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✓ 7 results · all verified · 5 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Bocksteins Steenrod Squares and Cohomology Operations — Examples

1 · Prerequisites

2 · Summary

The first calculation follows the degree-one class of real projective space through both coefficient sequences: its integral Bockstein generates the degree-two integral two-torsion, while its mod-two Bockstein is $S q^{1} (a) = a^{2}$ . Cartan's formula then turns the total square of the projective generator into the binomial formula $S q^{i} (a^{j}) = (\binom{j}{i}) a^{j + i}$ , with truncation on finite projective space. The analogous complex-projective calculation has only even squares: $S q^{2 i} (c^{j}) = (\binom{j}{i}) c^{j + i}$ , and every odd square vanishes.

The relation $S q^{1} S q^{1} = 0$ is checked directly through the integral and mod-four Bocksteins, independently of the general Adem theorem. The surface example then constructs the orientation-sign cocycle from local orientation transport. Its chain proof draws on the later local-coefficients treatment: the signed orientation zero-cochain is capped with the canonical twisted fundamental cycle, and the cap-boundary identity converts its even coboundary into the mod-four $S q^{1}$ pairing. This proves $v_{1} = w_{1}$ , along with the orientability criterion, without assuming Wu's formula as an input.

The final counterexamples separate three assertions that can otherwise look deceptively similar. Two degree-two classes can have the same zero top square but different lower $S q^{1}$ , so top squares do not determine the lower operations. Also $S q^{2}$ cannot be the mod-two reduction of an integral-valued operation: on a degree-three class of ${R P}^{\infty}$ it has a nonzero degree-five value although the integral degree-five cohomology group vanishes. This obstruction does not apply to $S q^{1}$ , whose integral-valued lift is the integral Bockstein.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Bockstein detects integral two-torsion in real projective space

Example

Assume AC. Let $n \geq 2$ be an integer and let $a \in H^{1} ({R P}^{n}; F_{2})$ be the unique nonzero class. For the integral and mod-two coefficient sequences, respectively, write

$β_{Z} : H^{1} ({R P}^{n}; F_{2}) ⟶ H^{2} ({R P}^{n}; Z), β_{2} : H^{1} ({R P}^{n}; F_{2}) ⟶ H^{2} ({R P}^{n}; F_{2}) .$

Then $β_{Z} (a)$ generates $H^{2} ({R P}^{n}; Z) ≅ Z / 2$ , and

$β_{2} (a) = S q^{1} (a) = a^{2} .$

In particular, $a^{2}$ is the nonzero element of $H^{2} ({R P}^{n}; F_{2})$ .

Facts & Assumptions

Given: An integer $n \geq 2$ , the space $X = {R P}^{n}$ , and its unique nonzero class $a \in H^{1} (X; F_{2})$ .

[F1]

For $0 \to Z \overset{2}{\to} Z \to F_{2} \to 0$ , Bockstein connecting operation represents the integral Bockstein by lifting a cocycle $c$ to an integral cochain $\hat{c}$ , writing $δ \hat{c} = 2 h$ , and taking $[h]$ .

[F2]

This Bockstein class is independent of the lift and representative by The Bockstein is independent of lift and representative.

[F3]

The choice-free integral clause of Real projective space cellular homology and the pinch map gives, for $n \geq 2$ ,

$H_{0} (X; Z) = Z, H_{1} (X; Z) = Z / 2, H_{2} (X; Z) = 0.$

[F4]

Under AC, Topological universal coefficient short exact sequence for cohomology gives the integral cohomology sequence with its Ext term computed in the first variable.

[F5]

Under AC, Mod-two cohomology ring of infinite real projective space gives

$H^{*} ({R P}^{\infty}; F_{2}) ≅ F_{2} [a_{\infty}], ∣ a_{\infty} ∣ = 1,$

and restriction to $X = {R P}^{n}$ is an isomorphism through degree $n$ . For $n \geq 2$ it sends $a_{\infty}$ to $a$ and $a_{\infty}^{2}$ to $a^{2}$ , so $a$ and $a^{2}$ are the unique nonzero classes in degrees one and two.

[F6]

The mod-two Bockstein is $S q^{1}$ by Sq^1 is the mod-two Bockstein.

[F7]

The top-square identity in Steenrod normalization, instability, suspension, and top square says $S q^{r} (x) = x ⌣ x$ for a class $x$ of degree $r$ .

[A1]

The Axiom of Choice is assumed for the present combined argument. [F1], [F2], [F4], and [F5] are cited under their published AC hypotheses; the integral cellular calculation in [F3], the canonical residue lift, and the finite Ext calculations below introduce no further choice.

Verification

technique · direct lift-and-divide calculation

1.1

Choose a mod-two singular cocycle $c$ representing $a$ . [given, F1, F2] Let $\hat{c}$ be its valuewise lift with values zero or one. Since $c$ is a cocycle, every value of $δ \hat{c}$ is even. Hence there is a unique integral cochain $h$ with $δ \hat{c} = 2 h$ , and [F1]--[F2] give $β_{Z} (a) = [h]$ .

1.2

For the mod-two coefficient sequence, the Bockstein is the square of $a$ . [F5, F6, F7] Indeed, [F6] and the degree-one instance of [F7] give

$β_{2} (a) = S q^{1} (a) = a ⌣ a = a^{2} .$

Since $n \geq 2$ , [F5]'s restriction isomorphism in degree two carries the nonzero polynomial class $a_{\infty}^{2}$ to $a^{2}$ , so $a^{2}$ is the unique nonzero degree-two mod-two class.

