Gysin long exact sequence of an oriented sphere bundle

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.