Leray–Hirsch, the Thom Isomorphism, and Gysin Sequences

Let $\gamma :b\to c$ be a path. Fiber transport is represented, up to the equivalence built into [F1], by a map $T_{\gamma }:F_b\to F_c$ lying over the path. The inclusion maps of the endpoint fibers into $E$ are homotopic after composing the first with $T_{\gamma }$. Therefore [F2] gives $T_{\gamma }^{\ast }(e_i{\mathrm{\mid }}_{F_c})=e_i{\mathrm{\mid }}_{F_b}$ for every $i$, in the variance convention of [F1].

Since the restricted $e_i$ are a basis in both endpoint stalks, step 1.1 says that the transport matrix sends every named basis vector to the corresponding named basis vector. It is therefore the identity matrix. This holds for every path class, so the basis identifies the local system with the constant graded $R$-module $H^{\ast }(F_b;R)$.

If $m=0$, the basis hypothesis says every stalk is the zero module and the conclusion is the constant zero system. The zero ring and zero-degree classes obey the same calculation. A path-connected base is not needed: the argument applies independently on each component on which the same finite list is a basis. No basis or representative is chosen; the list is part of the hypotheses. AC is used exactly through [A1] in [F1].

Put a summand $H^{q-\mathrm{\mid }{e}_i\mathrm{\mid }}(B;R)$ in filtration at least $p$ when its base class has Serre filtration at least $p$. By [F2], $p^{\ast }{a}_i$ has that filtration and cupping with the fixed class $e_i$ cannot decrease it. Hence the displayed $\mathrm{\Phi }$ is a filtered homomorphism, not merely a map invented on $E_{\mathrm{\infty }}$.

Fix total degree $q$. The first-quadrant bounds make both induced filtrations finite in that degree, while [F1] supplies exhaustivity, completeness, and the identification with the stable page. Under the hypothesis that $\mathrm{gr}\mathrm{\Phi }$ is an isomorphism, [F3] therefore says $\mathrm{\Phi }:H_{\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}}^q\to H^q(E;R)$ is an isomorphism.

Applying step 2.1 in every degree proves the graded-module assertion. The inverse is obtained by the filtered-isomorphism lemma from kernels and cokernels along the finite filtration; no complements or splitting maps are chosen. Empty summand families and the zero ring give zero maps between zero modules, a one-step filtration reduces to the assumed graded isomorphism, and filtration endpoints are covered by finiteness. AC is used exactly through [A1] in the cohomological Serre suppliers.

By [F1] and [F3], the $E_2$ page is $H^p(B;R){\otimes }_R{H}^q(F;R)$ in the named basis. Each $e_i$ comes from total-space cohomology, so its fiber restriction is represented by the fiber edge and is a permanent cycle. The product in [F3] therefore makes the map induced by $\mathrm{\Phi }$ on $E_2$ the basis map ${⨁}_i{H}^p(B;R)[-\mathrm{\mid }{e}_i\mathrm{\mid }]\to H^p(B;R){\otimes }_R{H}^q(F;R)$. It is an isomorphism in every bidegree.

A morphism of spectral sequences that is an isomorphism on one page is an isomorphism on all subsequent pages, so step 1.1 gives the associated-graded isomorphism at $E_{\mathrm{\infty }}$. Applying [F2] to the actual filtered cup-product map proves that $\mathrm{\Phi }$ is an isomorphism in every total degree.

The formula gives $\mathrm{\Phi }(c{a}_i)=p^{\ast }c⌣\mathrm{\Phi }(a_i)$, so it is an $H^{\ast }(B;R)$-module map. For a map of fibrations carrying every specified target $e_i$ to the corresponding source $e_i$, contravariance of pullback and cup naturality make the two displayed formulas commute; this is the asserted, choice-of-basis-relative naturality. If the finite list is empty, all fiber cohomology is zero and both sides are zero; one basis element, the zero ring, degree-zero classes, and filtration endpoints are included in steps 1.1–2.1. No splitting or basis is selected, and AC is used exactly through [A1].

