= Solution
Choose a generator $[S^n]$ of $\widetilde H_n(S^n;\mathbb Z)\cong\mathbb Z$. The <mapping degree> of a <continuous map> is the integer determined by
$$
\boxed{f_*[S^n]=(\deg f)[S^n].}
$$
For $n\geq1$ this is ordinary top-dimensional <homology>; using <reduced homology> also gives a convention for $n=0$. Domain and target use the same <orientation>, so reversing the chosen generator in both leaves the integer unchanged. <Homotopy invariance of homology> makes the <mapping degree> invariant under <homotopy>.
We interpret the <manifold criterion for a suspension> in the category of nonempty <topological manifolds> without boundary, and “<sphere>” as homeomorphic to the standard <sphere>. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit <homeomorphism> from its <suspension> to $S^{d+1}$ is
$$
[x,t]\longmapsto(\sin(\pi t)x,\cos(\pi t)).
$$
It is a <continuous map> that is a <bijection>: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The <compact-to-Hausdorff continuous bijection theorem> makes it a <homeomorphism>.
Conversely suppose $SX$ is a <topological manifold>. Its open part $X\times(0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is <contractible>, while $U\setminus\{v\}$ admits a <deformation retraction> onto $X$. The <long exact sequence in relative homology> and <excision> give the <local homology from a link> calculation
$$
H_i(SX,SX\setminus\{v\};\mathbb Z)\cong\widetilde H_{i-1}(X;\mathbb Z).
$$
A point of a $(d+1)$-dimensional <topological manifold> without boundary has <local homology> $\mathbb Z$ in degree $d+1$ and zero in all other degrees. Consequently
$$
\widetilde H_j(X;\mathbb Z)\cong
\begin{cases}\mathbb Z,&j=d,\\0,&j\neq d.\end{cases}
$$
Thus $X$ is an <integral homology sphere>. This conclusion alone does not identify its <homeomorphism> type.
If $d\geq2$, it is also <simply connected>. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By <compactness> of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times\{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor as
$$
V\setminus\{v\}\longrightarrow B\setminus\{v\}\longrightarrow U\setminus\{v\}.
$$
The composite induces an <isomorphism> of <fundamental groups>, since both cone punctures are products of $X$ with intervals and their inclusion is a <homotopy equivalence>. The middle punctured ball has the <homotopy type> of $S^d$ and is <simply connected> for $d\geq2$. Hence the composite is also zero on the <fundamental group>, forcing $\pi_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a <sphere>.
The <Hurewicz theorem>, applied successively to the vanishing lower <homology groups>, now gives $\pi_j(X)=0$ for $j<d$ and $\pi_d(X)\cong H_d(X)\cong\mathbb Z$. A representative of its generator is a <continuous map> $S^d\to X$ inducing an <isomorphism> in every <integral homology> degree. <Topological manifolds> have the <homotopy type> of <CW complexes>, so the <homological Whitehead theorem> makes it a <homotopy equivalence>. We have proved that $X$ is a <homotopy sphere>.
At this point the \b[topological classification theorem is essential]: the <topological generalized Poincare theorem> says that a closed <homotopy sphere> is homeomorphic to the standard <sphere>. Applying it completes the converse for $d\geq2$. For $d=1$, the local calculation makes $X$ a connected <compact> one-dimensional <manifold> without boundary, hence a <circle>. For $d=0$, a <compact> zero-dimensional <manifold> is a finite set, and $\widetilde H_0(X)\cong\mathbb Z$ forces exactly two points. Thus it is $S^0$. Altogether,
$$
\boxed{SX\text{ is a topological manifold without boundary}\iff X\cong S^d.}
$$
The last classification step is not a consequence of <homology> alone, and we make no assertion of <diffeomorphism> with a prescribed smooth structure. If the word <manifold> allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\leq1$, although the interval is not a <sphere>.
To construct <sphere maps of arbitrary integer degree>, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta\mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\deg h_m=m$.
Two cone neighbourhoods cover a <suspension>; their intersection admits a <deformation retraction> onto the original space. The reduced <Mayer–Vietoris theorem> gives a natural <suspension isomorphism>
$$
\widetilde H_{q+1}(SX)\cong\widetilde H_q(X).
$$
Naturality means that the <suspension> of a <continuous map> induces the same integer multiplier. This proves <degree under suspension>. Using the standard identification $S(S^k)\cong S^{k+1}$, set
$$
\boxed{f_{n,m}=S^{\,n-1}h_m:S^n\to S^n,\qquad \deg f_{n,m}=m\quad(n\geq1).}
$$
Thus every integer occurs in every positive <sphere> dimension, with negative degrees handled just as well as positive ones.
If “every $n$” includes zero, that final printed assertion is false. The generator of $\widetilde H_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving precisely
$$
\boxed{\deg f\in\{-1,0,1\}\quad\text{on }S^0.}
$$
These <degrees of maps of the zero-sphere> are the explicit exception to the arbitrary-degree construction.
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