Let
- $d\in\mathbb N$;
- $x^\ast\in\mathbb R^d$;
- $k\in\mathbb N$;
- $f:\mathbb R^d\to\mathbb R^k$ be Fréchet differentiable at $x^\ast$;
- $v^\ast:=f(x^\ast)$;
- $A:={\rm D}f(x)$;
- $M:=f(\mathbb R^d)$.
Assume there is a $C^1$-diffeomorphism from an open neighorhood $\Omega$ of $v^\ast$ in $M$ onto an open subset $U$ of $\mathbb H^k:=\mathbb R^{k-1}\times\{0\}$; i.e. $\Omega$ is a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^k$.
We can easily show that$^1$ $\mathcal R(A)\subseteq T_{v^\ast}\:M$. If $A$ is surjective, then this is even an equality.
Now assume $\operatorname{rank}A=k-1$. How can we show that $\mathcal R(A)=T_{v^\ast}\:\partial M$?
My idea is as follows: Assume the claim is wrong. Since $\dim\mathcal R(A)=k-1=\dim T_{v^\ast}\:\partial M$, it must hold $\mathcal R(A)\not\subseteq T_{v^\ast}\:\partial M$. So, there is a $v\in\mathcal R(A)\setminus T_{v^\ast}\:\partial M$. Since $v\in\mathcal R(A)$, there is a $h\in\mathbb R^d$ with $v=Ah$ and we easily see that $$\gamma(t):=f(x^\ast+th)\;\;\;\text{for }t\in\mathbb R$$ is a $C^1$-curve on $M$ through $x^\ast$ with $\gamma'(0)$. Thus, $$v\in T_{v^\ast}M\tag1.$$ By $(1)$ and $v\not\in T_{v^\ast}\:\partial M$, we can conclude that $v$ is either inward or outward pointing.
This should be a contradiction to the definition of $M$, since for small enough $t$, either $\gamma(t)$ or $-\gamma(t)$ should move outside the set $M=f(\mathbb R^d)$; which is clearly impossible by definition of $\gamma$.
But how can we argue rigorously?
$^1$ The tangent space $T_{v^\ast}\:M$ is the set of all $\gamma'(0)$, where $\gamma$ is a $C^1$-curve on $M$ though $v^\ast$.