In rank zero each fiber is the singleton zero vector, so [F1] gives $D(0_B)=B$ and $S(0_B)=\mathrm{\varnothing }$. By the based-quotient convention, $\mathrm{Th}(0_B)=B_{+}$.

For $B\times {\mathbb{R}}^n$ with the product metric, the norm depends only on the second coordinate, hence $(D,S)=(B\times D^n,B\times S^{n-1})$. Collapsing the second subspace gives $(B\times D^n)\mathrm{/}(B\times S^{n-1})\cong B_{+}\wedge (D^n\mathrm{/}{S}^{n-1})=B_{+}\wedge S^n$.

By definition $B_{+}\wedge S^n={\mathrm{\Sigma }}^n{B}_{+}$. Every displayed map sends $(b,v)$ by the identity formula and therefore commutes with pullback along a map $B^{\mathrm{\prime }}\to B$. For $B=\mathrm{\varnothing }$ both sides are the one-point based space; for $n=0$, $S^{-1}=\mathrm{\varnothing }$ and step 1.2 reduces to step 1.1; for $n=1$ the boundary consists of the two endpoints. Zero and unit radii and the quotient basepoint are preserved, and no choice is used.

For $n=0$, $(D^0,S^{-1})=(\ast ,\mathrm{\varnothing })$, so [F1]'s dimension axiom gives $R$ in degree zero and zero in every other degree. Its distinguished element is $1_R$.

The two points of $S^0$ have $H^0(S^0;R)=R\oplus R$ by finite additivity. The reduced group is the cokernel of the diagonal constants and is freely generated by $(1,0)$ modulo the diagonal, hence is $R$; all other reduced groups vanish.

Suppose $m\ge 1$ and the reduced cohomology of $S^{m-1}$ is $R$ in degree $m-1$ and zero otherwise. Cover $S^m$ by two slightly enlarged hemispheres. Each is contractible and their intersection deformation retracts onto the equator $S^{m-1}$. In the Mayer–Vietoris sequence [F2], the map on degree-zero constants is the diagonal-difference map and the higher groups of the hemispheres vanish. Exactness therefore makes the connector an isomorphism ${\tilde{H}}^{q-1}(S^{m-1};R)\cong {\tilde{H}}^q(S^m;R)$ in every $q$. Thus the asserted sphere calculation propagates from step 1.2.

For $n\ge 1$, $D^n$ is contractible and $S^{n-1}$ is nonempty. The pair sequence [F3], together with the sphere calculation of step 2.1, identifies $H^q(D^n,S^{n-1};R)$ naturally with ${\tilde{H}}^{q-1}(S^{n-1};R)$. It is therefore $R$ exactly at $q=n$ and zero otherwise. Following $1_R$ through the ordered hemisphere connectors fixes the stated coordinate generator; reversing one ordered coordinate reverses the first difference and hence negates that generator.

A linear automorphism is a homeomorphism of the pair, so functoriality in [F1] makes its top-degree action an $R$-module automorphism of the rank-one module calculated in step 3.1. Such an automorphism is multiplication by a unique unit: the image of $1$ is $u$, and the inverse image supplies $v$ with $uv=1$. At $n=0$ this is the identity. Empty boundary, zero ring, ranks zero and one, both hemisphere endpoints, repeated/degenerate cover pieces, and the two possible coordinate signs are all included above. Only a fixed finite cover and its canonical maps occur, so neither arbitrary additivity nor AC is used.

Projection to the second factor is a map of pairs, so $u={\mathrm{pr}}_2^{\ast }{v}_n$ is defined. Its restriction to every fiber is literally $v_n$; hence it is normalized by [F5].

For one interval, the pair sequence for $(B\times I,B\times \mathrm{\partial }I)$ has restriction $H^q(B;R)\to H^q(B;R)\oplus H^q(B;R)$ equal to the diagonal, by the two endpoint deformation retractions and finite additivity in [F3]. Exactness identifies the next relative group with the cokernel of that diagonal, and $(x,y)\mapsto y-x$ identifies the cokernel with $H^q(B;R)$; the following diagonal is injective. Thus the connecting map is an isomorphism $H^q(B;R)\cong H^{q+1}(B\times I,B\times \mathrm{\partial }I;R)$.

