Opposite-index nondegenerate zeros cancel in a ball

In the parametrization of [F1] the normalized field $u(y):=Y(y)\mathrm{/}\mathrm{\mid }Y(y)\mathrm{\mid }$ on the boundary sphere is smooth and its degree equals ${\mathrm{ind}}_{b^{-1}(p)}Y+{\mathrm{ind}}_{b^{-1}(q)}Y=0$ by [F1] and [F3].

Fix $0<r_0<1$ so that $Y\mathrm{\ne }0$ on the collar $\{r_0\le \mathrm{\parallel }y\mathrm{\parallel }\le 1\}$, put $u_0(v):=Y(r_{0}v)\mathrm{/}\mathrm{\mid }Y(r_{0}v)\mathrm{\mid }$, and note that the ball of radius $r_0$ contains the same two zeros, so [F1] gives $\mathrm{deg}{u}_0=0$ as well; by [F2] applied to $u_0$ there is a smooth nowhere-zero $F_0$ on ${\bar{B}}_{r_0}(0)$ with $F_0(y)=u_0(y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel })$ for $\frac{2}{3}{r}_0\le \mathrm{\parallel }y\mathrm{\parallel }\le r_0$. Put $\mathrm{\Psi }(v,s):=Y(sv)\mathrm{/}\mathrm{\mid }Y(sv)\mathrm{\mid }$ on $S^{n-1}\times [r_0,1]$, choose by [F4] a smooth function $\sigma :[r_0,1]\to [r_0,1]$ with $\sigma (s)=s$ for $s$ near $1$ and $\sigma (s)=r_0$ for $s$ near $r_0$, and define $F:=F_0$ on ${\bar{B}}_{r_0}(0)$ and $F(y):=\mathrm{\Psi }(y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel },\sigma (\mathrm{\parallel }y\mathrm{\parallel }))$ for $r_0\le \mathrm{\parallel }y\mathrm{\parallel }\le 1$: the two formulas agree on the sphere $\mathrm{\parallel }y\mathrm{\parallel }=r_0$, where both equal $u_0(y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel })$ and are independent of $\mathrm{\parallel }y\mathrm{\parallel }$ in a one-sided neighbourhood of it, so $F$ is smooth and nowhere zero on ${\bar{B}}_2(0,1)$, and $F(y)=Y(y)\mathrm{/}\mathrm{\mid }Y(y)\mathrm{\mid }$ on a collar $\{\mathrm{\parallel }y\mathrm{\parallel }\ge 1-\delta \}$ because $\sigma (s)=s$ there.

Choose by [F4] a smooth bump $\mu$ equal to $1$ on a neighbourhood of the collar $\{\mathrm{\parallel }y\mathrm{\parallel }\ge 1-\delta \}$ and supported in a slightly larger zero-free collar, extend $\mu \mathrm{\mid }Y\mathrm{\mid }$ smoothly by zero from that zero-free collar, and write $\psi :=\mu \mathrm{\mid }Y\mathrm{\mid }+(1-\mu )c$ with a positive constant $c$, and put $X^{\mathrm{\prime }\mathrm{\prime }}:=\psi F$; then $\psi$ is smooth and positive on the ball with $\psi =\mathrm{\mid }Y\mathrm{\mid }$ on $\{\mathrm{\parallel }y\mathrm{\parallel }\ge 1-\delta \}$, so $X^{\mathrm{\prime }\mathrm{\prime }}$ is a smooth nowhere-zero field on the ball, and on that collar $X^{\mathrm{\prime }\mathrm{\prime }}=\mathrm{\mid }Y\mathrm{\mid }\cdot Y\mathrm{/}\mathrm{\mid }Y\mathrm{\mid }=Y$. Therefore the field equal to $X^{\mathrm{\prime }\mathrm{\prime }}$ on $B$ (transported back by $db$) and to $X$ outside $\mathrm{int}B$ is smooth, agrees with $X$ on a neighbourhood of $\mathrm{\partial }B$ and outside $\mathrm{int}B$, and is nowhere zero on $B$.

The resulting smooth field $X^{\mathrm{\prime }}$ on $M$ therefore has no zero in $B$ and coincides with $X$ off $\mathrm{int}B$, so its zero set is exactly the zero set of $X$ outside $B$, as claimed.