= Solution
For the <sphere> <homology> calculation use <singular homology>, <homotopy invariance of homology>, and the reduced <Mayer–Vietoris sequence>, not <cellular homology>. For a point, its <chain group> for <singular homology> is $\mathbb Z$ in every nonnegative degree, and its <boundary operator> is multiplication by $\sum_{i=0}^q(-1)^i$, hence alternately zero and one in positive degrees. Its <homology> is therefore $\mathbb Z$ in degree zero and zero above. <Singular chains> of a disjoint union split into the summand <chain complexes>. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb Z)=\mathbb Z^2$, all positive groups vanish, and $\widetilde H_0(S^0;\mathbb Z)=\mathbb Z$. For $n\geq1$, remove the north and south poles to obtain an <open cover> $U,V$ of $S^n$. Each member is <contractible>, while $U\cap V$ has a <deformation retraction> to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced <exact sequence> therefore supplies
$$
\widetilde H_q(S^n;\mathbb Z)\cong\widetilde H_{q-1}(S^{n-1};\mathbb Z)\qquad(q\geq1).
$$
Also $S^n$ is connected, so its reduced degree-zero group vanishes. Induction gives
$$
\boxed{H_q(S^n;\mathbb Z)\cong\begin{cases}\mathbb Z,&q=0,n,\\0,&q\notin\{0,n\},\end{cases}\qquad n\geq1.}
$$
The separate $S^0$ calculation avoids conflating its two degree-zero generators.
Choose an <orientation> of $S^n$, with $n\geq1$, and let $[S^n]$ be its <fundamental class>. The <mapping degree> is the <integer> determined by $f_*[S^n]=\deg(f)[S^n]$. To define the <local degree> at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairs
$$
(B,B\setminus\{x_0\})\longrightarrow(S^n,S^n\setminus\{y_0\}).
$$
The degree-$n$ <relative homology> groups on both sides are $\mathbb Z$, by a <manifold chart> and <excision>. The <local orientation of a manifold> chooses their generators. The induced map multiplies these generators by an <integer>, denoted $\deg_{x_0}f$. <Excision> and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $Df_{x_0}$ is invertible, the <inverse function theorem> gives $\deg_{x_0}f=\operatorname{sign}\det Df_{x_0}$ relative to the <orientations>. Without isolation in the fiber this pointwise definition need not apply.
For the <degree as a sum of local degrees>, assume $y\in S^n$ has finite fiber $F=\{x_1,\ldots,x_s\}$. Disjoint coordinate neighborhoods and <excision> identify
$$
H_n(S^n,S^n\setminus F;\mathbb Z)\cong\bigoplus_{j=1}^s H_n(S^n,S^n\setminus\{x_j\};\mathbb Z).
$$
The absolute <fundamental class> maps to the tuple of local <orientation> generators. Under the relative map induced by $f$, the $j$th generator goes to $\deg_{x_j}f$ times the generator at $y$, and the resulting map from the <direct sum> adds these contributions. On the other hand, naturality says that first applying the absolute map $f_*$ and then passing to the local group at $y$ gives $\deg(f)$ times that generator. Thus
$$
\boxed{\deg(f)=\sum_{x\in f^{-1}(y)}\deg_x f.}
$$
If the fiber is empty the sum is zero: the map factors through the <contractible> punctured <sphere> and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a <regular value> of a smooth map gives the signed count of its inverse images. For $n=0$, define a <mapping degree> on the reduced generator $[p_+]-[p_-]$: the identity has <mapping degree> one, the swap of the two points minus one, and either constant map zero. With local <orientation> signs positive at $p_+$ and negative at $p_-$, the same sum formula holds. The connected-<sphere> convention above is the standard positive-dimensional one.
Back to article page