Normalize the interval generator as the connector of the boundary class $(0,1)$. Applying [F4]'s second connector formula to ${\pi }^{\ast }a$ and $(0,1)$ says that the isomorphism of step 1.2 sends $a$ to $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{\pi }^{\ast }a⌣v_1$. Multiplication by this degree-dependent sign is itself an automorphism, so cup product with $v_1$ is an isomorphism in every degree.

We first record the relative form needed for iteration. If $(X,A)$ is a pair with an explicit collar of $A$, the one-interval connector gives H^q(X,A;R)\xrightarrow{\cong}H^{q+1}\bigl(X\times I,\,A\times I\cup X\times\partial I;R\bigr). \tag{1} To verify this rather than assume it, use the collar to subdivide chains until every simplex in the union lies in one of its two members. The small-chain equivalence in [F3] then identifies the target relative cochain complex with the kernel in the termwise split restriction sequence for the pairs $(X\times I,X\times \mathrm{\partial }I)$ and $(A\times I,A\times \mathrm{\partial }I)$. The resulting cohomology sequence forms a natural exact ladder over the pair sequence of $(X,A)$. The absolute vertical maps for $X$ and $A$ are the one-interval isomorphisms of step 2.1, so [F3]'s five lemma makes (1) an isomorphism. Connector naturality in [F4] identifies (1), up to the same invertible degree sign, with relative cup by the interval generator.

Write the $n$-cube as an ordered product of intervals. Each pair $(B\times I^j,B\times \mathrm{\partial }{I}^j)$ has an explicit radial collar in the cube coordinate, so apply (1) successively for $j=0,\dots ,n-1$. The generator $v_n$ in [F2] is the iterated ordered connector generator. At the cochain level, the iterated relative products are represented by repeated Alexander--Whitney cup products; associativity in [F4] identifies either parenthesization with the single product ${\pi }^{\ast }a⌣(v_1⌣\cdots ⌣v_1)$. Connector naturality identifies the parenthesized fiber product with $v_n$, up to the product of the displayed invertible signs. Thus the composite is $a\mapsto {\pi }^{\ast }a⌣v_n$ up to a unit sign and is an isomorphism. Radial identification of cube and disk pairs, and [F1], give the stated target.

For $n=0$, $v_0=1_R$ and the map is the identity of $H^k(B;R)$. Empty $B$, the zero ring, $n=1$, negative/zero $k$, both interval endpoints, a constant class, and degenerate singular simplices all remain inside the exact pair calculation. Every sign is a unit and hence affects neither bijectivity nor normalization. The construction uses finitely many fixed connectors and no arbitrary family, so it is choice-free.

Put $D_T=D(\xi ){\mathrm{\mid }}_T$ and $S_T=S(\xi ){\mathrm{\mid }}_T$. Let $C_{\ast }^{\mathcal{U}}(D_{U\cup V},S_{U\cup V})$ be the quotient of $C_{\ast }(D_U)+C_{\ast }(D_V)$ by its sphere subcomplex. The small-chain subdivision of [F3] makes its inclusion in the ordinary relative chain complex a chain-homotopy equivalence. There is a degreewise split exact chain sequence $0\to C_{\ast }(D_W,S_W)\to C_{\ast }(D_U,S_U)\oplus C_{\ast }(D_V,S_V)\to C_{\ast }^{\mathcal{U}}(D_{U\cup V},S_{U\cup V})\to 0,$ where the last map is addition and the first is the signed pair of inclusions. Dualizing this split sequence gives a termwise exact cochain sequence whose first term is the small relative cochain complex and whose other maps are restriction and restriction-difference. Transporting its cohomology through the small-chain equivalence gives the ordinary relative Mayer–Vietoris sequence. The usual kernel/image chase in [F2] fixes its connecting map and signs.

