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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.

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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!

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  • $\begingroup$ 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
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    $\begingroup$ 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

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