \land β) = F^{*} α \land F^{*} β .$

[F3]

The exterior derivative commutes with pullback: For every smooth map $F : M \to N$ and every form $ω$ on $N$ , $d (F^{*} ω) = F^{*} (d ω) .$

Proof

technique · direct

1.1

Choose a coordinate neighbourhood $W$ of $p$ mapped diffeomorphically by $x$ onto a Euclidean open ball. In those coordinates $\tilde{ω} = (x^{- 1})^{*} (ω ∣_{W})$ is closed: naturality gives $d \tilde{ω} = (x^{- 1})^{*} d (ω ∣_{W}) = 0$ . The ball is star-shaped, so there is a smooth $(k - 1)$ -form $η$ there with $d η = \tilde{ω}$ .

F1F3given

2.1

Pulling back by $x$ , naturality of the exterior derivative gives $d (x^{*} η) = x^{*} d η = x^{*} (x^{- 1})^{*} (ω ∣_{W}) = ω ∣_{W}$ . Thus $x^{*} η$ is the requested local primitive. If the form is forced to be zero by dimension, the zero primitive works; on an empty manifold the assertion over all points is vacuous.

F2F3step 1.1∎

Source locator

Lee, Corollary 17.15, p.447, restricted explicitly to positive degree.

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De rham cohomology of a contractible smooth manifold

Statement

Assume countable choice. A nonempty contractible smooth manifold $M$ has $H_{d R}^{0} (M) ≅ R$ and $H_{d R}^{k} (M) = 0$ for every $k > 0$ .

Facts & Assumptions

Given: A nonempty contractible smooth manifold $M$ and countable choice.

[F1]

De rham cohomology is continuous homotopy invariant on smooth manifolds: Assume countable choice ${A C}_{ω}$ . Continuously homotopic smooth maps induce equal de Rham maps. Continuous homotopy equivalences between smooth manifolds induce inverse de Rham graded algebra maps via smooth representatives, independently of those representatives.

[F2]

Zero and out of range de rham cohomology: $H_{d R}^{k} (M) = 0$ if $k < 0$ or $k > \dim M$ . If $M = \emptyset$ , its cohomology vanishes in every degree.

[F3]

Zero th de rham cohomology is locally constant functions: $H_{d R}^{0} (M)$ is the algebra of locally constant real functions. For nonempty connected $M$ it is canonically $R$ .

[F4]

Nullhomotopic maps and contractible spaces: Let $f : X \to Y$ be continuous. The map $f$ is nullhomotopic if there is a point $y_{0} \in Y$ such that $f$ is homotopic to the constant map $c_{y_{0}} : X \to Y$ , $c_{y_{0}} (x) = y_{0}$ (def-homotopy-relative-and-path-homotopy). A nonempty topological space $X$ is contractible if every continuous map $f : X \to Y$ to every topological space $Y$ is nullhomotopic. This definition separates the property of the space from the particular map ${id}_{X}$ . The next corollary proves that it is equivalent to the familiar condition that the identity map be nullhomotopic.

Proof

technique · direct

1.1

Apply contractibility to the identity map to obtain a continuous homotopy from ${id}_{M}$ to a constant $c_{p}$ for some $p \in M$ . Let $a : M \to {p}$ and $b : {p} \to M$ be the unique map and inclusion. Then $a b = {id}_{{p}}$ and $b a = c_{p} ≃ {id}_{M}$ , so these are continuous homotopy inverses.

F4given

2.1

Continuous homotopy invariance identifies the de Rham groups with those of the point. Its zero-degree functions are precisely $R$ , and all positive-degree form spaces vanish by dimension. Hence its positive-degree cohomology vanishes and its degree-zero cohomology is $R$ , giving the asserted groups of $M$ .

F1F2F3step 1.1∎

Source locator

Lee, Theorem 17.13, pp.446–447; continuous smoothing is supplied by the preceding fully local corollary.

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Two open set de rham mayer vietoris cochain maps

Definition

For an open cover $M = U \cup V$ , put $W = U \cap V$ . The two-open-set de Rham maps are $r : Ω^{∙} (M) \to Ω^{∙} (U) \oplus Ω^{∙} (V)$ , $r ω = (ω ∣_{U}, ω ∣_{V})$ , and $s : Ω^{∙} (U) \oplus Ω^{∙} (V) \to Ω^{∙} (W)$ , $s (α, β) = β ∣_{W} - α ∣_{W}$ .

The complexes are De rham cochain complex. Restrictions are pullbacks along open inclusions, so Pullback is a morphism of de rham complexes gives $d r = r d$ and $d s = s d$ . Both maps are real linear. The middle differential acts componentwise. Empty opens have zero form spaces. The order second minus first fixes the sign of every connecting map below.

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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The de rham mayer vietoris sequence is exact at the first two terms

Statement

The sequence $0 \to Ω^{k} (M) \overset{r}{\to} Ω^{k} (U) \oplus Ω^{k} (V) \overset{s}{\to} Ω^{k} (U \cap V)$ is exact at the first two nonzero terms.

Facts & Assumptions

Given: An open cover $M = U \cup V$ and the maps $r, s$ just defined.

