Gysin sequence from a sphere-fiber Serre spectral sequence

The orientation identifies the only nonzero fiber-cohomology local systems as the constant copies of $R$ in rows $0$ and $n$. Hence [F1] gives $E_2^{a,0}=H^a(B;R)$ and $E_2^{a,n}=H^a(B;R)u$, with all other rows zero. For $2\le s\le n$, a differential from the top row has target in a row strictly between $0$ and $n$, while a differential from the bottom row has negative second coordinate. Thus these differentials vanish. On page $n+1$ the only possible nonzero differential is $d_{n+1}:E_{n+1}^{a,n}\to E_{n+1}^{a+n+1,0}$. All later differentials have a zero endpoint, so [F5] gives collapse at $E_{n+2}$.

Put $e=d_{n+1}(u)$. Since no earlier differential reaches $(n+1,0)$, [F3] identifies its target with $H^{n+1}(B;R)$, so $e$ is an actual base class. A bottom-row class $a$ has zero differential. The Leibniz rule in [F2], applied to the product $au$, gives $d_{n+1}(au)=(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}ae$, proving (2). Thus the only page differential is, up to the displayed unit sign, cup multiplication by $e$.

In total degree $k$, the two stable terms are therefore $E_{\mathrm{\infty }}^{k,0}=\mathrm{coker}(H^{k-n-1}(B;R)\stackrel{⌣e}{\to }{H}^k(B;R))$ and $E_{\mathrm{\infty }}^{k-n,n}=\ker (H^{k-n}(B;R)\stackrel{⌣e}{\to }{H}^{k+1}(B;R))\mathrm{.}$ The decreasing abutment filtration has no other nonzero quotient, so [F1] and [F4] give a natural short exact sequence $0⟶E_{\mathrm{\infty }}^{k,0}⟶H^k(E;R)⟶E_{\mathrm{\infty }}^{k-n,n}⟶0.\text{(3)}$ The left arrow is the bottom edge. Naturality of [F1] applied to the map of fibrations from $p$ to the identity fibration of $B$ identifies its composite from $H^k(B;R)$ with $p^{\ast }$. The right arrow followed by the orientation is, by definition, $p_{!}$.

Exactness of (3) says successively that the kernel of $p^{\ast }$ is the image of cup multiplication by $e$, the image of $p^{\ast }$ is the kernel of $p_{!}$, and the image of $p_{!}$ is the kernel of the next cup multiplication. Placing these short exact sequences for consecutive total degrees next to one another gives (1), with no appeal to a homological exact-couple connector. Formula (2) contributes $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}$ to every other displayed cup map; multiplying that arrow by the unit $-1$ produces the stated cup-$e$ convention without changing kernels or images.

For an orientation-preserving pullback, the coefficient generators correspond. Naturality of [F2] commutes with $d_{n+1}$, so the Euler class pulls back; naturality of the filtration, the bottom edge, and the top-row quotient in [F1] and [F4] commutes with $p^{\ast }$ and $p_{!}$. Hence the whole sequence is natural.

If $B$ is empty the path-connected hypothesis excludes the case; if the zero ring is allowed, every displayed group and map is zero and the unique element is the orientation generator. For $n=1$ the first possible differential is $d_2$, exactly as above. Negative cohomological degrees are zero, so (1) has valid endpoints for every integer $k$. The cases $e=0$, $a=0$, $a=1$, a zero or one-term kernel, and a one-cell base are included. Degenerate cochain representatives disappear on passage to cohomology. Both rows, both ends of (3), all three adjacent exactness assertions, and both orientations of every pullback square have been checked. AC is used only through [F1]–[F2]. There is no iff assertion and no splitting of (3) is claimed.