I have a question about an argument in a proof from J.P. May's "A Concise Course in Algebraic Topology" (page 196):
We use following notations: Let $\zeta: E \to B$ be a $n$-vector bundle and $Sph(E)$ the sphere bundle of $E$.
Denote by $T \zeta := Sph(E)/B$ the Thom space obtained by quotient out $B$ embedded as section (compare with constuction at p. 194).
NOW THE PROBLEM:
Applying Serre spectral sequence to the sphere bundle $Sph(E) \to B$ with fiber $S^n$ we deduce that $$H^*(Sph(E);R) \cong H^*(B;R) \otimes H^*(S^n;R)$$ as stated in the excerpt.
Then May says that we obtain $$\tilde{H}^*(T\zeta;R) \cong H^*(B;R) \otimes H^*(S^n;R)$$
by the canonical quotient map $ Sph(E) \to T \zeta$.
I don't understand this argument. Does $ H^*(Sph(E);R) \cong \tilde{H}^*(T\zeta;R)$ hold? Why?
Remark: Here the full book: A Concise Course in Algebraic Topology
No, that's not what he says. What he says is we obtain $\tilde{H}^*(T\zeta;R)\cong H^*(B;R)\otimes H^n(S^n;R)$. Note that $$H^*(Sph(\zeta))\cong H^*(B) \otimes H^*(S^n) \cong (H^*(B)\otimes H^0(S^n))\oplus (H^*(B)\otimes H^n(S^n))$$ (the coefficients of $R$ are omitted for brevity) and moreover that the restriction map $H^*(Sph(\zeta))\to H^*(B)$ can be identified with the projection onto the first summand $H^*(B)\otimes H^0(S^n)\cong H^*(B)$. It then follows from the long exact sequence on cohomology associated to the cofiber sequence $B\to Sph(\zeta)\to T\zeta$ that the map $Sph(\zeta)\to T\zeta$ induces an isomorphism between $\tilde{H}^*(T\zeta)$ and the second summand $H^*(B)\otimes H^n(S^n)$.