[F1]

Two open set de rham mayer vietoris cochain maps: For an open cover $M = U \cup V$ , put $W = U \cap V$ . The two-open-set de Rham maps are $r : Ω^{∙} (M) \to Ω^{∙} (U) \oplus Ω^{∙} (V)$ , $r ω = (ω ∣_{U}, ω ∣_{V})$ , and $s : Ω^{∙} (U) \oplus Ω^{∙} (V) \to Ω^{∙} (W)$ , $s (α, β) = β ∣_{W} - α ∣_{W}$ . The complexes are def-de-rham-cochain-complex. Restrictions are pullbacks along open inclusions, so prop-pullback-is-a-morphism-of-de-rham-complexes gives $d r = r d$ and $d s = s d$ . Both maps are real linear. The middle differential acts componentwise. Empty opens have zero form spaces. The order second minus first fixes the sign of every connecting map below.

[F2]

A smooth differential $k$ -form: Let $M$ be a smooth manifold and $k \geq 0$ . A smooth differential $k$ -form on $M$ is a smooth section of $⋀^{k} T^{*} M \to M$ . The space of such forms is denoted $Ω^{k} (M)$ , and $Ω^{0} (M) = C^{\infty} (M)$ .

Proof

technique · direct

1.1

If $r ω = 0$ , the form vanishes at each point because every point lies in $U$ or $V$ . Thus $r$ is injective. Also $s r ω = ω ∣_{U \cap V} - ω ∣_{U \cap V} = 0$ , so $im r \subseteq \ker s$ .

F1given

2.1

If $s (α, β) = 0$ , the two forms agree on the overlap. Define $ω_{p} = α_{p}$ for $p \in U$ and $ω_{p} = β_{p}$ for $p \in V$ . Agreement makes this unambiguous, and near every point it equals a smooth section, so it is smooth. Then $r ω = (α, β)$ , proving the reverse inclusion and exactness. The same definitions work for empty opens, including empty $M$ .

F1F2step 1.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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The de rham mayer vietoris difference map is surjective

Statement

Assume countable choice. The difference map $s : Ω^{k} (U) \oplus Ω^{k} (V) \to Ω^{k} (U \cap V)$ is surjective in every degree.

Facts & Assumptions

Given: An open cover $M = U \cup V$ , a smooth form $ω$ on $U \cap V$ , and countable choice.

[F1]

Two open set de rham mayer vietoris cochain maps: For an open cover $M = U \cup V$ , put $W = U \cap V$ . The two-open-set de Rham maps are $r : Ω^{∙} (M) \to Ω^{∙} (U) \oplus Ω^{∙} (V)$ , $r ω = (ω ∣_{U}, ω ∣_{V})$ , and $s : Ω^{∙} (U) \oplus Ω^{∙} (V) \to Ω^{∙} (W)$ , $s (α, β) = β ∣_{W} - α ∣_{W}$ . The complexes are def-de-rham-cochain-complex. Restrictions are pullbacks along open inclusions, so prop-pullback-is-a-morphism-of-de-rham-complexes gives $d r = r d$ and $d s = s d$ . Both maps are real linear. The middle differential acts componentwise. Empty opens have zero form spaces. The order second minus first fixes the sign of every connecting map below.

[F2]

Smooth partitions of unity exist on manifolds: Every open cover of a smooth manifold admits a smooth partition of unity subordinate to it.

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): The Axiom of Countable Choice, written ${A C}_{ω}$ , is the following statement. > For every family $(X_{n})_{n \in N}$ of nonempty sets indexed by > $N$ there is a function $f$ with domain $N$ such that > $f (n) \in X_{n}$ for every $n \in N$ . Equivalently, in the vocabulary of def-choice-function: every at most countable family of nonempty sets (def-countable) has a choice function.

Proof

technique · direct

1.1

The partition construction supplies a locally finite family $(ϕ_{i})$ with nonnegative smooth terms summing to one and each closed support contained in $U$ or $V$ . Its countable-choice implementation uses all admissible coordinate-ball tuples; a fixed countable basis and countable choice select covering tuple representatives. For unions $H_{r}$ of their first $r$ compact closures, take least larger indices giving $H_{r} \subseteq int H_{r'}$ . For the resulting exhaustion $K_{m}$ , the compact annulus $K_{m} ∖ int K_{m - 1}$ has a finite covering list of nested chart pairs inside selected balls and inside $int K_{m + 1} ∖ K_{m - 2}$ . Countable choice selects these finite lists and their countably many bumps. The annulus separation makes their supports locally finite, and division by their positive smooth sum gives $(ϕ_{i})$ . All eligible tuples are formed before selection; least-index recursion uses no dependent choice.

F2F3given

2.1

Assign $i$ to $U$ if $supp ϕ_{i} \subseteq U$ , and to $V$ otherwise. Set $ρ_{U} = \sum_{i assigned to U} ϕ_{i}$ and $ρ_{V} = \sum_{i assigned to V} ϕ_{i}$ . Locally these are finite smooth sums and they sum to one. Each grouped union of closed supports is closed by local finiteness and lies in its assigned open, so $supp ρ_{U} \subseteq U$ and $supp ρ_{V} \subseteq V$ .

step 1.1

3.1

Define $α = - ρ_{V} ω$ on $U \cap V$ and zero on $U ∖ supp ρ_{V}$ . These two open sets cover $U$ , and the expressions agree where they overlap; hence $α$ is a smooth form on $U$ . Similarly $β = ρ_{U} ω$ on the overlap and zero on $V ∖ supp ρ_{U}$ is smooth on $V$ . On the overlap, $β - α = (ρ_{U} + ρ_{V}) ω = ω$ , proving surjectivity. Empty overlap or out-of-range degree has only $ω = 0$ , lifted by $(0, 0)$ ; empty $M$ uses the empty family.

