Poincare-Hopf for closed manifolds

Reduction to nondegenerate zeros: by [F1] the zeros of $X$ are finite, and in pairwise disjoint small balls around them $X$ may be replaced by fields whose zeros in those balls are nondegenerate with the same index sum; the replacements paste smoothly with the unchanged field outside and produce a smooth field $X_0$ on $M$ with only nondegenerate zeros and ${\sum }_p{\mathrm{ind}}_p{X}_0={\sum }_p{\mathrm{ind}}_{p}X$. Since the right-hand side $\chi (M)$ does not involve $X$, it suffices to prove the identity for fields with nondegenerate zeros.

An invariant: embed $M$ properly in ${\mathbb{R}}^k$ with $k=2n+1$ by [F3] and let $N$ be a closed tubular neighbourhood given by normal addition (radius $\epsilon >0$ uniform by compactness of $M$) with nearest-point retraction $r:N\to M$. Define $w(z):=z-r(z)+X_0(r(z))$ for $z\in N$; the two summands are orthogonal, since $z-r(z)\perp T_{r(z)}M$ and $X_0(r(z))\in T_{r(z)}M$. Hence $w(z)=0$ iff $z=r(z)\in M$ and $X_0(z)=0$: the zeros of $w$ are exactly the zeros of $X_0$, viewed in $M\subseteq N$. On $\mathrm{\partial }N$ we have $\mathrm{\mid }z-r(z)\mathrm{\mid }=\epsilon$ and the outward normal is $(z-r(z))\mathrm{/}\epsilon$, so $⟨w(z),z-r(z)⟩={\epsilon }^2>0$: the field $w$ points strictly outward, and in particular $w\mathrm{\ne }0$ on $\mathrm{\partial }N$. At a zero $p\in M$ the derivative of $w$ as a map on $N\subseteq {\mathbb{R}}^k$ is $D{w}_p=D{X}_0{}_p\oplus I$ in the splitting $T_{p}N=T_{p}M\oplus T_p{M}^{\perp }$ (the map $z\mapsto z-r(z)$ has derivative the orthogonal projection onto the normal space), so $\det D{w}_p=\det (D{X}_0{)}_p$ and ${\mathrm{ind}}_{p}w={\mathrm{ind}}_p{X}_0$ by the determinant sign formula. Hopf's boundary lemma [F2] applied to $N$ therefore gives ${\sum }_{p:X_0(p)=0}{\mathrm{ind}}_p{X}_0=\mathrm{deg}(g:\mathrm{\partial }N\to S^{k-1}),$ a number depending only on $M$ (through its embedding and tube), not on $X_0$.

Evaluation: choose an excellent Morse function $f:M\to \mathbb{R}$ by [A1] and [F4] and a Riemannian metric $g$, and take $X_0:={\mathrm{grad}}_{g}f$, a field with only nondegenerate zeros. By [F4] each critical point $p$ contributes $(-1{)}^{\mathrm{ind}(p)}$, so the invariant of step 2.1 equals ${\sum }_p(-1{)}^{\mathrm{ind}(p)}=\chi (M)$ by [F5]; combining with steps 1.1 and 2.1 gives ${\sum }_p{\mathrm{ind}}_{p}X=\chi (M)$ for the original field $X$, which is therefore independent of $X$ and equal to $0$ when $X$ has no zeros.

If $M$ is disconnected, apply the identity on each component and add: the index sum splits over the components, and $\chi$ is additive over disjoint unions by [F5], so the same identity holds; the empty zero set is included (the empty alternating sum is $0$).