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.
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