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I got really stuck here for problem 2.3.8 on GP:

Suppose $m > 1$. Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and let $K \subset \mathbb{R}^n$ be compact. Show that for any $\epsilon > 0$ there exists $g: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $|f(x)-g(x)|<\epsilon$ for $x \in K$ and $dg_x \neq 0$ for all $x \in \mathbb{R}^n$.

Hint: Consider $f + Ax$ for $A \in \mathrm{Mat}_{m\times n}$. Define $F: \mathbb{R}^n \times \mathrm{Mat}_{m\times n} \rightarrow \mathrm{Mat}_{m\times n}$ by $F(x,A) = df_x + A$ and apply the regular value theorem.

My attempt: Consider \begin{eqnarray*} dF_{x,A}(0,B) &=& \lim_{t \rightarrow 0}\frac{F(x,A+tB)-F(x,A)}{t}\\ &=& \lim_{t \rightarrow 0}\frac{df_x + A+ tB-df_x - A}{t}\\ &=& B. \end{eqnarray*} Therefore, $\forall m \in \mathrm{Mat}_{m\times n}, \exists B \in \mathrm{Mat}_{m\times n}$, such that $dF_{x,A}(0,B) = m$. In particular, $B=m$. So $dF$ is surjective at the zero matrix {0}, and the dimension of the image of $dF$ is $m \times n$. Also, {0} has dimension 0 and. Finally, $\mathrm{Mat_{m \times n}}$ has dimension $m \times n$. Therefore dim Image $df_x$ + dim {0} = dim $T_y(Y)$. Then by the definition of transversality, Image$(df_x) + Ty(Z)=Ty(Y)$. Hence $F$ is transversal to {0}.

Therefore, all condition for Sard's theorem are met: {0} and $\mathrm{Mat}_{m \times n}$ does not have boundary, and both $F$ and $\partial F$ are transerveral to $Z$. Therefore, $F(x,A)$ is transverse to 0 for almost every $A$, hence $df_x^\prime \neq 0$.

Then I was not able to proceed.

Also, I am not quite sure about which regular value theorem do I am advised to use - is the local submersion theorem on P20? I couldn't find regular value theorem directly addressed on GP, however, I found:

Regular value theorem: If $b \in N$ is a regular value for $F$, then $F^{−1}(b)$ is a manifold of dimension $m − n$.

But I am not sure if this is the theorem I am advised to use, nor how to use it. Thank you very much.

1LiterTears
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  • http://math.stackexchange.com/questions/419887/does-mathrmmat-m-times-n-have-boundary/419894?noredirect=1#comment897488_419894, @Henry T. Horton explains: You should read the conditions in GP68 as "only X is allowed to have boundary, all other manifolds in question must be boundaryless." So X does not have to have a nonempty boundary. This is really a condition imposed on Y, Z, and S, since we are saying they cannot have boundary. X can have any boundary possible, including the empty boundary! – 1LiterTears Jun 14 '13 at 04:41

1 Answers1

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F is a submersion: For fixed $p \in \mathbb{R}^n $, let

$F^p : Mat \rightarrow Mat, A \mapsto F(p,A)$.

Then

$d(F^p) _A= id|_{\mathbb{R}^{nm}}$ for all $A \in Mat$. But we have $dF_{(p,A)} \circ d\iota _A = d(F^p)_A$, where $\iota$ is the inclusion $Mat \rightarrow \mathbb{R}^n \times Mat, A \mapsto p \times A$. Therefore $dF_{(p,A)}$ is surjective. $\Box$

Now the "transversality theorem" on page 68, with $Z= \{0\}$ implies that for almost every $A\in Mat$, we have $F_A :\mathbb{R}^n \rightarrow Mat$ transversal to $\{0\}$. Therefore we may choose such an $A$ with $|Ax| < \epsilon$ for all $x \in K$. Now transversality to $\{0\}$ means $d(F_A)_p:T_p(\mathbb{R}^n)\rightarrow T_0(Mat)$ is surjective for all $p\in F_A^{-1}(0)$. But $T_0(Mat)$ has dimension $nm>n$ (since $m>1$) so that $F_A^{-1}(0)$ must be empty. Since $d(f+A)_x=F_A(x)$, this completes the proof. $\Box \Box$

Let me know if you'd like more details

Tim kinsella
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  • Hi Tim, thank you so much. But I am not quite following your argument - would you mind give me more details? To start with, could you explain what do you mean by $F^p : Mat \rightarrow Mat, A \mapsto F(p,A)$? – 1LiterTears Jun 12 '13 at 17:12
  • Also, thanks a lot for pointing me to GP 68, but I could quite employ it in this proof. So {0} is boundaryless, right? Then, we need to show $F$ transversal with {0} in order to use this theorem. Then I get lost... – 1LiterTears Jun 12 '13 at 17:14
  • I got lost because the Transversality theorem calls for only $X$ has boundary. But $\mathbb{R}^n$ is boundaryless. Also, how could {0} be a submanifold of Mat$_{m\times n}$ - unless it represents the 0 matrix? – 1LiterTears Jun 12 '13 at 17:43
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    $F^p$ is the function you get from $F$ when you hold the first variable constant (the constant p). I.e. $F^p(A) := F(p,A)$. Yes ${0}$ is a set consisting of the zero matrix. it is a 0 dimensional submanifold of Mat. "I got lost because..." : actually manifolds without boundary are also manifolds with boundary; their boundaries just happen to be empty. – Tim kinsella Jun 12 '13 at 19:29
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    "Then we need to show F transversal" : A submersion is transversal to any submanifold of its codomain – Tim kinsella Jun 12 '13 at 19:30
  • Hi Tim, this is surprising, why Manifolds without boundary are also manifolds with empty boundary? Then we can just say all boundaryless manifolds are manifolds with empty boundary? – 1LiterTears Jun 12 '13 at 19:46
  • Also, if $F^p(A) := F(p,A)$, then it is from Mat$\times \mathbb{R}^n$ to Mat..? – 1LiterTears Jun 12 '13 at 19:48
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    "Why manifolds without boundary..." Look again at the definition of manifold with boundary. "Also, if F^p(A)..." Do you know how, for instance, partial derivatives are defined? – Tim kinsella Jun 12 '13 at 19:52
  • Thanks - so, could you give me an example of boundaryless space that, unlike manifold, is NOT a space with empty boundary? Thanks again. – 1LiterTears Jun 12 '13 at 20:02
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    I'm not really sure what you're asking; the word "boundary" will mean something else when you're not talking about manifolds. – Tim kinsella Jun 12 '13 at 20:31
  • Hi @Tim, sorry - let me reformulate my question: so all manifold has boundary. The boundaryless manifold has empty boundary. =)? – 1LiterTears Jun 13 '13 at 00:06
  • let us continue this discussion in chat – 1LiterTears Jun 13 '13 at 02:36