Compatibility says $(u_U,u_V)$ lies in the kernel of the relative difference map. Exactness in step 1.1 supplies $u\in H^n(D_{U\cup V},S_{U\cup V};R)$ restricting to both local classes. Each fiber lies in at least one open set, so these restrictions show that $u$ is normalized.

For every $k$, place the base Mayer–Vietoris sequence from [F2] over the relative sequence of step 1.1 shifted by $n$, and use cup product with $u$ vertically. Restrictions commute by relative naturality in [F4]. For the Mayer–Vietoris connector, choose a cochain on one member lifting a difference cocycle. The lower connector is represented by its coboundary. The Leibniz rule in [F4], together with $\delta u=0$, says that the coboundary of the lifted cochain cupped with $u$ is its coboundary cupped with $u$, up to the fixed degree sign. Thus the connector square commutes with that unit sign by a direct calculation in the termwise split small-cochain sequences of step 1.1. The four outer vertical maps at the $U$, $V$, and $W$ terms are the stipulated local Thom isomorphisms. Hence [F5] makes the middle map $H^k(U\cup V;R)\to H^{k+n}(D_{U\cup V},S_{U\cup V};R)$ an isomorphism.

If $u^{\mathrm{\prime }}$ is another normalized gluing class, the isomorphism in step 3.1 writes $u-u^{\mathrm{\prime }}={\pi }^{\ast }a⌣u$ for a unique $a\in H^0(U\cup V;R)$. Restriction to any fiber evaluates $a$ at its basepoint times the orientation generator; normalization makes this zero, so $a$ vanishes on every component and $u=u^{\mathrm{\prime }}$. If $U$, $V$, or $W$ is empty, the sequence reduces to an identity or disjoint finite additivity and the same argument applies. Rank zero, the zero ring, a one-set cover, zero classes, both difference-map endpoints, and every graded connector sign are included. Exactness supplies one class for one compatible pair; it does not choose an indexed family, so no AC is used.

The base case $m=0$ has empty base: the unique zero relative class is normalized and the map between zero cohomology groups is an isomorphism. For $m=1$, the trivial-bundle case contained in [F1] supplies the normalized class and isomorphism.

Assume as induction hypothesis that the claim holds on $W_j=U_1\cup \cdots \cup U_j$ for some $1\le j<m$. It holds on $U_{j+1}$ because that restriction is trivial. On $W_j\cap U_{j+1}$, the two restricted classes are both normalized for the same supplied orientation and are equal by the uniqueness clause of [F1]; their cup maps are isomorphisms by restriction to the trivializing open $U_{j+1}$.

Apply [F1] to the two opens $W_j$ and $U_{j+1}$. It gives a unique normalized Thom class and isomorphism on $W_{j+1}$. Thus the induction hypothesis propagates, and after the finite final index it holds on $W_m=B$.

The argument needs neither a shrink nor the numeration: openness and finite triviality suffice. Any finite cover comes with some finite enumeration as part of the witness that it is finite, and fixing that one witness is not AC. Repeated or empty members, empty intersections, $m=0,1$, rank zero, the zero ring, and the first and last induction endpoints are covered by [F1] and steps 1.1–2.1. Uniqueness makes the output independent of the chosen enumeration.

Choose the metric from [F0]. Over a CW base, filter the relative cochain complex $C^{\ast }(D(\xi ),S(\xi );R)$ by inverse images of the base skeleta. In the cellwise calculation in [F2], quotient every disk-bundle chain group by its sphere-bundle subcomplex. Subdivision and fibration lifting in [F1] preserve that subcomplex, so the identical exact-couple argument has fiber term $H^q(D^n,S^{n-1};R)$ and yields the displayed relative $E_2$ page with the same convergence bounds.

By [F3], those fiber groups vanish unless $q=n$, where they form the orientation local system. The supplied orientation identifies that system with the constant system $R$. Hence every differential has a zero source or target, $E_2=E_{\mathrm{\infty }}$, and in total degree $k+n$ there is exactly one filtration quotient, $E_{\mathrm{\infty }}^{k,n}=H^k(B;R)$. Thus there is no additive extension to split.

