Let $n\geq 1$ and let $\pi:E\to B$ be a fiber bundle with fiber $S^n$ with $B$ simply connected. We will fix an orientation of $p$. An easy analysis of the Serre spectral sequence shows that there is an exact sequence
$$\cdots \to H^p(B)\xrightarrow{d_{n+1}} H^{p+n+1}(B)\xrightarrow{\pi^*} H^{p+n+1}(E)\xrightarrow{\phi} H^{p+1}(B)\xrightarrow{d_{n+1}} \cdots.$$
This is the Gysin sequence of $\pi$.
There is another derivation of the Gysin sequence, which uses the Thom isomorphism. Let $\pi':E'\to B$ denote the disk bundle associated with $\pi$. Our orientation of $\pi$ determines the Thom class $t\in H^{n+1}(E',E)$, i.e., the cohomology class whose image in $H^{n+1}(\pi^{\prime-1}(b),\pi^{-1}(b))$ is the preferred generator for each $b\in B$. The Thom isomorphism theorem states that the map
$$H^p(B)\to H^{p+n+1}(E',E),\,b\mapsto \pi^{\prime\ast}(b)\smallsmile t$$
is an isomorphism of abelian groups for all $p$. Using this isomorphism, the long exact sequence of pairs $(E',E)$, and the observation that $p'$ is a (weak) homotopy equivalence, we obtain another exact sequence
$$\cdots \to H^p(B)\xrightarrow{e\smallsmile} H^{p+n+1}(B)\xrightarrow{\pi^*} H^{p+n+1}(E)\xrightarrow{\phi'} H^{p+1}(B)\xrightarrow{e\smallsmile} \cdots.$$
Here $e$ denotes the image of $t$ under the composite $H^n(E',E)\to H^n(E)\cong H^n(B)$, and $\phi'$ denotes the composite $H^{p+n+1}(E)\xrightarrow{\delta}H^{p+n+2}(E',E)\cong H^{p+1}(B).$
In Switzer's book Algebraic Topology-Homology and Homotopy, it is stated as an exercise (15.52) that the two exact sequences coincide; that is, $d_{n+1}=e\smallsmile$ and $\phi=\phi'$. Does anyone know how to prove this? I've thought about this for a while, but to no avail. Any help and comments are appreciated. Thanks in advance.
Remark
- Using the compatibility of the Serre spectral sequence with products, we can show that $d_{n+1}$ is given by the left multiplication by $z=d_{n+1}(1)\in H^{n+1}(B)$. So to prove that $d_{n+1}=e\smallsmile,$ it suffices to show that $z=e$. However, it is not clear to me why this is the case.