F1step 2.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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Short exact mayer vietoris sequence of de rham complexes

Statement

Under countable choice, $0 \to Ω^{∙} (M) \overset{r}{\to} Ω^{∙} (U) \oplus Ω^{∙} (V) \overset{s}{\to} Ω^{∙} (U \cap V) \to 0$ is short exact as a sequence of real cochain complexes.

Facts & Assumptions

Given: An open cover $M = U \cup V$ and countable choice.

[F1]

The de rham mayer vietoris sequence is exact at the first two terms: The sequence $0 \to Ω^{k} (M) \overset{r}{\to} Ω^{k} (U) \oplus Ω^{k} (V) \overset{s}{\to} Ω^{k} (U \cap V)$ is exact at the first two nonzero terms.

[F2]

The de rham mayer vietoris difference map is surjective: Assume countable choice. The difference map $s : Ω^{k} (U) \oplus Ω^{k} (V) \to Ω^{k} (U \cap V)$ is surjective in every degree.

[F3]

Short exact sequence of complexes: A short exact sequence of complexes is a sequence of chain maps $0 \to A_{∙} \to B_{∙} \to C_{∙} \to 0$ that is exact in each degree as a sequence in the ambient abelian category.

Proof

technique · direct

1.1

In every degree, the first lemma gives injectivity of $r$ and $\ker s = im r$ , and the second gives surjectivity of $s$ . Hence each degree is a short exact sequence of real vector spaces.

F1F2given

2.1

The maps $r, s$ commute with the differentials because they are the restriction cochain maps of those lemmas. Reindexing by $C_{n} = C^{- n}$ turns this into a sequence of chain maps exact in every degree, precisely the definition of a short exact sequence of complexes. Thus it is the claimed cochain version, including zero terms.

F3step 1.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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Mayer vietoris sequence in de rham cohomology

Statement

Under countable choice the de Rham Mayer–Vietoris sequence is exact: $\dots \to H^{k} (M) \overset{r^{*}}{\to} H^{k} (U) \oplus H^{k} (V) \overset{s^{*}}{\to} H^{k} (U \cap V) \overset{δ}{\to} H^{k + 1} (M) \to \dots$ , beginning with $0 \to H^{0} (M)$ .

Facts & Assumptions

Given: An open cover $M = U \cup V$ and countable choice.

[F1]

Short exact mayer vietoris sequence of de rham complexes: Under countable choice, $0 \to Ω^{∙} (M) \overset{r}{\to} Ω^{∙} (U) \oplus Ω^{∙} (V) \overset{s}{\to} Ω^{∙} (U \cap V) \to 0$ is short exact as a sequence of real cochain complexes.

[F2]

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.

[F3]

The long exact sequence in cohomology: Let $0 \to A^{∙} \to B^{∙} \to C^{∙} \to 0$ be a short exact sequence of cochain complexes in an abelian category. Then there is a natural exact sequence $\dots \to H^{n} (A) \to H^{n} (B) \to H^{n} (C) \overset{\partial^{n}}{\to} H^{n + 1} (A) \to H^{n + 1} (B) \to H^{n + 1} (C) \to \dots .$

Proof

technique · direct

1.1

The short exact sequence of de Rham cochain complexes satisfies the hypotheses of the cohomology long exact sequence theorem in the abelian category of real vector spaces. It gives the connecting map from overlap degree $k$ to global degree $k + 1$ , with no additional differential sign.

F1F3given

2.1

For the middle complex, $d (α, β) = (d α, d β)$ , so its cycle space is $Z^{k} (U) \oplus Z^{k} (V)$ and its boundary space is $B^{k} (U) \oplus B^{k} (V)$ . The quotient map sends $[(α, β)]$ to $([α], [β])$ , bijectively: a pair maps to zero exactly when both entries have primitives. Thus its cohomology is the displayed direct sum. All negative-degree cohomology is zero by the definition of forms, giving the stated initial zero.

F2step 1.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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Explicit de rham mayer vietoris connecting class

Statement

Under countable choice, let $δ$ be the Mayer–Vietoris connecting homomorphism obtained from the short exact cochain sequence in F3 by reindexing $C_{n} = C^{- n}$ and using the connector convention in F4. For a closed $k$ -form $ω$ on $U \cap V$ , $δ [ω] = [η]$ , where $η ∣_{U} = d (- ρ_{V} ω)$ and $η ∣_{V} = d (ρ_{U} ω)$ , with products smoothly extended by zero as in the lift construction. This class is independent of partition, lift and representative.

Facts & Assumptions

Given: Assume countable choice. A closed overlap form $ω$ and the partition lift $(α, β) = (- ρ_{V} ω, ρ_{U} ω)$ .

[F1]

Mayer vietoris sequence in de rham cohomology: Under countable choice the de Rham Mayer–Vietoris sequence is exact: $\dots \to H^{k} (M) \overset{r^{*}}{\to} H^{k} (U) \oplus H^{k} (V) \overset{s^{*}}{\to} H^{k} (U \cap V) \overset{δ}{\to} H^{k + 1} (M) \to \dots$ , beginning with $0 \to H^{0} (M)$ . Its proof obtains this $δ$ by applying the long exact sequence theorem to the short exact de Rham cochain sequence in F3, with no additional differential sign.