In total degree $n$, the unit section $1\in H^0(B;R)=E_2^{0,n}$ survives and, through convergence, defines a class $u\in H^n(D(\xi ),S(\xi );R)$. The cellwise edge restriction sends it to the chosen generator in every fiber, so [F5] makes it a normalized Thom class. By [F2] and [F4], multiplication by this permanent edge class sends $a\in E_2^{k,0}(B)$ to $a\in E_2^{k,n}$ under the orientation identification. Since both source and target have a single filtration quotient, the abutment edge is exactly $a\mapsto {\pi }^{\ast }a⌣u$ and is an isomorphism.

Now let $B$ be paracompact Hausdorff of CW type and choose a homotopy equivalence $f:K\to B$ from a CW complex, part of the CW-type hypothesis. Apply steps 1.1–3.1 to $f^{\ast }\xi$. A homotopy inverse and [F6] identify the iterated pullbacks with the original bundle; the induced radial bundle maps are homotopy inverse maps of disk/sphere pairs. Ordinary and relative homotopy invariance therefore identify the two Thom maps and transport the normalized class and isomorphism back to $B$.

For $n=0$ the only row is $q=0$, $u=1$, and the edge is the identity. Empty bases are handled componentwise by zero groups; disconnected bases use the componentwise construction and AC already assumed in [A1] for the cohomological comparison. The zero ring, a point base, the first and last filtration pieces, identity pullback, and both homotopy-equivalence composites are included. AC is used exactly through [F0], [F2], and [F6]; the one-row collapse and relative cell quotient add no choice.

Choose the metric by [F1]. With the orientation supplied, [F2] gives a normalized class $u_{\xi }$ and identifies the only relative Serre row with $H^k(B;R)$. Its edge is the displayed cup-product map and is an isomorphism in every degree.

For the unoriented calculation, use the $E_2$ formula and finite convergence stated in [F2]. By [F3], every row is zero except $q=n$, and that row is exactly ${\mathcal{O}}_R(\xi )$ before any orientation is selected. Thus the sequence collapses with one filtration quotient and gives the canonical isomorphism $H^k(B;{\mathcal{O}}_R(\xi ))\cong H^{k+n}(D(\xi ),S(\xi );R)$. A global untwisted Thom class is neither chosen nor asserted in this clause.

If $u$ and $u^{\mathrm{\prime }}$ are normalized, step 1.1 writes $u-u^{\mathrm{\prime }}={\pi }^{\ast }a⌣u_{\xi }$ for a unique $a\in H^0(B;R)$. Fiber restriction gives $a(b)o_b=0$ for every $b$; since $o_b$ is a free rank-one generator, $a(b)=0$. Thus $a=0$ componentwise and $u=u^{\mathrm{\prime }}$.

Under a supplied finite trivializing cover, [F4] constructs the same unique normalized class and cup isomorphism by finite Mayer–Vietoris. Uniqueness from step 2.1 identifies it with the Serre class, so this is genuinely a special case and not an additional hypothesis on the general theorem.

For $n=0$, $u=1$ and both maps are identities; on an empty base they are the unique maps of zero groups. Point and disconnected bases, the zero ring, degree-zero/negative input, the only Serre row and both filtration endpoints are covered by [F2]. AC is used exactly through [F1], [F2], and the local-coefficient cohomology interface [F3]; the finite-cover proof [F4] uses none.

Equip $f^{\ast }\xi$ with the pulled-back metric $f^{\ast }h$, defined by $\mathrm{\parallel }(b^{\mathrm{\prime }},v){\mathrm{\parallel }}_{f^{\ast }h}=\mathrm{\parallel }v{\mathrm{\parallel }}_h$. The canonical bundle map of [F2] preserves this norm exactly, so the defining inequalities and equalities restrict it to a continuous map of pairs $(D(f^{\ast }\xi ),S(f^{\ast }\xi ))⟶(D(\xi ),S(\xi ))\mathrm{.}$ Pull $u_{\xi }$ back along this pair map. On the fiber over $b^{\mathrm{\prime }}$, functoriality identifies its restriction with the restriction of $u_{\xi }$ on the fiber over $f(b^{\mathrm{\prime }})$. Because the pullback orientation was specified to preserve that generator, the pulled-back class is normalized. Uniqueness in [F1] gives $f^{\ast }{u}_{\xi }=u_{f^{\ast }\xi }$.