1.3

The two required integral cohomology groups follow from the A-level cellular calculation. [F3, F4] In UCT degree one, the outside terms are

${Ext}_{Z}^{1} (Z, Z) = 0, {Hom}_{Z} (Z / 2, Z) = 0.$

The first equality holds because $Z$ is free, and the second because $Z$ is torsion-free. Hence $H^{1} (X; Z) = 0$ . In degree two, the Hom term is zero because $H_{2} (X; Z) = 0$ , while the free resolution

$0 ⟶ Z \overset{2}{\to} Z ⟶ Z / 2 ⟶ 0$

computes ${Ext}_{Z}^{1} (Z / 2, Z)$ as the cokernel of multiplication by two on $Z$ , namely $Z / 2$ . Exactness therefore gives $H^{2} (X; Z) ≅ Z / 2$ .

2.1

The integral class $[h]$ is nonzero. [F5, step 1.1, step 1.3] Suppose instead that $h = δ k$ for an integral degree-one cochain $k$ . Then

$δ (\hat{c} - 2 k) = 2 h - 2 δ k = 0.$

By $H^{1} (X; Z) = 0$ in step 1.3, there is an integral degree-zero cochain $ℓ$ with $\hat{c} - 2 k = δ ℓ$ . Reduction modulo two would give $c = δ (ℓ m o d 2)$ , contrary to the nonzero class $a = [c]$ in [F5].

3.1

The integral Bockstein class is a generator. [step 1.3, step 2.1] Indeed, step 1.3 identifies $H^{2} (X; Z)$ with a group having exactly one nonzero element, step 2.1 makes $β_{Z} (a)$ that element and therefore a generator of its integral two-torsion.

4.1

The endpoint and excluded cases introduce no missing assertion. [F3, F4, F5, A1, step 1.1, step 1.2, step 1.3, step 2.1, step 3.1] The endpoint $n = 2$ is included: the two degree-two groups in step 1.3 and [F5] are still nonzero, whereas $n = 0, 1$ are excluded because the promised integral degree-two target is absent. The zero degree-one class is explicitly excluded because its Bockstein is zero and cannot generate. Real projective spaces are nonempty, and their point and degree-zero cases do not enter the claim. The calculation uses ordinary singular cochains, including degenerate simplices, and makes no cellular-to-singular cochain identification. AC occurs through [F1], [F2], [F4], and [F5]; the zero/one lift itself is canonical. No biconditional or converse is asserted. ∎

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Steenrod squares on real projective space

Example

Assume AC, and write $H^{*} ({R P}^{\infty}; F_{2}) = F_{2} [a]$ with $∣ a ∣ = 1$ . For all integers $i, j \geq 0$ ,

$S q^{i} (a^{j}) = (\binom{j}{i}) a^{j + i},$

where the binomial coefficient is reduced modulo two. If $a_{n}$ is the named degree-one class on ${R P}^{n}$ , the finite-dimensional formula is

$S q^{i} (a_{n}^{j}) = (\binom{j}{i}) a_{n}^{j + i} in F_{2} [a_{n}] / (a_{n}^{n + 1}) .$

Thus the right side vanishes when $i > j$ or $j + i > n$ .

Facts & Assumptions

Given: Integers $i, j, n \geq 0$ , with $n$ used for the finite-dimensional assertion.

[F1]

Under AC, Mod-two cohomology ring of infinite real projective space gives the polynomial ring on $a$ . For $n \geq 1$ , restriction to ${R P}^{n}$ preserves the degree-one generator; for $n = 0$ it sends $a$ to zero.

[F2]

Total Steenrod square defines $S q (x) = \sum_{r = 0}^{\deg x} S q^{r} (x)$ on a homogeneous class, a finite sum.

[F3]

Steenrod normalization, instability, suspension, and top square gives $S q^{0} x = x$ , $S q^{r} x = 0$ for $r > \deg x$ , and $S q^{\deg x} x = x ⌣ x$ .

[F4]

Cartan formula for Steenrod squares gives $S q^{k} (x y) = \sum_{r + s = k} S q^{r} (x) S q^{s} (y)$ , with only finitely many nonzero terms.

[F5]

Real projective space cellular homology and the pinch map constructs ${R P}^{n}$ with one cell in every degree from zero through $n$ , and its integral cellular boundary coefficients are zero or two. Axiomatic cellular boundaries are integral incidence matrices with coefficients says that this integral incidence matrix acts on arbitrary coefficients. Cellular homology computes singular homology compares the resulting mod-two cellular homology with singular homology, and, under AC, Cohomology over a field is dual to homology over that field computes the corresponding singular cohomology.

[F6]

Steenrod squares commute with pullback by Steenrod squares are well-defined and natural.

[F7]

Pullback preserves cup products and powers by Cup product is natural, unital and associative.

[A1]

The Axiom of Choice is assumed exactly through the ring supplier in [F1] and field duality in [F5]. Reducing the finite cellular boundary and the binomial calculation make no further choices.

Verification

technique · total Cartan followed by a finite binomial expansion

1.1

The total square of the degree-one generator is $S q (a) = a + a^{2}$ . [F2, F3] Indeed, $S q^{0} (a) = a$ , the top square is $S q^{1} (a) = a^{2}$ , and instability removes every higher component.

2.1

The total square is multiplicative on the powers of $a$ . [F2, F4, step 1.1] All component sums are finite, so summing [F4] over $k$ gives

$S q (x y) = \sum_{k} \sum_{r + s = k} S q^{r} (x) S q^{s} (y) = S q (x) S q (y) .$

Induction on the finite integer $j$ , beginning with $S q (1) = 1$ , therefore gives $S q (a^{j}) = S q (a)^{j}$ .

3.1

The infinite-dimensional formula follows by coefficient comparison. [F1, step 1.1, step 2.1] The ordinary binomial theorem over $F_{2}$ gives

$S q (a^{j}) = (a + a^{2})^{j} = \sum_{r = 0}^{j} (\binom{j}{r}) a^{j + r} .$

The homogeneous component of degree $j + i$ on the left is $S q^{i} (a^{j})$ . The component on the right is $(\binom{j}{i}) a^{j + i}$ when $0 \leq i \leq j$ , and is zero when $i > j$ , which agrees with the usual zero convention for that binomial coefficient.