[F2]

The de rham mayer vietoris difference map is surjective: Assume countable choice. The difference map $s : Ω^{k} (U) \oplus Ω^{k} (V) \to Ω^{k} (U \cap V)$ is surjective in every degree.

[F3]

Short exact mayer vietoris sequence of de rham complexes: Under countable choice, $0 \to Ω^{∙} (M) \overset{r}{\to} Ω^{∙} (U) \oplus Ω^{∙} (V) \overset{s}{\to} Ω^{∙} (U \cap V) \to 0$ is short exact as a sequence of real cochain complexes, with $s (α, β) = β - α$ .

[F4]

Elementwise formula for the connecting map in module categories: Let $R$ be a ring and let $0 \to A_{∙} \overset{i}{\to} B_{∙} \overset{p}{\to} C_{∙} \to 0$ be a short exact sequence of chain complexes of left $R$ -modules. If $[c] \in H_{n} (C)$ is represented by a cycle $c \in C_{n}$ , choose a lift $b \in B_{n}$ with $p_{n} (b) = c$ , and let $a \in A_{n - 1}$ be the unique element satisfying $i_{n - 1} (a) = d_{n}^{B} (b) .$ Then $\partial_{n} ([c]) = [a] \in H_{n - 1} (A) .$ This class is independent of the chosen lift $b$ and of the chosen cycle representative $c$ .

Proof

technique · direct

1.1

On the overlap, $d β - d α = d (β - α) = d ω = 0$ . Thus the two smooth forms $d α, d β$ glue to $η$ . On each open $d η = d^{2} α$ or $d^{2} β$ , hence is zero. The graded product rule also gives $η ∣_{U \cap V} = d ρ_{U} \land ω = - d ρ_{V} \land ω$ , fixing the sign.

F2F3given

2.1

Reindex the short exact sequence F3 as chain complexes with $C_{n} = C^{- n}$ . By F1 its connecting homomorphism is the displayed $δ$ , with no added sign. The lift $(α, β)$ in degree $- k$ has differential $r η$ in degree $- k - 1$ , so F4 gives $δ [ω] = [η]$ in $H^{k + 1} (M)$ . Its independence of lift and cycle representative applies over the ring $R$ ; any other partition supplies another lift, so partition independence follows too.

F1F3F4step 1.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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Naturality of de rham mayer vietoris for maps of covered manifolds

Statement

Assume countable choice. For smooth $F : M \to N$ with $F (U) \subseteq U'$ and $F (V) \subseteq V'$ , pullback gives a contravariant commutative ladder of the two Mayer–Vietoris sequences; in particular $δ_{M} F_{U \cap V}^{*} = F^{*} δ_{N}$ .

Facts & Assumptions

Given: Assume countable choice. Open covers $M = U \cup V$ , $N = U' \cup V'$ and the stated smooth covered map $F$ .

[F1]

Mayer vietoris sequence in de rham cohomology: Under countable choice the de Rham Mayer–Vietoris sequence is exact: $\dots \to H^{k} (M) \overset{r^{*}}{\to} H^{k} (U) \oplus H^{k} (V) \overset{s^{*}}{\to} H^{k} (U \cap V) \overset{δ}{\to} H^{k + 1} (M) \to \dots$ , beginning with $0 \to H^{0} (M)$ .

[F2]

Explicit de rham mayer vietoris connecting class: Under countable choice, for a closed $k$ -form $ω$ on $U \cap V$ , $δ [ω] = [η]$ , where $η ∣_{U} = d (- ρ_{V} ω)$ and $η ∣_{V} = d (ρ_{U} ω)$ , with products smoothly extended by zero as in the lift construction. This class is independent of partition, lift and representative.

[F3]

Pullback of forms is smooth functorial and preserves wedges: For a smooth map $F : M \to N$ , pullback sends smooth differential forms on $N$ to smooth differential forms on $M$ , is functorial, and satisfies $F^{*} (α \land β) = F^{*} α \land F^{*} β .$

[F4]

Naturality of the homology connecting morphism: A morphism of short exact sequences of complexes induces a commutative square $\begin{array}{ccc} H_{n} (C) & \overset{\partial_{n}}{\to} & H_{n - 1} (A) \\ ↓ & ↓ \\ H_{n} (C^{'}) & \overset{\partial_{n}^{'}}{\to} & H_{n - 1} (A^{'}) \end{array}$ for every $n \in Z$ .

[F5]

Pullback is a morphism of de rham complexes: A smooth map $F : M \to N$ induces a degree-zero real cochain map $F^{*} : Ω^{∙} (N) \to Ω^{∙} (M)$ .

Proof

technique · direct

1.1

Restriction of a pullback is pullback by the restricted map. Therefore $r_{M} F^{*} = (F_{U}^{*} \oplus F_{V}^{*}) r_{N}$ and $s_{M} (F_{U}^{*} \oplus F_{V}^{*}) = F_{U \cap V}^{*} s_{N}$ , with the second identity using $β - α$ on both sides. These are cochain squares, since all restrictions and pullbacks commute with $d$ .

F1F3F5given

2.1

These squares are a morphism from the short exact sequence for $N$ to that for $M$ . Reindexing by degree negation and applying naturality of the homology connector gives the displayed connecting square in degree $k$ to $k + 1$ . Concretely, a lift $(α', β')$ of a closed overlap form on $N$ pulls back to a lift on $M$ , and its glued derivative is $F^{*} η'$ ; the connector formula gives exactly the same sign. Partition preservation is unnecessary, since the class is lift independent.

