 Solution

Because $S^n$ is simply connected for $n\ge 2$, the homological Serre spectral sequence has constant coefficients and only two nonzero columns:Its only possible nonzero differential isUsing the orientation generator of $H_n(S^n)$ to identify both columns with $H_{\ast }(F)$ defines the Wang homomorphismThe kernel and cokernel descriptions of the two surviving columns splice with the filtration of $H_{\ast }(E)$ to give the Wang sequence over a sphere

Let $F_m$ be the homotopy fiber of a degree-$m$ map $f_m:S^n\to S^n$. Apply this sequence to the fibrationAt the bottom, the map between the two copies of $H_n(S^n)$ is multiplication by $m$. It follows thatThe first torsion group can also be seen from ${\pi }_{n-1}(F_m)\cong \mathbb{Z}/m$ and the Hurewicz theorem; the Wang map then propagates it periodically.

The universal coefficient theorem for cohomology givesEvery product of two positive-degree classes is zero. For $n>2$, this follows immediately because the sum of two degrees of the form $k(n-1)+1$ is not of that form. For $n=2$, degree counting does not suffice, since $H^j(F_m)\cong \mathbb{Z}/m$ for every $j\ge 2$. Use instead the multiplicative cohomological Serre spectral sequence forIts transgression in degree one is multiplication by $m$. Every positive integral cohomology class that survives lies in filtration two, while the product of two such classes lies in filtration four; the base has dimension two, so filtration four is zero. Thus the reduced cohomology is a square-zero ideal, andas a graded ring, with zero multiplication on the second summand.