4.1

Restriction gives exactly the truncated finite formula. [F1, F5, F6, F7, step 3.1] Reduction modulo two turns every boundary coefficient in [F5] into zero. Thus cellular comparison and field duality give one copy of $F_{2}$ in cohomological degrees $0, \dots, n$ and zero above degree $n$ . By [F1] and [F7], the restrictions $a_{n}^{k} = i_{n}^{*} (a^{k})$ are nonzero for $0 \leq k \leq n$ . Consequently these powers form every nonzero graded piece and

$H^{*} ({R P}^{n}; F_{2}) = F_{2} [a_{n}] / (a_{n}^{n + 1}) .$

Naturality [F6]--[F7] and [F1] give

$S q^{i} (a_{n}^{j}) = S q^{i} (i_{n}^{*} a^{j}) = i_{n}^{*} S q^{i} (a^{j}) = (\binom{j}{i}) a_{n}^{j + i} .$

The quotient in [F5] makes this zero when $j + i > n$ . If $j > n$ , both sides are already zero: the input power vanishes, while $j + i > n$ for every $i \geq 0$ .

5.1

The endpoint and choice conventions agree with the formulas. [F1, F2, F3, F5, F6, A1, step 1.1, step 2.1, step 3.1, step 4.1] For $j = 0$ , the formula says $S q^{0} (1) = 1$ and all positive squares of the unit vanish. For $i = 0$ it says $S q^{0} (a^{j}) = a^{j}$ ; for $i = j$ it is the top-square identity $S q^{j} (a^{j}) = a^{2 j}$ ; and for $i > j$ it is instability. The case $n = 0$ is the point: $a_{0} = 0$ and only its zeroth power survives. Projective spaces are nonempty, zero classes map to zero, and degenerate singular simplices are already included in the natural operations of [F6]. AC is used only by the two ring suppliers [F1] and [F5]; every sum and induction here is finite. The formula is an equality, not either direction of a biconditional. ∎

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Steenrod squares on complex projective space mod two

Example

Assume AC, and write $H^{*} ({C P}^{\infty}; F_{2}) = F_{2} [c]$ with $∣ c ∣ = 2$ . For all integers $i, j \geq 0$ ,

$S q^{2 i} (c^{j}) = (\binom{j}{i}) c^{j + i},$

with the coefficient reduced modulo two, while $S q^{r} (c^{j}) = 0$ for every odd integer $r \geq 0$ . For the named class $c_{n}$ on ${C P}^{n}$ , the same formulas hold in $F_{2} [c_{n}] / (c_{n}^{n + 1})$ ; in particular the even formula vanishes when $i > j$ or $j + i > n$ .

Facts & Assumptions

Given: Integers $i, j, n \geq 0$ and an odd integer $r \geq 0$ , with $n$ used for the finite-dimensional assertion.

[F1]

Under AC, Mod-two cohomology rings of complex projective spaces gives the finite and infinite polynomial rings on the degree-two classes $c_{n}, c$ , makes skeletal restrictions preserve them, and makes all odd cohomology groups zero.

[F2]

Total Steenrod square defines $S q (x) = \sum_{s = 0}^{\deg x} S q^{s} (x)$ as a finite sum.

[F3]

Steenrod normalization, instability, suspension, and top square gives $S q^{0} x = x$ , $S q^{s} x = 0$ for $s > \deg x$ , and $S q^{\deg x} x = x ⌣ x$ .

[F4]

Cartan formula for Steenrod squares gives the finite component formula $S q^{k} (x y) = \sum_{s + t = k} S q^{s} (x) S q^{t} (y)$ .

[F5]

Steenrod squares commute with pullback by Steenrod squares are well-defined and natural.

[F6]

Pullback preserves products and powers by Cup product is natural, unital and associative.

[A1]

The Axiom of Choice is assumed exactly through [F1].

Verification

technique · total Cartan and comparison of homogeneous degrees

1.1

The total square of the degree-two generator is $S q (c) = c + c^{2}$ . [F1, F2, F3] Normalization gives the degree-two term $S q^{0} (c) = c$ , and the top square gives the degree-four term $S q^{2} (c) = c^{2}$ . The intermediate class $S q^{1} (c)$ lies in the zero group $H^{3} ({C P}^{\infty}; F_{2})$ from [F1], and instability removes every higher component.

2.1

The total square is multiplicative on powers of $c$ . [F2, F4, step 1.1] Summing the finite Cartan identities and regrouping their finite terms gives

$S q (x y) = \sum_{k} \sum_{s + t = k} S q^{s} (x) S q^{t} (y) = S q (x) S q (y) .$

Starting with $S q (1) = 1$ , finite induction yields $S q (c^{j}) = S q (c)^{j}$ .

3.1

Homogeneous components give both the even formula and odd vanishing. [F1, step 1.1, step 2.1] The binomial theorem gives

$S q (c^{j}) = (c + c^{2})^{j} = \sum_{q = 0}^{j} (\binom{j}{q}) c^{j + q} .$

Every term on the right has degree $2 j + 2 q$ . The degree- $2 j + 2 i$ component is therefore $(\binom{j}{i}) c^{j + i}$ , and every component of degree $2 j + r$ for odd $r$ is zero. When $i > j$ , the relevant binomial coefficient is zero, agreeing with instability since $2 i > 2 j$ .

4.1

Restriction gives the finite formulas and their truncation. [F1, F5, F6, step 3.1] For $i_{n} : {C P}^{n} ↪ {C P}^{\infty}$ , naturality gives

$S q^{s} (c_{n}^{j}) = S q^{s} (i_{n}^{*} c^{j}) = i_{n}^{*} S q^{s} (c^{j}) .$

Facts [F1] and [F6] identify all powers under this pullback. Thus step 3.1 restricts to the two claimed formulas, and the relation $c_{n}^{n + 1} = 0$ makes the even right side zero when $j + i > n$ . If $j > n$ , the input and every displayed right side already vanish.