F2F4step 1.1∎

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: $s (α, β) = β - α$ .

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De rham cohomology of a finite disjoint union is the direct sum

Statement

For a finite disjoint union $M = ∐_{j = 1}^{m} M_{j}$ , restrictions give $H_{d R}^{k} (M) ≅ ⨁_{j = 1}^{m} H_{d R}^{k} (M_{j})$ .

Facts & Assumptions

Given: A finite family of smooth manifolds and any integer $k$ .

[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.

Proof

technique · direct

1.1

A form on $M$ is uniquely a tuple of forms on the open components $M_{j}$ : define its value componentwise, which is smooth locally. Its derivative is componentwise too, so $Z^{k} (M) = ⨁_{j} Z^{k} (M_{j})$ . A tuple of exact forms has a tuple of primitives, obtained by finite choice, so $B^{k} (M) = ⨁_{j} B^{k} (M_{j})$ .

F1given

2.1

The resulting map on quotient classes is onto, since a finite tuple of classes has a finite tuple of closed representatives. Its kernel consists precisely of tuples with all entries exact, which step 1.1 identifies with $B^{k} (M)$ . Thus it is an isomorphism. For $m = 0$ both sides are zero, and for $m = 1$ it is the identity.

F1step 1.1∎

Source locator

Lee, Proposition 17.5, pp.442–443; the local statement is finite only, where products and sums coincide and all witness selection is finite.

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De rham cohomology of spheres

Statement

Assume countable choice. For $n \geq 1$ , $H_{d R}^{k} (S^{n})$ is $R$ in degrees $0, n$ and zero otherwise. For $S^{0}$ it is $R^{2}$ in degree zero and zero otherwise.

Facts & Assumptions

Given: The unit sphere $S^{n} \subset R^{n + 1}$ and countable choice.

[F1]

Mayer vietoris sequence in de rham cohomology: Under countable choice the de Rham Mayer–Vietoris sequence is exact: $\dots \to H^{k} (M) \overset{r^{*}}{\to} H^{k} (U) \oplus H^{k} (V) \overset{s^{*}}{\to} H^{k} (U \cap V) \overset{δ}{\to} H^{k + 1} (M) \to \dots$ , beginning with $0 \to H^{0} (M)$ .

[F2]

Poincare lemma for differential forms on star shaped domains: Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

[F3]

De rham cohomology is smooth homotopy invariant: A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

[F4]

Zero th de rham cohomology is locally constant functions: $H_{d R}^{0} (M)$ is the algebra of locally constant real functions. For nonempty connected $M$ it is canonically $R$ .

[F5]

De rham cohomology of a finite disjoint union is the direct sum: For a finite disjoint union $M = ∐_{j = 1}^{m} M_{j}$ , restrictions give $H_{d R}^{k} (M) ≅ ⨁_{j = 1}^{m} H_{d R}^{k} (M_{j})$ .

Proof

technique · direct

1.1

For $n \geq 1$ , write points as $(x', z)$ and remove the poles to form $U, V$ . Stereographic coordinates $x' / (1 - z)$ and $x' / (1 + z)$ identify these opens with $R^{n}$ ; the first inverse is $y \mapsto (2 y / (1 + ∣ y ∣^{2}), (∣ y ∣^{2} - 1) / (1 + ∣ y ∣^{2}))$ , and changing the sign of the last coordinate gives the second. Substitution verifies both inverses. Positive-degree cohomology of each open vanishes by Poincaré, and its degree-zero group is $R$ .

F2F4given

2.1

The overlap is diffeomorphic to $S^{n - 1} \times R$ by $(x', z) \mapsto (x' / ∣ x' ∣, z / ∣ x' ∣)$ , with inverse $(u, t) \mapsto (u / \sqrt{1 + t^{2}}, t / \sqrt{1 + t^{2}})$ . The homotopy $(u, t, s) \mapsto (u, (1 - s) t)$ retracts it smoothly onto $S^{n - 1}$ . For $n = 1$ the overlap is two contractible components, so its $H^{0}$ is $R^{2}$ and its positive groups vanish. The map on $H^{0}$ is $(a, b) \mapsto (b - a, b - a)$ ; its kernel is the diagonal and its cokernel is $R$ , via $(c, d) \mapsto d - c$ . Exactness therefore gives $H^{0} (S^{1}) = H^{1} (S^{1}) = R$ . In degrees $k \geq 2$ the form spaces on this one-manifold vanish, so its cohomology also vanishes.

F1F3F5step 1.1

3.1

For $n \geq 2$ the overlap is connected: $S^{n - 1}$ is path connected, since non-antipodal points join by normalized line segments and antipodal points join through one perpendicular unit vector. Thus the degree-zero difference map is the surjection $(a, b) \mapsto b - a$ onto $R$ . Exactness gives $H^{0} (S^{n}) = R$ and $H^{1} (S^{n}) = 0$ . For $k \geq 2$ both adjacent positive-degree groups of $U, V$ vanish, so exactness gives $H^{k} (S^{n}) ≅ H^{k - 1} (S^{n - 1})$ . Repeatedly applying this identity reaches the circle calculation or degree one, proving all asserted positive degrees. Negative degrees vanish by the complex convention. Finally $S^{0}$ is two points, each with only $Ω^{0} = R$ , and finite disjoint union gives its stated groups.

