Critical images of proper local Fredholm restrictions are nowhere dense

In normal-form coordinates, $DP$ is onto exactly when $D_{v}g$ is onto. Surjectivity of a linear map between the finite-dimensional spaces $K,Q$ is an open condition on its matrix entries. Thus $A:=C\cap \mathrm{Crit}(P)$ is closed in $C$.

The image $B:=P(A)$ is closed in $V$. Indeed, if $y_i\in B$ converges to $y\in V$, choose $x_i\in A$ with $P(x_i)=y_i$. The set consisting of $y$ and the $y_i$ is compact in the metrizable chart $V$. Properness makes its inverse image in $C$ compact. A subsequence of $x_i$ therefore converges in $C$ to $x\in A$, and continuity gives $P(x)=y$. Sequential closedness equals closedness in the metrizable chart.

If $Q=\{0\}$, every $D_{v}g$ is onto and $B=\mathrm{\varnothing }$. Otherwise suppose a nonempty target-coordinate open rectangle $U_0\times Q_0$ lay inside $B$. Fix $u\in U_0$. For the $C^h$ map $g_u:v\mapsto g(u,v)$ on the open finite-dimensional kernel-coordinate domain, every $q\in Q_0$ would be a critical value: each $(u,q)\in B$ comes from some $(u,v)\in C$ with $D_{v}g(u,v)$ not onto. If $h=\mathrm{\infty }$, choose any finite integer $r>\max \{\dim K-\dim Q,0\}$; otherwise take $r=h$. Then [F1] applies to the $C^r$ slice and says its critical values are null, so they cannot contain the nonempty open set $Q_0$. Thus $B$ has empty interior in $V$, and step 2.1 makes it nowhere dense there.

Since $V$ is open in $N$, the closure in $N$ of a relatively nowhere dense subset of $V$ has empty interior: any open set in that closure must meet $V$ (the boundary of an open set has empty interior), contradicting relative nowhere density. Hence $B$ is nowhere dense in $N$ as well.