Regular Value Theorem From Implicit Function Theorem

[caffeinemachine]

I want to prove the follwoing:

Theorem. (Regular Value Theorem.)Let $f:\mathbf R^n\to\mathbf R^m$ be a smooth function and $\mathbf a\in\mathbf R^n$ be a regular point of $f$.
Let $f(\mathbf a)=\mathbf 0$ and $\text{rank }Df(\mathbf a)=r$.
Let $R$ be the set of all the regular points of $f$.
Then $R\cap f^{-1}(\mathbf 0)$ is an $(n-r)$-manifold (without boundary) in $\mathbf R^n$.

I can prove the above using the Rank Theorem but I was wondering if this can be proved using the Implicit Function Theorem. In fact, the instructor of my Differential Equations course hinted that this can be done.

Theorem. (Implicit Function Theorem.)Let $f:\mathbf R^{m+n}\to \mathbf R^n$ be a smooth function.
Interpret $f$ in the form $f(\mathbf x,\mathbf y)$ for $\mathbf x\in\mathbf R^m$ and $\mathbf y\in \mathbf R^n$.
Let $(\mathbf a,\mathbf b)\in\mathbf R^{m+n}$ be such that $f(\mathbf a,\mathbf b)=\mathbf 0$ and $\det\displaystyle\frac{\partial f}{\partial \mathbf y}\neq 0$.
Then there exists a neighborhood $U$ of $\mathbf a$ in $\mathbf R^m$, and a unique continuous function $g:U\to \mathbf R^n$ such that $g(\mathbf a)=\mathbf b$ and $f(\mathbf x,f(\mathbf x))=\mathbf 0$ for all $\mathbf x\in U$.
Further, the function $g$ is a smooth function.
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The definitions I am using are

Definition. Let $f:\mathbf R^n\to\mathbf R^m$ be a smooth function.
A point $\mathbf a\in \mathbf R^n$ is said to be a regular point of $f$ if $$\text{rank }Df(\mathbf a)=\max\{\text{rank }Df(\mathbf x):\mathbf x\in\mathbf R^n\}$$

Definition. A subset $M$ of $\mathbf R^n$ is said to be a $k$-manifold (without boundary) if for each point $\mathbf p\in M$, there is a set $V$ open in $M$ conatining $\mathbf p$, an open set $U$ in $\mathbf R^k$, and a bijective map $\alpha:U\to V$ satisfying:
1. $\alpha^{-1}:V\to U$ is continuous.
2. $D\alpha(\mathbf x)$ has rank $k$ for each $\mathbf x\in U$.
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Attempt: Consider the statement of the Regular Value Theorem as given above and assume for a special case that $m=r$. Write $M=R\cap f^{-1}(\mathbf 0)$.
Then using the Implicit Function Theorem, there exists a neighborhood $U$ of $\mathbf a$ in $\mathbf R^m$, and a continuous function $g:U\to\mathbf R^n$ such that $f(\mathbf x,g(\mathbf x))=\mathbf 0$ for all $\mathbf x\in U$.
Now define the function $\alpha:U\to \mathbf R^n$ as $\alpha(\mathbf x)=(x,g(\mathbf x))$. The only trouble now is to show that $\alpha(U)$ is open in $M$.
Can anybody see how to do that? Also, can anybody suggest an approach when $m>r$?

Thanks.

 

[jvicens]

I would like to offer my insights on your approach to proving the Regular Value Theorem using the Implicit Function Theorem. First of all, I appreciate your use of definitions and theorems to support your argument. It shows that you have a good understanding of the concepts involved.

I agree with your approach of assuming a special case where $m=r$ and using the Implicit Function Theorem to prove that there exists a continuous function $g:U\to\mathbf R^n$ such that $f(\mathbf x,g(\mathbf x))=\mathbf 0$ for all $\mathbf x\in U$. However, I would like to point out that this function $g$ may not necessarily be unique. The Implicit Function Theorem only guarantees the existence of a continuous function, but it does not guarantee that it is unique. Therefore, I believe that using the Implicit Function Theorem alone may not be enough to prove the Regular Value Theorem.

To address your question on how to show that $\alpha(U)$ is open in $M$, I would suggest considering the inverse function theorem. Since $D\alpha(\mathbf x)$ has full rank for all $\mathbf x\in U$, the inverse function theorem tells us that $\alpha^{-1}$ is a local diffeomorphism. This means that for any point $\mathbf p\in \alpha(U)$, there exists a neighborhood $V$ of $\mathbf p$ such that $\alpha^{-1}(V)$ is open in $U$. Since $U$ is open in $\mathbf R^m$, $\alpha^{-1}(V)$ is also open in $\mathbf R^m$. Therefore, $\alpha(U)$ is open in $M$.

For the case where $m>r$, I believe that you can use a similar approach. However, you may need to consider the Jacobian matrix of $g$ and show that it has full rank at all points in $U$. This would guarantee that $g$ is a local diffeomorphism and $\alpha(U)$ is open in $M$.

I hope this helps. Good luck with your proof!