 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 ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, 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 suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. 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 pairsThe 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 ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ 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 $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ 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 $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf 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.