5.1

The boundary and choice conventions are complete. [F1, F2, F3, F5, F6, A1, step 1.1, step 2.1, step 3.1, step 4.1] For $j = 0$ , only $S q^{0} (1) = 1$ survives. For $i = 0$ the formula is $S q^{0} (c^{j}) = c^{j}$ ; for $i = j$ it is the top square $S q^{2 j} (c^{j}) = c^{2 j}$ ; and $i > j$ is zero. Odd indices include $r = 1$ and are zero even before finite truncation. The point case $n = 0$ , the first truncated exponent $j + i = n + 1$ , zero inputs, and nonemptiness are explicit. Degenerate singular simplices are included in the natural operations [F5]. AC is inherited only from [F1], while every sum and induction here is finite. No biconditional or converse is asserted. ∎

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The relation Sq^1Sq^1=0

Example

For every space $X$ , every integer $n \geq 0$ , and every $x \in H^{n} (X; F_{2})$ ,

$S q^{1} S q^{1} (x) = 0.$

This is the first positive Adem relation, but the calculation below does not use the general Adem theorem and assumes no form of choice.

Facts & Assumptions

Given: A space $X$ , an integer $n \geq 0$ , and $x \in H^{n} (X; F_{2})$ .

[F1]

Bockstein connecting operation defines the integral Bockstein $\tilde{β}$ from $0 \to Z \overset{2}{\to} Z \to F_{2} \to 0$ and the mod-two Bockstein $β$ from $0 \to F_{2} \overset{2}{\to} Z / 4 \to F_{2} \to 0$ ; zero/one residue lifts make both constructions choice-free.

[F2]

Sq^1 is the mod-two Bockstein gives $S q^{1} = β$ on every mod-two cohomology group without AC.

Verification

technique · compare the two cyclic Bocksteins on cochains and compose

1.1

Reduction modulo two satisfies $β = ρ \tilde{β}$ . [given, F1] Let $c$ be a mod-two cocycle representing a class $u$ , and let $\hat{c}$ be its integer zero/one lift. Since $δ c = 0$ , every value of $δ \hat{c}$ is even, so there is a unique integer cochain $h$ with

$δ \hat{c} = 2 h .$

Also $δ h = 0$ , because $2 δ h = δ^{2} \hat{c} = 0$ and integer cochains are torsion-free. Thus $\tilde{β} (u) = [h]$ . Reducing $\hat{c}$ modulo four gives a lift of $c$ for the mod-two coefficient sequence, and its coboundary is $2 (h m o d 2)$ in $Z / 4$ . Hence $β (u) = [h m o d 2] = ρ \tilde{β} (u)$ .

1.2

The integral Bockstein kills reduced integral classes: $\tilde{β} ρ = 0$ . [F1] If $z$ is an integral cocycle, then $z$ itself is an integer lift of its mod-two reduction. Its coboundary is zero, so the lift/divide definition gives $\tilde{β} (ρ [z]) = 0$ .

2.1

The mod-two Bockstein squares to zero. [step 1.1, step 1.2] For every mod-two class $u$ ,

$β^{2} (u) = ρ \tilde{β} ρ \tilde{β} (u) = 0.$

3.1

Substitution of $S q^{1} = β$ proves the claim. [F2, step 2.1] Apply [F2] first to $x$ and then to the class $β (x)$ :

$S q^{1} S q^{1} (x) = β (β (x)) = β^{2} (x) = 0.$

4.1

The boundary and choice cases introduce no exceptions. [F1, F2, step 1.1, step 1.2, step 2.1, step 3.1] For the empty space, a zero class, or a point in degree zero, every displayed positive-degree output is zero. The first allowed degree $n = 0$ is included, and there is no upper endpoint. Integer multiplication by two is injective even when a cochain group is zero, so the division argument is unique; ordinary singular cochains include degenerate simplices. Every lift used above is the specified residue lift or the already given cocycle $z$ , so no choice principle is spent. The proof establishes an equality, not either direction of a biconditional. ∎

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Wu classes of a closed surface

Example

Assume AC, and let $M$ be a nonempty closed connected topological surface; thus $M$ is compact, boundaryless, and two-dimensional. Choose a generator $e_{x}$ of the integral orientation stalk $O_{x}$ at every $x \in M$ . For a singular one-simplex $σ$ from $v_{0}$ to $v_{1}$ , define $ϵ (σ) \in F_{2}$ by

T_{σ} (e_{v_{0}}) = (- 1)^{ϵ (σ)} e_{v_{1}} .

The verification below proves that $ϵ$ is a cocycle and that its class is independent of the chosen generators. Write $w_{1} (M) = [ϵ] \in H^{1} (M; F_{2})$ . Then the Wu classes of $M$ are

v_{0} (M) = 1, v_{1} (M) = w_{1} (M), v_{i} (M) = 0 (i > 1) .

Moreover, $v_{1} (M) = 0$ exactly when $M$ is orientable.

Facts & Assumptions

Given: The surface $M$ , the family $(e_{x})_{x \in M}$ , and the cochain $ϵ$ specified above.

[F1]

Orientation local system and orientation cover defines the infinite cyclic stalks $O_{x}$ , their path transport, and the two-sheeted orientation cover. Transport is unchanged by endpoint-fixed homotopy and respects path concatenation.

[F2]

The declared supplier prop-the-manifold-orientation-system-is-a-local-system regards $O_{M}$ as a covariant integral local system and identifies a continuous generator section with an orientation.

[F3]

The declared supplier def-singular-and-cellular-chain-complexes-with-local-coefficients places a local coefficient at the first vertex and, on a one-simplex, gives

(δ c) (σ) = T_{σ}^{- 1} (c (v_{1})) - c (v_{0}) .

The same definition gives local chains as direct sums, hence as finite chains. The local differentials square to zero by the declared supplier lem-twisted-boundaries-square-to-zero-and-are-independent-of-lift-bases.

