In my hand out of manifold, I found the following lemma but there is no proof there:

Let $U\subseteq\mathbb{R}^m$ be open and pick some $a\in U$. Suppose that $f:U\mapsto \mathbb{R}^n$ is a smooth function such that $df_a$ is full rank $m\leq n$. Then there exists an open subset $V$ of $\mathbb{R}^n$ with $f(a)\in V$, an open subset $U^\prime\subseteq U$ with $a\in U^\prime$ and $f(U^\prime)\subseteq V$, an open subset $O\subseteq \mathbb{R}^{n-m}$, and a diffeomorphism $\theta:V\mapsto U^\prime\times O$ such that $$ \theta(f(x_1,\dots,x_m))=(x_1,\dots,x_m,0,\dots,0) $$ for all $x_1,\dots,x_m)\in U^\prime$.

I tried to prove the above lemma as follows: Since $f$ smooth so continuous. Fix $\epsilon>0$, then we can choose $\delta>0$ such that $f(N_{a}(\delta))\subset N_{f(a)}(\epsilon)$. Moreover, since $df_a$ is full rank then we can choose this $\epsilon$ small enough such that $f|_{N_{a}(\delta)}$ (function $f$ restricted on $N_{a}(\delta)$) is injective. I guess that we can choose $V=N_{f(a)}(\epsilon)$ and $U^\prime=N_{a}(\delta)$. Is my guess correct? Moreover, how to prove the existence of $\theta$ and define $O$? Thank you in advance.


This theorem is a variant of the Implicit/Inverse function theorems. If you look up the proofs of those theorems, I think you'll find that this one follows easily. You've certainly made generally correct first few steps in your attempt at a proof, although there are some problems.

You write:

"can choose $\delta >0$ such that $N_a(\delta) \subset N_{f(a)}(\epsilon)$."

That should be:

"can choose $\delta >0$ such that $f(N_a(\delta)) \subset N_{f(a)}(\epsilon)$."

Also, you should choose $\delta$ so small that $N_a(\delta) \subset U$, or you'll have a hard time proving $U' \subset U$.

For the rest of the proof --- look up the implicit and inverse function theorems. (Or read on.)

Added in response to comments: a proof from Munkres: $$ \newcommand{\RR}{{\mathbb R}}$$ Thm: Let $U$ be an open subset of $\RR^m$; let $f : U \to \RR^n$ be a map of class $C^r$ such that $f$ has rank $m$ at the origin, and $f(0) = 0$. Then there is a $C^r$ diffeomorphism $g$ of a neighborhood of the origin in $\RR^n$ onto another such that $$ gf(x^1, \ldots, x^m) = (x^1, \ldots, x^m, 0, \ldots, 0). $$

Note that in your case, $r = \infty$, and rather than having $f(0) = 0$, you've got a more general $f$. You can fix this by defining $h(x) = f(x+a) - f(a)$; the function $h$ now satisfies the hypotheses of Munkres, and you can work from there. Also note that Munkres uses superscripts for coordinate-indexing.

Proof: We may assume that the submatrix $\dfrac{\partial (f^1, \ldots, f^m)}{\partial(x^1, \ldots, x^m)}$ of $Df$ is nonsingular at zero, since this condition may be obtained by following $f$ byu a suitable non-singular linear map.

Define $F : U \times \RR^{n-m} \to \RR^n$ by the equation $$ F(x^1, \ldots, x^n) = f(x^1, , \ldots, x^m) + (0,\ldots, 0, x^{m+1}, \ldots, x^n). $$ Then $F$ has rank $n$ at zero, for $DF$ has the block form $$ \begin{bmatrix} & | & 0\\ \partial f/ \partial x & | & \\ & | & I \end{bmatrix} $$

Hence $F$ has a local inverse $g$ (Ed: By the inverse function theorem). Now $g$ is a $C^r$ diffeeomorphism of a neighborhood of the origin onto itself, and $$ gf(x^1, \ldots, x^m) = g(F(x^1, \ldots, x^m, 0, \ldots, 0)) = (x^1, \ldots, x^m, 0, \ldots, 0). $$

And that's the proof!

  • $\begingroup$ OK @John. Thanks for your correction $\endgroup$ – Jlamprong Feb 18 '14 at 16:54
  • $\begingroup$ I have looked up the proof of implicit function theorem. But, there is no a method in constructing $ \theta$. Do you have advice? $\endgroup$ – Jlamprong Feb 18 '14 at 22:45
  • $\begingroup$ Actually, this is closer to the inverse function theorem. $\theta$ is, more or less, the inverse of $f$, except that it's defined on a neighborhood of the image of $f$, and carries things in that neighborhood near the image of $f$ back to a product neighborhood with $U'$. Hmm. That's a lot like the proof of the tubular neighborhood theorem, which I cannot recall right now. BTW, in your statement of the theorem, you have $f(U') \subset U$, but it should be $f(U') \subset V$, I believe. Now I've found a proof (in Munkres' Elementary Differential Topology, p. 14). I'll insert it above. $\endgroup$ – John Hughes Feb 18 '14 at 23:19
  • $\begingroup$ Oh, OK @John. Thank you very much for your helps and time $\endgroup$ – Jlamprong Feb 18 '14 at 23:24
  • 1
    $\begingroup$ Well, your hypothesis says that $df_a$ is full rank, so the upper-left block of $DF(0)$ is a rank-$n$ matrix of size $n \times n$. If you apply Gram-Schmidt to the matrix from right to left, you never change the rank, but you clear out the lower left $(n-m) \times m$ block. The resulting matrix has an $m \times m$ full-rank matrix at the upper left, and and $(n-m) \times (n-m)$ identity at the lower right. Its columns are evidently linearly independent. $\endgroup$ – John Hughes Feb 20 '14 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.