F1F4F5step 1.1step 2.1∎

Source locator

Lee, Theorem 17.21, pp.450–451. The local proof replaces Lee’s fundamental-group input by the explicit degree-zero Mayer–Vietoris maps and avoids any later punctured-space computation.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham cohomology of punctured euclidean space

Statement

Under countable choice, $R^{n} ∖ {0}$ has cohomology $R$ in degrees $0, n - 1$ only for $n \geq 2$ . For $n = 1$ it has $R^{2}$ in degree zero only, and for $n = 0$ all groups vanish.

Facts & Assumptions

Given: A nonnegative integer $n$ and countable choice.

[F1]

De rham cohomology of spheres: Assume countable choice. For $n \geq 1$ , $H_{d R}^{k} (S^{n})$ is $R$ in degrees $0, n$ and zero otherwise. For $S^{0}$ it is $R^{2}$ in degree zero and zero otherwise.

[F2]

De rham cohomology is smooth homotopy invariant: A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

[F3]

De rham cohomology of a finite disjoint union is the direct sum: For a finite disjoint union $M = ∐_{j = 1}^{m} M_{j}$ , restrictions give $H_{d R}^{k} (M) ≅ ⨁_{j = 1}^{m} H_{d R}^{k} (M_{j})$ .

[F4]

Poincare lemma for differential forms on star shaped domains: Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

Proof

technique · direct

1.1

For $n \geq 2$ , set $r (x) = x / ∣ x ∣$ and $i : S^{n - 1} ↪ R^{n} ∖ {0}$ . Then $r i = id$ and $F (x, t) = ((1 - t) + t / ∣ x ∣) x$ is a smooth homotopy from the identity to $i r$ . Its scalar coefficient is strictly positive for $0 \leq t \leq 1$ , so it never reaches zero. Smooth homotopy invariance and the sphere computation give the groups asserted.

F1F2given

2.1

For $n = 1$ , the two half-lines are star-shaped and connected, so each has only $H^{0} = R$ ; their finite disjoint union gives $R^{2}$ . For $n = 0$ the punctured space is empty, all its form spaces are zero, and every cohomology group is zero.

F3F4given∎

Source locator

Lee, Corollary 17.23, p.451, with the low-dimensional cases computed explicitly.

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The de rham cohomology class of a form is defined without closedness

Statement

False claim: every smooth differential form represents a de Rham cohomology class.

Facts & Assumptions

Given: The smooth one-form $α = x d y$ on $R^{2}$ .

[F1]

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^{k} (M) = \ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))$ and $B^{k} (M) = im (d : Ω^{k - 1} (M) \to Ω^{k} (M))$ . A form is closed if it belongs to $Z^{k}$ and exact if it belongs to $B^{k}$ . If $ω = d η$ , then $d ω = d^{2} η = 0$ by thm-the-exterior-derivative-squares-to-zero, so $B^{k} \subseteq Z^{k}$ . In particular $B^{0} = 0$ , since $Ω^{- 1} = 0$ . The zero form is both closed and exact in every degree.

[F2]

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.

[F3]

The local coordinate formula for the exterior derivative: Let $(U, x^{1}, \dots, x^{n})$ be a smooth chart on a smooth manifold and $ω$ a smooth $k$ -form on $U$ , with $k \geq 0$ . Summing over increasing $k$ -tuples $I$ , and writing $d x^{I} = d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , if $ω = \sum_{I} ω_{I} d x^{I}$ , then $d ω = \sum_{I} d ω_{I} \land d x^{I} .$

Refutation

technique · direct

1.1

The coordinate formula gives $d α = d x \land d y$ , whose value on $(\partial_{x}, \partial_{y})$ is $1$ . Thus $d α \neq 0$ .

F3given

2.1

A class in $H^{1}$ must be represented by an element of $Z^{1} = \ker d$ . The displayed smooth form is outside that numerator, so $[α]$ is not a de Rham class.

F1F2step 1.1∎

Source locator

Lee, p.441, definition of the cycle quotient; the witness is calculated locally.

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Every smooth form is either closed or exact

Statement

False claim: every smooth form is either closed or exact.

Facts & Assumptions

Given: The smooth form $α = x d y$ on $R^{2}$ .

[F1]

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^{k} (M) = \ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))$ and $B^{k} (M) = im (d : Ω^{k - 1} (M) \to Ω^{k} (M))$ . A form is closed if it belongs to $Z^{k}$ and exact if it belongs to $B^{k}$ . If $ω = d η$ , then $d ω = d^{2} η = 0$ by thm-the-exterior-derivative-squares-to-zero, so $B^{k} \subseteq Z^{k}$ . In particular $B^{0} = 0$ , since $Ω^{- 1} = 0$ . The zero form is both closed and exact in every degree.

[F2]

The local coordinate formula for the exterior derivative: Let $(U, x^{1}, \dots, x^{n})$ be a smooth chart on a smooth manifold and $ω$ a smooth $k$ -form on $U$ , with $k \geq 0$ . Summing over increasing $k$ -tuples $I$ , and writing $d x^{I} = d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , if $ω = \sum_{I} ω_{I} d x^{I}$ , then $d ω = \sum_{I} d ω_{I} \land d x^{I} .$

[F3]

The exterior derivative squares to zero: For every differential form $ω$ , $d (d ω) = 0$ .

