= Solution
Use the unit $\eta$ just constructed and replace it by the inclusion $A=S^{r-1}\hookrightarrow Y=\operatorname{Cyl}(\eta)$ into its <mapping cylinder>; $Y\simeq\Omega S^r$. Put $m=2r-3$. The <homology> isomorphisms from question 4(ii) and the relative <exact sequence> give $H_j(Y,A;\mathbb Z)=0$ for $j\leq m$.
Both $A$ and $Y$ are simply connected for $r\geq3$. For the degree-two start, the absolute <Hurewicz theorem> identifies their $\pi_2$ with $H_2$; naturality and the <homology> isomorphism show $\eta_*$ is an isomorphism on $\pi_2$, so $\pi_2(Y,A)=0$. The relative <fundamental group> also vanishes. If there were a first nonzero <relative homotopy group> in some degree $3\leq k\leq m$, the <Relative Hurewicz theorem> for a $(k-1)$-connected simply connected pair would identify it with $H_k(Y,A)$, which is zero. Induction therefore gives $\pi_j(Y,A)=0$ for all $j\leq m$.
The relative <homotopy> sequence now makes $\eta_*:\pi_i(S^{r-1})\to\pi_i(\Omega S^r)$ an isomorphism when $i+1\leq m$, and a surjection at $i=m$. Under loop-suspension adjunction, the adjoint of $\eta\circ a$ is $\Sigma a$, so this particular isomorphism is the <suspension> homomorphism. Consequently the requested <Freudenthal suspension theorem> is
$$
\boxed{\pi_i(S^{r-1})\xrightarrow{\ \Sigma\ }\pi_{i+1}(S^r)\text{ is an isomorphism for }r\geq3,\ i\leq2r-4.}
$$
For $i=0$ this is read as the identification of connected-component sets; all degree-one groups here also vanish. The proof is the <homology comparison proof of Freudenthal suspension>.
To compute the specified unstable groups, $S^1$ has contractible <universal cover>, so $\pi_2(S^1)=0$. The <Hopf fibration> $S^1\to S^3\to S^2$ has exact segment $0=\pi_3S^1\to\pi_3S^3\to\pi_3S^2\to\pi_2S^1=0$, giving $\pi_3S^2=\mathbb Z$. The supplied starting value is $\pi_4S^3=\mathbb Z/2$. To pass from the group over $S^{n-1}$ to that over $S^n$, use $r=n,i=n$; the condition $n\leq2n-4$ holds exactly when $n\geq4$. Thus every later step is an isomorphism, and
$$
\boxed{\pi_{n+1}(S^n)=\begin{cases}0,&n=1,\\\mathbb Z,&n=2,\\\mathbb Z/2,&n\geq3.\end{cases}}
$$
This is the <first homotopy group above the dimension of a sphere>. The stable range deliberately excludes the step from $n=2$ to $n=3$, so the different groups there do not conflict with the theorem.
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