Poincare-Hopf with outward-pointing boundary

Given: A compact smooth $n$-manifold $M$, $n\ge 1$, and a smooth field $X$ with only isolated zeros, strictly outward along $\mathrm{\partial }M$.

If $\mathrm{\partial }M=\mathrm{\varnothing }$, the statement is Poincare-Hopf for closed manifolds; if $n$ is even and $\mathrm{\partial }M\mathrm{\ne }\mathrm{\varnothing }$, it is The index sum of an outward field on an even-dimensional manifold.

Products of a boundary chart of $M$ with an endpoint half-interval give normal quadrant charts; on boundaryless interiors the ordinary product theorem applies. For odd $n$, the product $W:=M\times [0,1]$, with its two codimension-two corner strata $\mathrm{\partial }M\times \{0\}$ and $\mathrm{\partial }M\times \{1\}$ rounded by the standard corner-rounding convention, is a compact smooth $(n+1)$-manifold with boundary (an even-dimensional one); its boundary is the rounded version of $\mathrm{\partial }M\times [0,1]\cup M\times \{0\}\cup M\times \{1\}$, and the rounding changes only a collar of the corner strata, so $W$ is homotopy equivalent to $M$ (the rounded product is a deformation retract of the original product: in each inward normal quadrant, slide $(r,s)$ along $(1,1)$ to the first point of the retained rounded region. The required nonnegative displacement is continuous because the rounding profile is monotone and transverse to $(1,1)$, and is zero on the retained region. Multiplying that displacement by a homotopy parameter gives a deformation fixing the rounded region, supported in the corner collar. The normal formulas agree along the corner stratum. Thus the rounded product is homotopy equivalent to $M$, so homology is unchanged and $\chi (W)=\chi (M)$) (Products of smooth manifolds have a canonical product smooth structure, Attaching a smooth handle with corner rounding, Smooth handle attachment is independent of corner rounding up to diffeomorphism, Homotopy equivalences induce isomorphisms on singular homology, Euler characteristic of a compact manifold).

A product-type zero is nondegenerate with the product index: if $X$ has a nondegenerate zero at $p$ and $\psi (t)=t-\frac{1}{2}$ has its simple zero at $t_0=\frac{1}{2}$, then $Z(x,t):=(X(x),\psi (t){\mathrm{\partial }}_t)$ has a nondegenerate zero at $(p,t_0)$ with ${\mathrm{ind}}_{(p,t_0)}Z={\mathrm{ind}}_{p}X\cdot \mathrm{sign}{\psi }^{\mathrm{\prime }}(t_0)={\mathrm{ind}}_{p}X$; the linearization is block diagonal with blocks $D{X}_p$ and ${\psi }^{\mathrm{\prime }}(t_0)$ (The index of a nondegenerate vector-field zero, Nondegenerate zero of a vector field).

The field $Z$ is strictly outward along $\mathrm{\partial }W$: on $\mathrm{\partial }M\times [0,1]$ the outward normal of $W$ is the outward normal of $\mathrm{\partial }M$ in $M$ and the inward boundary defining coordinate $r$ satisfies $dr(Z)=dr(X)<0$; on $M\times \{0\}$ the outward normal is $-{\mathrm{\partial }}_t$ and $⟨Z,-{\mathrm{\partial }}_t⟩=-\psi (0)=\frac{1}{2}>0$; on $M\times \{1\}$ the outward normal is $+{\mathrm{\partial }}_t$ and $⟨Z,{\mathrm{\partial }}_t⟩=\psi (1)=\frac{1}{2}>0$ (Inward, outward, and boundary-tangent vectors). Near a lower corner use inward coordinates $r\ge 0$, $s=t\ge 0$; near an upper corner use $r\ge 0$, $s=1-t\ge 0$. In a sufficiently small uniform corner neighbourhood, both $dr(Z)<0$ and $ds(Z)<0$. Choose the standard monotone rounding whose outward conormal is $-adr-bds$, where $a,b\ge 0$ and $a+b>0$. Its evaluation on $Z$ is strictly positive, so $Z$ stays strictly outward on every rounded face as well. The rounding is supported away from all zeros and from $t=1\mathrm{/}2$.

Reduction to nondegenerate zeros of $X$ by The local index is additive under a transverse perturbation can be performed inside the interior of $M$, leaving a neighbourhood of $\mathrm{\partial }M$ fixed, hence preserving strict outwardness.