[F4]

The declared supplier lem-canonical-twisted-fundamental-classes-over-compact-subsets gives the canonical class $[M]^{t w} \in H_{2} (M; O_{M})$ whose local value at $x$ is $o_{x} \otimes o_{x}$ , independently of the sign of the generator $o_{x}$ .

[F5]

The declared supplier def-cup-and-cap-products-with-local-coefficient-pairings defines the cohomology-first local cap product and fixes its chain sign:

\partial (ϕ \cap c) = (- 1)^{p} (ϕ \cap \partial c - (δ ϕ) \cap c) (ϕ \in C^{p}) .

[F6]

Fundamental class of a compact oriented manifold characterizes the ordinary mod-two fundamental class by its nonzero value in every local top-homology stalk.

[F7]

Bockstein connecting operation defines the mod-two Bockstein from $0 \to F_{2} \overset{2}{\to} Z / 4 \to F_{2} \to 0$ by lifting a cocycle and dividing its coboundary. The canonical zero/one lift is choice-free. Sq^1 is the mod-two Bockstein identifies this operation with $S q^{1}$ .

[F8]

Singular cochain complex with coefficients uses the positive ordinary coboundary convention $(δ a) (z) = a (\partial z)$ .

[F9]

Steenrod normalization, instability, suspension, and top square gives $S q^{0} x = x$ , $S q^{k} x = 0$ for $k > \deg x$ , and $S q^{\deg x} (x) = x ⌣ x$ .

[F10]

Wu classes of a closed manifold defines $v_{i} (M)$ as the unique class representing the $S q^{i}$ functional under the mod-two cup pairing, and sets it to zero outside $0 \leq i \leq \dim M$ .

[A1]

The Axiom of Choice is used directly once: from the nonempty two-element set of generators of every stalk $O_{x}$ , it supplies the set-indexed family $(e_{x})_{x \in M}$ . It is also inherited through [F10]'s perfect-pairing result. No later family of representatives, paths, charts, or primitives is chosen.

Verification

Proof technique: compare the mod-four cohomology Bockstein pairing with the integral homology lift-and-divide cycle obtained from the twisted fundamental cycle.

1.1

The edge signs form a cocycle. [F1, F2, A1, given] For a singular two-simplex, write $ϵ_{i j}$ for the sign on its affine edge from vertex $i$ to vertex $j$ . The $02$ edge is homotopic relative to its endpoints to the $01$ edge followed by the $12$ edge. Functoriality of orientation transport therefore gives

ϵ_{02} = ϵ_{01} + ϵ_{12} in F_{2} .

With the positive singular coboundary, $(δ ϵ) (012) = ϵ_{12} - ϵ_{02} + ϵ_{01} = 0$ . Hence $ϵ$ is a cocycle. A degenerate edge has identity transport and therefore sign zero, consistently with this calculation.

2.1

The cohomology class does not depend on the generator family. [F1, step 1.1] Any other family has the form $e_{x}^{'} = (- 1)^{t (x)} e_{x}$ for a unique ordinary zero-cochain $t : M \to F_{2}$ . Its edge signs satisfy

ϵ^{'} (σ) = ϵ (σ) + t (v_{1}) - t (v_{0}) = ϵ (σ) + (δ t) (σ) .

Thus $[ϵ^{'}] = [ϵ]$ , so $w_{1} (M)$ is well defined.

3.1

The signed generator cochain has an even coboundary whose half reduces to $ϵ$ . [F1, F3, step 1.1, step 2.1] Define the local zero-cochain $c \in C^{0} (M; O_{M})$ by $c (x) = e_{x}$ . Since $T_{σ} (e_{v_{0}}) = (- 1)^{ϵ (σ)} e_{v_{1}}$ , the formula in [F3] gives

(δ c) (σ) = ((- 1)^{ϵ (σ)} - 1) e_{v_{0}} .

Consequently there is a unique local one-cochain $b$ with $δ c = 2 b$ : $b (σ) = 0$ when $ϵ (σ) = 0$ and $b (σ) = - e_{v_{0}}$ when $ϵ (σ) = 1$ . Since local cochain groups are products of infinite cyclic groups, they have no two-torsion. Thus $2 δ b = δ^{2} c = 0$ implies $δ b = 0$ .

There is a canonical morphism of local systems $r : O_{M} \to {\underline{F}}_{2}$ : if $e$ is either generator, $r (n e) = n m o d 2$ . The formula is independent of replacing $e$ by $- e$ , and orientation transport changes a generator only by sign, so it commutes with transport. The displayed values of $b$ give $r (b) = ϵ$ .

4.1

Cap the twisted fundamental cycle with the generator cochain. [F3, F4, F5, F6, step 3.1] There is a canonical local-coefficient pairing

q : O_{M} \otimes O_{M} ⟶ \underline{Z}, q_{x} (n e, m e) = n m,

where $e$ is either generator of $O_{x}$ . Simultaneously replacing $e$ by $- e$ leaves $n m$ unchanged, and simultaneous orientation transport does the same, so $q$ is well defined and transport-compatible.

Choose one finite twisted cycle $C$ representing $[M]^{t w}$ ; this is a single existential witness, not a family of choices. Applying $r$ to its coefficients gives an ordinary mod-two cycle $\overline{C}$ . At every point, the canonical local value $o_{x} \otimes o_{x}$ from [F4] maps to the unique nonzero mod-two local orientation. The uniqueness in [F6] therefore gives $[\overline{C}] = [M]_{2}$ .

Put $Z = c \cap_{q} C \in C_{2} (M; Z)$ . On each simplex, reduction modulo two turns $q$ into multiplication in $F_{2}$ , turns $r (c)$ into the constant zero-cochain $1$ , and turns $C$ into $\overline{C}$ . Hence

\overline{Z} = 1 \cap \overline{C} = \overline{C} .

Thus $Z$ is an integral lift of the mod-two fundamental cycle.

