The index sum of an outward field is the Gauss degree

Given: A compact smooth $m$-manifold with boundary $N\subset {\mathbb{R}}^m$, oriented by the ambient orientation of ${\mathbb{R}}^m$, and a smooth field $Y$ on $N$ with $Y\mathrm{\ne }0$ on $\mathrm{\partial }N$ and only isolated zeros.

The zeros of $Y$ are finitely many: the zero set is closed, and an infinite closed discrete subset of the compact space $N$ would have an accumulation point $x\in N$ at which continuity gives $Y(x)=0$ while every neighbourhood of $x$ contains other zeros, contradicting isolatedness. (Isolated zero and local index of a vector field)

The index of an isolated zero is the degree of $x\mapsto Y(x)\mathrm{/}\mathrm{\mid }Y(x)\mathrm{\mid }$ on a small sphere around the zero, with the standard orientations, and for $m=1$ the reduced degree of that map $S^0\to S^0$ (Isolated zero and local index of a vector field, The reduced degree of a map into the 0-sphere).

For $m\ge 2$, a regular value exists by Morse-Sard for smooth manifolds and Regular values have null complement and are dense. For a proper smooth map $F:C\to S^{m-1}$ from a nonempty connected closed oriented $(m-1)$-manifold $C$ and a top form $\omega$ on $S^{m-1}$, ${\int }_C{F}^{\ast }\omega =\mathrm{deg}(F){\int }_{S^{m-1}}\omega ,$ where the degree is the closed-manifold degree of Degree of a map between oriented closed manifolds, equal to the compact-support cohomological degree of Regular-value formula for degree, and where a normalized volume form with integral one exists (Positive volume form on an oriented manifold, Integral of a compactly supported top form, Degree is well defined and independent of the normalized top form).

Under ${\mathrm{A}\mathrm{C}}_{\omega }$, manifold Stokes holds for a compact oriented manifold with boundary and a smooth $(m-1)$-form $\alpha$: ${\int }_{\mathrm{\partial }{N}^{\mathrm{\prime }}}\alpha ={\int }_{N^{\mathrm{\prime }}}d\alpha$, the boundary carrying the induced boundary orientation (The general Stokes theorem, Induced boundary orientation).

For $m=1$: $\mathrm{\partial }N$ consists of finitely many points with signs $\epsilon (x)$, and ${\sum }_{x\in \mathrm{\partial }N}\epsilon (x)=0$; the reduced degree of a map $h:\mathrm{\partial }N\to S^0$ is $\frac{1}{2}{\sum }_{x\in \mathrm{\partial }N}\epsilon (x)h(x)$ (Oriented boundary counts of a compact oriented 1-manifold cancel, The reduced degree of a map into the 0-sphere).

If $Y$ is strictly outward on $\mathrm{\partial }N$, with outward unit normal $g$ (Inward, outward, and boundary-tangent vectors), then $⟨Y\mathrm{/}\mathrm{\mid }Y\mathrm{\mid },g⟩>0$ pointwise, so $t\mapsto (tY\mathrm{/}\mathrm{\mid }Y\mathrm{\mid }+(1-t)g)\mathrm{/}\mathrm{\mid }\cdots \mathrm{\mid }$ is a homotopy from $Y\mathrm{/}\mathrm{\mid }Y\mathrm{\mid }$ to $g$; homotopic maps have equal degree, and for $m=1$ a homotopy $S^0\times [0,1]\to S^0$ is constant in the time variable, so the reduced degrees agree (Degree is invariant under proper smooth homotopy, Reduced degree into the 0-sphere is homotopy invariant and multiplicative).