If $u$ and $v$ are any normalized Thom classes for one supplied orientation, the uniqueness clause of [F1] gives $u=v$. Equivalently, the Thom isomorphism writes $u-v=a⌣u_{\xi }$ with $a\in H^0(B;R)$, and fiber normalization forces $a$ to vanish on every component.

Over $\mathbb{Z}$, replacing every orientation generator $o_b$ by $-o_b$ makes $-u_{\xi }$ restrict to the new generator on every fiber. It is therefore normalized for the reversed orientation, and step 1.2 makes it that orientation's unique Thom class.

For the empty base the unique class pulls back to itself; in rank zero the unit orientation reverses to $-1$ and the same calculation applies. Point bases, identity maps, zero classes, and both pullback-square composites are literal instances of step 1.1. In characteristic two the two signs coincide, but the asserted reversal clause is integral. AC is used exactly through [A1] in [F1], and pulling back the supplied class makes no selection.

Give $\xi \times \eta$ the sum metric. The product pair $(D\xi \times D\eta ,(S\xi \times D\eta )\cup (D\xi \times S\eta ))$ is the unit pair for the maximum norm. On every nonzero fiber vector $z=(v,w)$, the formula $z\mapsto (\max \{\mathrm{\parallel }v\mathrm{\parallel },\mathrm{\parallel }w\mathrm{\parallel }\}\mathrm{/}\sqrt{\mathrm{\parallel }v{\mathrm{\parallel }}^2+\mathrm{\parallel }w{\mathrm{\parallel }}^2})z$, extended by zero, is a base-preserving homeomorphism to the sum-metric disk/sphere pair; its inverse uses the reciprocal radial ratio.

Form $u_{\xi }\times u_{\eta }$ with [F4] on the product pair and transport it across step 1.1. On the fiber over $(b,c)$ its restriction is the ordered product of the two normalized generators. The connector normalization in [F3] makes this exactly the ordered rank-$(n+m)$ generator. Thus the transported class is normalized, and [F1] identifies it with $u_{\xi \times \eta }$.

For bundles over $B$, [F2] identifies $\xi \oplus \eta$ with the pullback of $\xi \times \eta$ along $\mathrm{\Delta }:B\to B\times B$. By [F1], its Thom class is ${\mathrm{\Delta }}^{\ast }(u_{\xi }\times u_{\eta })$. The cochain definition in [F4] pulls this external product back to $u_{\xi }⌣u_{\eta }$, proving the Whitney-sum formula.

Swapping the two ordered fiber blocks crosses $n$ degree-one coordinate connectors past $m$ such connectors. The signed relative product rule in [F4] contributes $(-1)$ for each of the $nm$ crossings, so the generator and Thom class change by $(-1{)}^{nm}$. This is precisely the orientation of the block permutation.

If either base is empty the external pair and both classes are zero; ranks zero and one reduce respectively to the unit and one connector. The zero ring, zero class, identity diagonal, equal bundles, the zero vector in the radial map, maximum/sum unit boundaries, and both factor orders are all covered above. The radial ratio is locally bounded at zero and the map fixes zero. AC is used exactly through [A1] in [F1]; all product and sign formulas are finite and choice-free.

The pair sequence [F1] contains $H^k(D,S)\stackrel{j^{\ast }}{\to }{H}^k(D)\to H^k(S)\to H^{k+1}(D,S)$. Radial retraction identifies $H^k(D)$ with $H^k(B)$, and the Thom isomorphism [F2] identifies $H^k(D,S)$ with $H^{k-n}(B)$. Under these identifications the middle restriction is $p^{\ast }$ and [F2]–[F3] identify $j^{\ast }$ with $a\mapsto a⌣e_{\mathrm{T}\mathrm{h}}(\xi )$.

