Separating critical values far from the boundary

Choose disjoint relatively compact interior neighbourhoods $U_i$ of the finitely many critical points $p_i$, contained in $W\setminus C$, and smaller neighbourhoods $V_i$ with closures in $U_i$. By [F1], on the boundaryless interior choose bumps ${\rho }_i$ equal to one near ${\bar{V}}_i$ with support in $U_i$, and extend them by zero to $W$. Set $K=W\setminus {\bigcup }_i{V}_i$, a compact set with no critical point.

At each point of $K$, including boundary points, some component of $df$ in a chart is nonzero. Consider all smaller chart neighbourhoods with compact cores on which such a component has absolute value at least a positive number. Their interiors cover $K$, and compactness selects finitely many. Let $m>0$ be the least of these finitely many positive bounds, and bound all derivatives of the ${\rho }_i$ in these coordinate directions on the compact cores. If $K$ is empty, all bounds are vacuous.

Choose arbitrarily small ${\lambda }_i$ such that $f(p_i)+{\lambda }_i$ are pairwise distinct, $g=f+{\sum }_i{\lambda }_i{\rho }_i\in \mathcal{U}$, and each coordinate derivative change on the finite chart cores is less than $m\mathrm{/}2$. Such coefficients exist because the distinct-value conditions exclude finitely many hyperplanes and [A2] makes all smallness conditions open near zero. On $V_i$, $g=f+{\lambda }_i$, preserving the critical points and Hessians; on $K$ one chosen component at every point remains nonzero.

Thus $g$ has exactly the original nondegenerate critical points and has distinct critical values. All bump supports miss $C$, so $g=f$ on $C$. With no critical points, use $g=f$. This proves the assertion without selecting a global metric or silently assuming its choice principle.