Refutation

technique · direct

1.1

Its derivative is $d α = d x \land d y \neq 0$ , since evaluation on the coordinate basis gives $1$ . Hence $α$ is not closed.

F1F2given

2.1

If $α = d η$ were exact, then $d α = d^{2} η = 0$ , contradicting step 1.1. Thus this form is neither closed nor exact, refuting the disjunction.

F1F3step 1.1∎

Source locator

Lee, p.441, exact forms are closed; the nonclosed witness is computed directly.

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Continuously homotopic smooth maps can be inserted directly into the differential form homotopy operator

Statement

False claim: an arbitrary continuous homotopy between smooth maps can be inserted directly into the differential-form homotopy operator.

Facts & Assumptions

Given: $M$ is a point, $N = R$ , and $H (t) = ∣ t - 1 / 2 ∣$ on $[0, 1]$ .

[F1]

Integration along the unit interval for a differential form: Let $ω \in Ω^{k} (M \times [0, 1])$ be smooth up to the endpoints. For $k \geq 1$ , its interval integral is the $(k - 1)$ -form $K ω = \int_{0}^{1} β_{t} d t$ , where $ω = α_{t} + d t \land β_{t}$ and both families are tangential to $M$ . Set $K = 0$ on degree zero and on zero terms. Use the product structure of prop-products-of-smooth-manifolds-have-a-canonical-product-smooth-structure, restricted from $M \times R$ . The families are intrinsically $α_{t} = i_{t}^{*} ω$ and $β_{t} = i_{t}^{*} (ι_{\partial_{t}} ω)$ , using def-interior-product-of-a-form-by-a-vector-field; evaluation on tangential tuples and on $(\partial_{t}, v_{1}, \dots, v_{k - 1})$ proves existence and uniqueness of the decomposition. The integral is in the fixed finite-dimensional fibre $⋀^{k - 1} T_{x}^{*} M$ . Coefficients have smooth local extensions across endpoints. thm-differentiation-under-the-integral-sign-on-a-compact-rectangle supplies parameter differentiation; coordinate independence and full smoothness are proved in lem-the-interval-homotopy-operator-is-coordinate-independent.

[F2]

De rham homotopy formula for a smooth homotopy: If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

Refutation

technique · direct

1.1

The function $H$ is continuous, and $H (0) = H (1) = 1 / 2$ are smooth maps from a point. At $t = 1 / 2$ the left derivative is $- 1$ and the right derivative is $1$ , so $H$ has no differential there.

givenalgebra

2.1

The operator for a homotopy is $K H^{*}$ on smooth forms, and pullback of $d y$ requires $d H$ at every point. At the midpoint this pullback is undefined as a smooth differential form. The smooth-homotopy formula therefore cannot accept this particular continuous homotopy directly.

F1F2step 1.1∎

Source locator

Lee, Lemma 17.9 and Proposition 17.10, pp.444–445: the operator acts on smooth pullbacks; the cusp is a direct witness to the missing hypothesis.

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The poincare lemma says every closed form is globally exact

Statement

False claim: the Poincaré lemma makes every closed positive-degree form globally exact on every smooth manifold.

Facts & Assumptions

Given: $α = (- y d x + x d y) / (x^{2} + y^{2})$ on $R^{2} ∖ {0}$ .

[F1]

Poincare lemma for differential forms on star shaped domains: Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

[F2]

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^{k} (M) = \ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))$ and $B^{k} (M) = im (d : Ω^{k - 1} (M) \to Ω^{k} (M))$ . A form is closed if it belongs to $Z^{k}$ and exact if it belongs to $B^{k}$ . If $ω = d η$ , then $d ω = d^{2} η = 0$ by thm-the-exterior-derivative-squares-to-zero, so $B^{k} \subseteq Z^{k}$ . In particular $B^{0} = 0$ , since $Ω^{- 1} = 0$ . The zero form is both closed and exact in every degree.

[F3]

The local coordinate formula for the exterior derivative: Let $(U, x^{1}, \dots, x^{n})$ be a smooth chart on a smooth manifold and $ω$ a smooth $k$ -form on $U$ , with $k \geq 0$ . Summing over increasing $k$ -tuples $I$ , and writing $d x^{I} = d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , if $ω = \sum_{I} ω_{I} d x^{I}$ , then $d ω = \sum_{I} d ω_{I} \land d x^{I} .$

[F4]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: Let $a < b$ . Suppose $G : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . If $f : [a, b] \to R$ is Riemann integrable and $f (x) = G^{'} (x) (a < x < b),$ then $\int_{a}^{b} f = G (b) - G (a) .$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

Refutation

technique · direct

1.1

Put $r^{2} = x^{2} + y^{2}$ . The coefficients are smooth since $r^{2} > 0$ , and $\partial_{x} (x / r^{2}) = (y^{2} - x^{2}) / r^{4} = \partial_{y} (- y / r^{2})$ . Thus $d α = 0$ .

F2F3given

2.1

For $γ (t) = (\cos t, \sin t)$ , $0 \leq t \leq 2 π$ , substitution gives $γ^{*} α = (\sin^{2} t + \cos^{2} t) d t = d t$ . If $α = d f$ , the chain rule and fundamental theorem would give $2 π = \int_{0}^{2 π} γ^{*} α = f (γ (2 π)) - f (γ (0)) = 0$ . This contradiction proves nonexactness. The Poincaré lemma has a star-shaped-domain hypothesis, which this global witness does not satisfy.

