Prove Implicit Function Theorem directly from Constant Rank Theorem

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For reference:

($\textbf{Constant Rank Theorem}$)

Suppose $U_0\subset\mathbb{R}^m$ is open and $F:U_0\rightarrow \mathbb{R}^n$ is a $C^r$ map with constant rank $k$ (that is, its Jacobian matrix has constant rank $k$ on $U_0$). Then for any $p\in U_0$ there exist $C^r$ charts $(U,\phi)$ for $\mathbb{R}^m$ and $(V,\psi)$ for $\mathbb{R}^n$, with $p\in U$ and $F(U)\subset V$, such that $$\psi\circ F\circ \phi^{-1}(u_1,\ldots,u_k,u_{k+1},\ldots,u_m)=(u_1,\ldots,u_k,0,\ldots,0)$$

($\textbf{Implicit Function Theorem}$)

Suppose $m>n$, $U_0\subset\mathbb{R}^m$ is open and $F:U_0\rightarrow \mathbb{R}^n$ is a $C^r$ map. Let $p=(a,b)\in U_0$, $a\in\mathbb{R}^n$, $b\in\mathbb{R}^{m-n}$, with $F(p)=q$, and let the Jacobian matrix of $F$ with respect to the first $n$ coordinates of $\mathbb{R}^m$ be invertible at $p$. Then there exist neighborhoods $N_a\subset\mathbb{R}^n$ of $a$ and $N_b\subset\mathbb{R}^{m-n}$ of $b$ and a $C^r$ function $h:N_b\rightarrow N_a$ that assigns to each $y\in N_b$ the unique $x\in N_a$ satisfying $F(x,y)=q$.

($\textbf{Inverse Function Theorem}$)

Suppose $U_0\subset\mathbb{R}^n$ is open and $F:U_0\rightarrow \mathbb{R}^n$ is a $C^r$ map. Let $p\in U_0$ and let the Jacobian matrix of $F$ be invertible at $p$. Then there exist neighborhoods $U\subset\mathbb{R}^n$ of $p$ and $V\subset\mathbb{R}^{n}$ of $F(p)$ such that $F|_U:U\rightarrow V$ is a $C^r$ diffeomorphism.

I understand how to derive the Inverse Function Theorem from the Constant Rank Theorem: the rank of $F$ stays maximal on some neighborhood of $p$. Then, applying the Constant Rank Theorem, $\psi\circ F\circ \phi^{-1}$ becomes the identity on $\phi(U)$ and we get $F|_U:U\rightarrow F(U)$ is a $C^r$ diffeomorphism, with $F(U)\subset\mathbb{R}^n$ open.

My question is: how do we derive the Implicit Function Theorem directly from the Constant Rank Theorem? I do not want to go Constant Rank => Inverse => Implicit Function, since that is well-known. I know that the Jacobian matrix of $F$ with respect to the first $n$ coordinates of $\mathbb{R}^m$ stays invertible around $p$ (constant rank $k=n$), and applying the Constant Rank Thm. gives $$\psi\circ F\circ \phi^{-1}(u_1,\ldots,u_n,u_{n+1},\ldots,u_m)=(u_1,\ldots,u_n),$$ a projection map onto the first $n$ coordinates. But I don't see how to go from there to arrive at the implicit function $h$. Thanks a lot!