5.1

The cap-boundary sign produces the correct lift-and-divide cycle. [F3, F5, step 3.1, step 4.1] Since $C$ is a cycle and $c$ has degree zero, the exact convention in [F5] gives

\partial Z = c \cap_{q} \partial C - (δ c) \cap_{q} C = - 2 (b \cap_{q} C) .

Set $W = - b \cap_{q} C$ . Then $\partial Z = 2 W$ . The ordinary singular chain group is free abelian on the singular simplices, so $2 \partial W = \partial^{2} Z = 0$ implies $\partial W = 0$ . After reducing modulo two, the minus sign disappears and step 3.1 gives

\overline{W} = ϵ \cap \overline{C} .

6.1

The mod-four $S q^{1}$ pairing is evaluation on $\overline{W}$ . [F7, F8, step 4.1, step 5.1] Let $x \in H^{1} (M; F_{2})$ , represent it by a cocycle $a$ , and let $\hat{a}$ be its canonical integer zero/one lift. There is a unique integer two-cochain $h$ such that $δ \hat{a} = 2 h$ . It is a cocycle because integer cochains have no two-torsion. Reducing $\hat{a}$ modulo four shows from [F7] that

S q^{1} (x) = [h m o d 2] .

The positive coboundary convention and $\partial Z = 2 W$ give the exact integer calculation

2 h (Z) = (δ \hat{a}) (Z) = \hat{a} (\partial Z) = 2 \hat{a} (W) .

Canceling $2$ in $Z$ and then reducing modulo two yields

⟨ S q^{1} (x), [M]_{2} ⟩ = ⟨ x, [\overline{W}] ⟩ .

7.1

Cap-cup adjunction identifies the orientation class. [F5, step 3.1, step 4.1, step 5.1, step 6.1] For the cohomology-first cap convention, evaluating $a$ on $ϵ \cap \overline{C}$ is exactly the Alexander--Whitney evaluation of $ϵ ⌣ a$ on $\overline{C}$ : $ϵ$ reads the front edge and $a$ reads the retained back edge. Therefore

\begin{aligned} ⟨ S q^{1} (x), [M]_{2} ⟩ & = ⟨ x, w_{1} (M) \cap [M]_{2} ⟩ \\ = ⟨ w_{1} (M) ⌣ x, [M]_{2} ⟩ . \end{aligned}

By [F9], this also states the surface self-intersection identity $⟨ x ⌣ x, [M]_{2} ⟩ = ⟨ w_{1} (M) ⌣ x, [M]_{2} ⟩$ ; it was derived from the chain calculation, not assumed as Wu's formula.

8.1

The degree-one Wu class is $w_{1} (M)$ . [F10, step 7.1] The identity in step 7.1 holds for every $x \in H^{1} (M; F_{2}) = H^{2 - 1} (M; F_{2})$ . By the defining uniqueness of the degree-one Wu class in [F10], it follows that $v_{1} (M) = w_{1} (M)$ .

9.1

The remaining Wu classes have the asserted values. [F9, F10, step 8.1] For $i = 0$ , [F9] makes the defining functional $x \mapsto ⟨ S q^{0} x, [M]_{2} ⟩$ equal to evaluation on the fundamental class, which is represented by the unit; uniqueness in [F10] gives $v_{0} (M) = 1$ . For $i = 2$ , the test classes in [F10] have degree zero, so [F9] gives $S q^{2} x = 0$ for all of them. The zero class represents this zero functional, and uniqueness gives $v_{2} (M) = 0$ . Indices $i > 2$ are zero by the out-of-range convention in [F10]. Hence $v_{i} (M) = 0$ for every $i > 1$ .

10.1

Vanishing of $w_{1} (M)$ is equivalent to orientability. [F1, F2, step 2.1, step 9.1] If $M$ is oriented, let $s_{x}$ be its continuous generator section. Write $s_{x} = (- 1)^{t (x)} e_{x}$ . Transport preserves $s$ , so the generator-change calculation in step 2.1 gives $ϵ = δ t$ and hence $w_{1} (M) = 0$ .

Conversely, if $w_{1} (M) = 0$ , choose an ordinary zero-cochain $t$ with $ϵ = δ t$ and set $e_{x}^{'} = (- 1)^{t (x)} e_{x}$ . Step 2.1 shows that all edge signs for $(e_{x}^{'})$ vanish. Thus transport along every singular path carries its initial $e^{'}$ to its terminal $e^{'}$ . Around any point, take a path-connected orientation-chart ball and the basic local orientation section whose value at that point is $e^{'}$ . Transport inside the ball generates that section, so the path-transport property makes it equal to $e^{'}$ throughout the ball. Hence $x \mapsto e_{x}^{'}$ is locally continuous, and therefore is a global section of the orientation cover. By [F2], it orients $M$ . Since $v_{1} = w_{1}$ by step 8.1, this proves the final biconditional.

11.1

Boundary and choice cases are explicit. [F3, F4, F5, F7, F8, F10, A1, step 1.1, step 3.1, step 4.1, step 5.1, step 6.1, step 7.1, step 8.1, step 9.1, step 10.1] The hypothesis excludes the empty and disconnected cases and fixes dimension two; closed excludes manifold boundary. Zero classes $x$ are included in step 6.1. The degree endpoints $i = 0, 1, 2$ and all out-of-range indices were handled in step 9.1. Degenerate simplices remain in the unnormalized singular complexes; their ordinary and local boundary formulas are the ones used above. The two divisions by $2$ are unique because the relevant integral cochain and chain groups are torsion-free. The zero/one lift of $a$ , the reductions $r$ , and the pairing $q$ are canonical. Apart from the one pointwise generator-family selection declared in [A1], only the single cycle representative $C$ , the single primitive $t$ under the hypothesis $w_{1} = 0$ , and one chart at a time are chosen; these are ordinary existential instantiations, not further uses of AC. ∎

Remarks

The five suppliers used in [F2]--[F5] are homed on the later page local-coefficients-twisted-homology-and-duality (batch 5): this examples page precedes that page in the reading order, and the batch-5 manifest already whitelists the target page under the examples page's forwardRefs. All five items are declared in deps here, so the dependency graph is complete; because their page is later, they are named by ID in [F2]--[F5] rather than linked, since a body hyperlink to later material must be declared as a forward reference and Step-5b resolution removes that declaration. Rehoming this example to local-coefficients-twisted-homology-and-duality-examples (an owner-only reading-order change) would make every citation backward and restore the links.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14Open item page →

Top squares do not determine lower squares

Statement refuted

The identity $S q^{∣ z ∣} (z) = z^{2}$ does not determine the lower Steenrod squares, even when the degree and the value of the top square are fixed.

