= Solution
\b[The degree and its local signs.] Orient $S^n$ and let $[S^n]$ generate its top <reduced homology> $\widetilde H_n(S^n;\mathbb Z)\cong\mathbb Z$. The <mapping degree> is the integer characterized by
$$
\boxed{f_*[S^n]=(\deg f)[S^n].}
$$
For $n\geq1$ this is the usual top <homology> definition using the <fundamental class>; using <reduced homology> also covers $S^0$.
Suppose $f$ is smooth and $p$ is a <regular value>. Each $x\in f^{-1}(p)$ has invertible tangent map $Df_x$, so the <inverse function theorem> makes $f^{-1}(p)$ discrete. It is also closed in the compact <sphere>, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a <local diffeomorphism>. The <Excision theorem> identifies the source local <homology> with the direct sum of one copy of $\mathbb Z$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $Df_x$ preserves or reverses orientation. The map from the global <fundamental class> to these local orientation classes therefore proves the <degree as a sum of local degrees> formula:
$$
\boxed{\deg f=\sum_{x\in f^{-1}(p)}\epsilon_x,\qquad
\epsilon_x=\operatorname{sgn}\det Df_x.}
$$
Here the <determinant> is computed in positively oriented tangent bases. Thus the <mapping degree> counts inverse images with signs, rather than just their cardinality.
\b[The quotient map.] Put $m=n+1$ and write $f=F|_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The <connecting homomorphism>
$$
\partial:H_m(D^m,S^{m-1};\mathbb Z)\longrightarrow
\widetilde H_{m-1}(S^{m-1};\mathbb Z)
$$
is an isomorphism: the disk has zero positive <reduced homology>. It sends the relative <fundamental class> $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on <relative homology> induced by $F$ multiplies this class by $d$, naturality gives
$$
f_*[S^{m-1}]=f_*\partial[D^m,S^{m-1}]
=\partial F_*[D^m,S^{m-1}]=d[S^{m-1}].
$$
Thus $d=\deg f$. Collapsing the boundary gives a <sphere> $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top <reduced homology> with the top <relative homology> of the disk pair. Give this quotient <sphere> the orientation determined by that identification. The relation $qF=\widetilde Fq$ now shows
$$
\boxed{\deg\widetilde F=\deg(F|_{S^n}).}
$$
This is the <quotient-sphere degree identity>.
\b[The graph intersection.] Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times\{0\}$ by its first factor, and orient the <graph of a function> $\Gamma$ by $x\mapsto(x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, <transverse intersection> means that $DF_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.
For the <smooth intersection number> use the ordered tangent spaces $T_xN$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrix
$$
\begin{pmatrix}I&I\\0&DF_x\end{pmatrix}.
$$
Its <determinant> is $\det DF_x$, so the intersection sign is $\epsilon_x=\operatorname{sgn}\det DF_x$. By the same local <Excision theorem> argument, now in <relative homology> at the interior point $0$, the sum of these signs equals the multiplier of $F_*$ on $H_m(D^m,S^{m-1};\mathbb Z)$. Therefore the <graph intersection formula for mapping degree> is
$$
\boxed{\deg(F|_{S^n})=N\cdot\Gamma
=\sum_{x:F(x)=0}\operatorname{sgn}\det DF_x.}
$$
If the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1)^m$; the order above specifies the appropriate convention.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-15-signed-intersections.png]
{title=Three transverse graph intersections with signs plus, minus, plus and total degree one}
{height=360}
The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the <graph intersection formula for mapping degree>: its three zeros have signs $+1,-1,+1$, and its endpoint map has <mapping degree> $1$ on <reduced homology>.
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