Let $n\ge 2$ and let $z$ be the normalized class of the sphere fiber. In the filtered-cochain definition [F4], survival of $z$ to page $n$ means choosing extensions over successive base skeleta whose coboundaries vanish through filtration $n-1$; $d_{n}z$ is represented by the remaining base-degree-$n$ coboundary. Perform the same extensions inside the disk bundle. The fiber pair connector sends $z$ to the normalized disk-pair generator, so the resulting relative class is $u_{\xi }$. Mapping it relative-to-absolute produces $j^{\ast }{u}_{\xi }$, and radial retraction produces $e_{\mathrm{T}\mathrm{h}}(\xi )$. Thus the very cochain that represents $d_{n}z$ represents $e_{\mathrm{T}\mathrm{h}}(\xi )$, with the positive pair connector and ordered fiber generator fixing the sign.

Substitute the identifications of step 1.1 at every degree of [F1]. Define ${\mathrm{\partial }}_G$ as the pair connector followed by the inverse Thom isomorphism. Exactness and naturality are preserved by isomorphism, giving the displayed long exact sequence and its pullback-natural ladders.

The sphere fiber has dimension $n-1\ge 1$, so [F5] applies with its symbol $n$ replaced by $n-1$. Its only differential is precisely the $d_n$ of step 1.2, and its cup map is multiplication by that transgression. Hence its Serre sequence agrees term-for-term with step 2.1 and the two Euler classes are equal.

For $n=0$, $S(\xi )=\mathrm{\varnothing }$, $e=1$, and the sequence alternates an identity with zero groups. For $n=1$, the pair sequence remains valid (an oriented line is covered directly), while [F5] is inapplicable because $S^0$ is not path-connected. Empty bases, the zero ring, zero/unit Euler classes, negative degrees, both pair-sequence endpoints, the connector sign and identity pullbacks are included in steps 1.1–3.1. AC is used exactly through [A1]; rewriting the supplied pair sequence adds no choice. No splitting or converse is asserted.

In an orientation-preserving pullback square, [F1] identifies the pulled-back Thom class. Pullback commutes with ${\pi }^{\ast }$, relative cup product and the relative-to-absolute pair map, so the Thom isomorphism and [F3]'s Gysin map commute. Pulling back along the zero section gives Euler naturality, and [F4] supplies the resulting natural ladder of exact sequences.

For the composable zero sections, functoriality in [F4] identifies the restriction of the second normal bundle to $B$ as $s_{\xi }^{\ast }\eta$. The supplied iterated tubular identification orders the first normal directions as those of $\xi$ and the second as those of $s_{\xi }^{\ast }\eta$, hence identifies the composite normal bundle with $\xi \oplus s_{\xi }^{\ast }\eta$.

Apply [F3] twice to $a\in H^{\ast }(B;R)$. Under the iterated pair identification, the result is the relative-to-absolute image of ${\pi }^{\ast }a$ cupped first with $u_{\xi }$ and then with the pullback of $u_{\eta }$. Associativity makes this cup product ${\pi }^{\ast }a⌣(u_{\xi }⌣u_{s_{\xi }^{\ast }\eta })$. By [F1]–[F2], the parenthesized class is the Thom class of the ordered normal sum from step 1.2, so [F3] identifies the result with $(s_{\eta }\circ s_{\xi }{)}_{!}(a)$.

Empty bases and zero rings give unique zero maps. A rank-zero section has Thom class and Euler class $1$ and acts as an identity, so either rank may be zero; ranks one and point bases require no change. Identity pullback squares, zero inputs, both orders of the normal sum, both stages of the composite, and every Gysin-sequence endpoint are covered by steps 1.1–2.1. Reversing the order would introduce the Koszul sign from [F2], which is why the order is stated. AC is used exactly through [A1]; the two-stage calculation is finite and makes no choices.