More explicitly, assume AC, base ${R P}^{2}$ at its zero-cell, and let $a$ be its nonzero class in $H^{1} ({R P}^{2}; F_{2})$ . If

$x = σ a \in {\tilde{H}}^{2} (Σ {R P}^{2}; F_{2})$

under the standard reduced cohomology-suspension isomorphism, and if $y$ is the nonzero class in $H^{2} (S^{2}; F_{2})$ , then

$x^{2} = 0 = y^{2}, S q^{1} (x) \neq 0, S q^{1} (y) = 0.$

Facts & Assumptions

Given: The based projective plane, its class $a$ , and the classes $x, y$ specified above.

[F1]

Bockstein detects integral two-torsion in real projective space states that, under AC, $S q^{1} (a) = a^{2}$ and that this class is nonzero.

[F2]

Under AC, Mod-two cohomology ring of infinite real projective space gives the infinite polynomial generator and says restriction to ${R P}^{2}$ is an isomorphism through degree two. The one-cell-per-degree construction and the integral incidence coefficients zero or two come from Real projective space cellular homology and the pinch map. By Axiomatic cellular boundaries are integral incidence matrices with coefficients, these coefficients act on $F_{2}$ , so they all vanish; then Cellular homology computes singular homology and field duality [F5] give $H^{q} ({R P}^{2}; F_{2}) = 0$ for $q > 2$ . In particular $a^{2}$ is nonzero and $a^{3} = 0$ .

[F3]

Steenrod normalization, instability, suspension, and top square gives $S q^{2} (z) = z^{2}$ for every degree-two class, makes $S q^{1}$ commute with the standard reduced cohomology suspension.

[F4]

Homology of spheres computes $H_{k} (S^{2}; F_{2})$ as $F_{2}$ for $k = 0, 2$ and zero otherwise.

[F5]

Cohomology over a field is dual to homology over that field turns [F4] into the corresponding mod-two cohomology calculation, under AC.

[A1]

The Axiom of Choice is used exactly through [F1], the infinite-ring and field-duality clauses in [F2], and [F5]. The cone-pair suspension and the Steenrod calculation in [F3] add no use of choice.

Counterexample

1.1

By definition, the standard reduced cohomology suspension $σ : {\tilde{H}}^{q} (Z; F_{2}) \to {\tilde{H}}^{q + 1} (Σ Z; F_{2})$ is an isomorphism; [F3] fixes this same standard suspension in its stability formula. In particular $x = σ a$ is nonzero.

2.1

The first square distinguishes the two classes. [F1, F2, F3, F4, F5, step 1.1] Stability and [F1] give

$S q^{1} (x) = S q^{1} (σ a) = σ S q^{1} (a) = σ (a^{2}) .$

The class $a^{2}$ is nonzero by [F1] (and explicitly by the ring [F2]); the degree-two instance of the isomorphism in step 1.1 therefore makes $S q^{1} (x)$ nonzero. On the other hand [F4]--[F5] give $H^{3} (S^{2}; F_{2}) = 0$ , so $S q^{1} (y) = 0$ .

2.2

Both top squares vanish. [F2, F3, F4, F5, step 1.1] The degree-three projective group is zero by [F2], so the degree-three instance of the suspension isomorphism gives ${\tilde{H}}^{4} (Σ {R P}^{2}; F_{2}) = 0$ . Thus [F3] gives

$x^{2} = S q^{2} (x) = 0.$

Likewise [F4]--[F5] give $H^{4} (S^{2}; F_{2}) = 0$ , whence $y^{2} = S q^{2} (y) = 0$ .

3.1

These computations refute determination by the top-square formula. [F3, step 2.1, step 2.2] The two nonzero degree-two classes have the same top-square value, namely zero, but different $S q^{1}$ values. Therefore knowing only $S q^{∣ z ∣} (z) = z^{2}$ cannot recover all lower squares.

4.1

The boundary and choice cases do not hide an exception. [F1, F2, F3, F4, F5, A1, step 1.1, step 2.1, step 2.2, step 3.1] Both spaces and both displayed input classes are nonempty and nonzero; the unit and zero classes are not the witnesses. The degree endpoint is exactly $∣ x ∣ = ∣ y ∣ = 2$ , so $S q^{1}$ is genuinely lower and $S q^{2}$ is genuinely top. The vanishing statements come from zero target groups, not from omitting degenerate singular simplices. AC is inherited exactly from the projective and field-duality computations [F1], [F2], and [F5]; suspension and all remaining calculations are choice-free. This is an explicit pair of witnesses, not either direction of a biconditional. ∎

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Steenrod squares do not all lift integrally

Statement refuted

It is false that every Steenrod square has a natural integral-valued lift. More precisely, assume AC. There is no degree-two cohomology operation

$Θ_{n} : H^{n} (-; F_{2}) ⟶ H^{n + 2} (-; Z)$

whose composite with coefficient reduction $ρ : H^{n + 2} (-; Z) \to H^{n + 2} (-; F_{2})$ is $S q^{2}$ for every space, degree, and class. In fact, the obstruction below rules out such a lift even before naturality is used.

This does not apply to $S q^{1}$ : the integral Bockstein is an integral-valued lift of $S q^{1}$ . The failed degree-one scaffold argument is therefore not used.

Facts & Assumptions

Given: A hypothetical family $Θ_{n}$ with the displayed lifting property.