F1F4step 1.1∎

Source locator

Lee, formula (17.1), p.441; direct coordinate differentiation and the fundamental theorem prove the obstruction without importing a later example.

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The mayer vietoris sequence is obtained by restricting forms without a partition of unity

Statement

Assume countable choice. Invalid proposed proof: restrictions alone establish full Mayer–Vietoris exactness, without a proof that the overlap difference map is surjective. In particular, extending an arbitrary overlap form unchanged to a prescribed cover member is not a valid general lift construction.

Facts & Assumptions

Given: Assume countable choice. $M = R$ , $U = (- \infty, 1)$ , $V = (0, \infty)$ , and the smooth overlap function $ω (x) = 1 / x$ on $(0, 1)$ .

[F1]

Two open set de rham mayer vietoris cochain maps: For an open cover $M = U \cup V$ , put $W = U \cap V$ . The two-open-set de Rham maps are $r : Ω^{∙} (M) \to Ω^{∙} (U) \oplus Ω^{∙} (V)$ , $r ω = (ω ∣_{U}, ω ∣_{V})$ , and $s : Ω^{∙} (U) \oplus Ω^{∙} (V) \to Ω^{∙} (W)$ , $s (α, β) = β ∣_{W} - α ∣_{W}$ . The complexes are def-de-rham-cochain-complex. Restrictions are pullbacks along open inclusions, so prop-pullback-is-a-morphism-of-de-rham-complexes gives $d r = r d$ and $d s = s d$ . Both maps are real linear. The middle differential acts componentwise. Empty opens have zero form spaces. The order second minus first fixes the sign of every connecting map below.

[F2]

The de rham mayer vietoris difference map is surjective: Assume countable choice. The difference map $s : Ω^{k} (U) \oplus Ω^{k} (V) \to Ω^{k} (U \cap V)$ is surjective in every degree.

Refutation

technique · direct

1.1

If $ω$ extended unchanged to a smooth function on $U$ , that extension would be continuous at $0$ . But $ω (1 / n) = n$ for all integers $n > 1$ , while $1 / n \to 0$ , so no continuous extension exists. Thus the naive unchanged extension recipe fails for an explicit smooth overlap form.

givenalgebra

2.1

The map $s$ requires a difference of two restricted forms, not either unchanged extension alone. The cutoff construction produces such a pair for this form (and every other form) under countable choice. Hence the actual sequence is exact, but the proposed recipe omits its essential lifting argument. This refutes that recipe, not the existence of other proofs of Mayer–Vietoris.

F1F2step 1.1∎

Source locator

Lee, proof of Theorem 17.20, p.463, where the cutoff lift is constructed explicitly.

False statementConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

De rham cohomology is a covariant functor

Statement

False claim: the pullback construction makes de Rham cohomology covariant.

Facts & Assumptions

Given: On the discrete three-point manifold $X = {1, 2, 3}$ , let $F$ swap $1, 2$ and $G$ swap $2, 3$ . Let $u$ be the indicator of ${1}$ .

[F1]

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

Refutation

technique · direct

1.1

Every function on $X$ is smooth and closed, with no nonzero degree-zero boundaries, so $u$ represents itself in $H^{0}$ . Pullback is composition. Hence $(F^{*} G^{*} u) (2) = u (G (F (2))) = u (1) = 1$ , while $(G^{*} F^{*} u) (2) = u (F (G (2))) = u (3) = 0$ .

F1given

2.1

Thus these pullback operators do not commute, and $(G \circ F)^{*} = F^{*} G^{*}$ cannot be replaced by $G^{*} F^{*}$ . In general $F : M \to N$ induces $F^{*} : H^{∙} (N) \to H^{∙} (M)$ , with the reversed source and target, exactly as the contravariant functor theorem states.

F1step 1.1∎

Source locator

Lee, Proposition 17.2(a) and Corollary 17.3, p.442; explicit noncommuting permutation pullbacks supply the witness.

The De Rham Complex Homotopy and Mayer Vietoris

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

De rham cochain complex

Definition

Source locator

Closed and exact differential forms

Definition

Source locator

De rham cohomology

Definition

Source locator

Zero and out of range de rham cohomology

Statement

Facts & Assumptions

Proof

Source locator

Zero th de rham cohomology is locally constant functions

Statement

Facts & Assumptions

Proof

Source locator

Pullback is a morphism of de rham complexes

Statement

Facts & Assumptions

Proof

Source locator

Pullback induces a well defined map on de rham cohomology

Statement

Facts & Assumptions

Proof

Source locator

De rham cohomology is a contravariant functor

Statement

Facts & Assumptions

Proof

Source locator

Wedge with a closed form preserves exactness classes

Statement

Facts & Assumptions

Proof

Source locator

Wedge product descends to de rham cohomology

Statement

Facts & Assumptions

Proof

Source locator

De rham cohomology ring

Definition

Source locator

Pullback is a homomorphism of de rham cohomology algebras

Statement

Facts & Assumptions

Proof

Source locator

Integration along the unit interval for a differential form

Definition

Source locator

The interval homotopy operator is coordinate independent

Statement

Facts & Assumptions

Proof

Source locator

De rham homotopy formula on a product

Statement

Facts & Assumptions

Proof

Source locator

De rham homotopy formula for a smooth homotopy

Statement

Facts & Assumptions

Proof

Source locator

Smoothly homotopic maps induce the same de rham map

Statement

Facts & Assumptions

Proof

Source locator