= Solution
Use the <based path space and the path-loop fibration> $\Omega S^r\to PS^r\to S^r$, where paths begin at a fixed basepoint and the projection evaluates their final endpoint. The total space is contractible: the <homotopy> $\gamma(t)\mapsto\gamma(st)$, $s\downarrow0$, contracts every path to the constant path. The fiber is path-connected because $S^r$ is simply connected. Hence $H_0(\Omega S^r)=\mathbb Z$ and all positive <homology> of $PS^r$ vanishes.
The <exact sequence> from part (i), for $i\geq2$, gives $H_{i-r}(\Omega S^r)\cong H_{i-1}(\Omega S^r)$. Put $d=r-1$. Taking negative-degree <homology> to be zero, this says $H_j\cong H_{j-d}$ for every $j\geq1$. Starting with $H_0=\mathbb Z$ determines every group:
$$
\boxed{H_j(\Omega S^r;\mathbb Z)=\begin{cases}\mathbb Z,&j\geq0\text{ and }d\mid j,\\0,&\text{otherwise}.\end{cases}}
$$
There are no torsion groups hidden in the recurrence, since each step is an actual integral isomorphism. This is the <integral homology of the loop space of a sphere>.
Choose the unit map $\eta:S^{r-1}\to\Omega S^r$ adjoint to the identity under $\Sigma S^{r-1}\cong S^r$. Loop adjunction gives $\pi_j(\Omega S^r)\cong\pi_{j+1}(S^r)$, so the <loop space> is $(r-2)$-connected and $\pi_{r-1}(\Omega S^r)\cong\mathbb Z$. The adjoint of $\eta$ is the degree-one identity, so $\eta$ sends the generator of $\pi_{r-1}(S^{r-1})$ to that generator. The <Hurewicz theorem> for both $(r-2)$-connected spaces, and its naturality, then show that $\eta_*$ is an isomorphism on $H_{r-1}$. It also is on $H_0$, since both spaces are connected. All other positive degrees below $2r-2$ vanish in both spaces. Thus
$$
\boxed{\eta_*:H_j(S^{r-1};\mathbb Z)\xrightarrow{\cong}H_j(\Omega S^r;\mathbb Z),\qquad j\leq2r-3.}
$$
This proves the required existence with a specific map, without using the <suspension> theorem in advance.
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