[F1]

Steenrod squares on real projective space gives, under AC, $H^{*} ({R P}^{\infty}; F_{2}) = F_{2} [a]$ , $∣ a ∣ = 1$ , and $S q^{i} (a^{j}) = (\binom{j}{i}) a^{j + i}$ with coefficients reduced modulo two.

[F2]

On each standard finite skeleton from [F1], Real projective space cellular homology and the pinch map gives one cell in every dimension up to its top dimension and integral incidence maps $d_{j} = 2$ for positive even $j$ and $d_{j} = 0$ for odd $j$ . Axiomatic cellular boundaries are integral incidence matrices with coefficients identifies these incidence matrices with the cellular differential, and Cellular homology computes singular homology applies to the resulting infinite CW complex.

[F3]

Topological universal coefficient short exact sequence for cohomology gives, under AC, the exact sequence

$0 ⟶ {Ext}_{Z}^{1} (H_{n - 1} (X; Z), G) ⟶ H^{n} (X; G) ⟶ {Hom}_{Z} (H_{n} (X; Z), G) ⟶ 0.$

[F4]

Singular cohomology with coefficients makes the coefficient map $Z \to F_{2}$ induce the reduction homomorphism $ρ$ used in the statement.

[F5]

Bockstein connecting operation gives the integral and mod-two Bocksteins by canonical cyclic residue lifts.

[F6]

Assuming AC, Bocksteins are natural and stable makes these Bocksteins natural cohomology operations.

[F7]

Sq^1 is the mod-two Bockstein identifies the mod-two Bockstein with $S q^{1}$ .

[A1]

The Axiom of Choice is used exactly through [F1], [F3], and [F6]. The cellular homology and cyclic residue-lift calculations add no choice.

Counterexample

1.1

Choose the explicit mod-two input $u = a^{3}$ . [F1] It is nonzero in the polynomial ring and has degree three. Substitution of $i = 2, j = 3$ in [F1] gives

$S q^{2} (u) = (\binom{3}{2}) a^{5} = a^{5} \neq 0,$

since $(\binom{3}{2}) = 3 = 1$ in $F_{2}$ and every power of the polynomial generator is nonzero.

1.2

The required integral target is zero. [F1, F2, F3] The standard skeletal inclusions in [F1] preserve the cells in [F2], and a cellular differential in degree $j$ is already determined on the finite skeleton ${R P}^{j}$ . Hence their union gives the infinite integral cellular complex with the same alternating differentials. In particular, $d_{4} = 2$ , $d_{5} = 0$ , and $d_{6} = 2$ . Therefore

$H_{4} ({R P}^{\infty}; Z) = \ker (2 : Z \to Z) = 0$

and

$H_{5} ({R P}^{\infty}; Z) = Z / 2 Z .$

At $n = 5$ with $G = Z$ , [F3] therefore becomes

$0 ⟶ {Ext}_{Z}^{1} (0, Z) ⟶ H^{5} ({R P}^{\infty}; Z) ⟶ {Hom}_{Z} (Z / 2, Z) ⟶ 0.$

The left group is zero from the zero projective resolution. The right group is zero because the image in the torsion-free group $Z$ of an element killed by two must be zero. Exactness therefore gives $H^{5} ({R P}^{\infty}; Z) = 0$ .

1.3

The degree-one exception really has an integral natural lift. [F4, F5, F6, F7] For a mod-two cocycle $c$ , let $\hat{c}$ be its canonical valuewise zero/one integer lift and write $δ \hat{c} = 2 h$ . By [F5], the integral Bockstein is $[h]$ . The cochain $\hat{c} m o d 4$ is a lift through the mod-four coefficient sequence, and its coboundary is $2 h m o d 4$ . Pulling back along $2 : F_{2} \to Z / 4$ therefore gives $h m o d 2$ , so

$β_{2} ([c]) = ρ \tilde{β} ([c]) .$

By [F6] both sides are natural operations, and [F7] identifies the left side with $S q^{1}$ . Thus the integral Bockstein is precisely the promised integral-valued lift of $S q^{1}$ .

2.1

The lifting equation fails on $u$ . [F1, F4, step 1.1, step 1.2] Step 1.2 forces $Θ_{3} (u) = 0$ , so coefficient reduction gives $ρ Θ_{3} (u) = 0$ . Step 1.1 gives $S q^{2} (u) = a^{5} \neq 0$ . Hence $ρ Θ_{3} (u) \neq S q^{2} (u)$ , contradicting the defining property of $Θ$ . This single value rules out the lift without invoking naturality.

3.1

The boundary, qualification, and choice cases are explicit. [F1, F2, F3, F4, F5, F6, F7, A1, step 1.1, step 1.2, step 1.3, step 2.1] The witness space is nonempty and the input $a^{3}$ and failed output $a^{5}$ are nonzero; the zero class and unit are not witnesses. The indices $i = 2, j = 3$ lie inside the projective-space formula rather than an instability or truncation range, while the integral vanishing is calculated in the exact target degree five. Degenerate singular simplices remain in [F3]--[F5]. AC is inherited exactly from [F1], [F3], and [F6]. The item asserts nonexistence of one kind of lift and makes no biconditional claim; step 1.3 proves rather than merely asserts the integral Bockstein lift of $S q^{1}$ . ∎

Bocksteins Steenrod Squares and Cohomology Operations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Bockstein detects integral two-torsion in real projective space

Example

Facts & Assumptions

Verification

Steenrod squares on real projective space

Example

Facts & Assumptions

Verification

Steenrod squares on complex projective space mod two

Example

Facts & Assumptions

Verification

The relation Sq^1Sq^1=0

Example

Facts & Assumptions

Verification

Wu classes of a closed surface

Example

Facts & Assumptions

Verification

Remarks

Top squares do not determine lower squares

Statement refuted

Facts & Assumptions

Counterexample

Steenrod squares do not all lift integrally

Statement refuted

Facts & Assumptions

Counterexample

Sources