Nearby slices of a regular germ have the same zero count

By [L1], choose a representative on a neighbourhood of $0$ for which $f(0,\zeta )={\zeta }^{d}h(\zeta )$ and $h(0)\mathrm{\ne }0$. By continuity from [L2], after shrinking there is $r>0$ such that $h(\zeta )\mathrm{\ne }0$ on $\mathrm{\mid }\zeta \mathrm{\mid }\le r$. Hence the central slice has no zero on $\mathrm{\mid }\zeta \mathrm{\mid }=r$ and exactly the order-$d$ zero at $\zeta =0$ inside $\mathrm{\mid }\zeta \mathrm{\mid }<r$.

The compact set $\{0\}\times \{\mathrm{\mid }\zeta \mathrm{\mid }=r\}$ is contained in the domain of the chosen representative, and step 1.1 gives $\mathrm{\mid }f(0,\zeta )\mathrm{\mid }>0$ there. By continuity from [L2], after shrinking the $z^{\mathrm{\prime }}$-neighbourhood to some $V$ we have $\mathrm{\mid }f(z^{\mathrm{\prime }},\zeta )-f(0,\zeta )\mathrm{\mid }<\mathrm{\mid }f(0,\zeta )\mathrm{\mid }(z^{\mathrm{\prime }}\in V,\mathrm{\mid }\zeta \mathrm{\mid }=r)\mathrm{.}$ Then [L3] applied to the functions $\zeta \mapsto f(z^{\mathrm{\prime }},\zeta )$ and $\zeta \mapsto f(0,\zeta )$ shows that each nearby slice has the same zero count $d$ inside $\mathrm{\mid }\zeta \mathrm{\mid }<r$. The strict boundary inequality also makes $f(z^{\mathrm{\prime }},\zeta )\mathrm{\ne }0$ on $\mathrm{\mid }\zeta \mathrm{\mid }=r$.