I have spent hours on this problem without getting far. The actual statement is as follows:
Suppose that $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a smooth map, $n>1$, and let $K \subset \mathbb{R}^{n}$ be compact and $\epsilon > 0$. Show that there exists a map $f':\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ such that $df'_{x}$ is never zero, but $|f - f'| < \epsilon$ on $K$.
In the book we are given a hint to start with $F:\mathbb{R}^{n} \times M(n) \rightarrow M(n)$ defined by $F(x, A) = df_x + A$ and show that it is a submersion, then find an A in M(n) such that $F_{A} \pitchfork \{0\}$.
First of all, when it says that $df'_{x}$ is never zero, it means that for all $x$, $df'_{x}$ is never the zero matrix right?
For the life of me I can't figure out how to connect the hint to the problem. At this point I think I have spent so much time thinking about it, that I can't get myself out of the little box I have put myself in. The only thing I can think of is that we want to find some $A \in M(n)$ such that $df_{x} + A$ is never the zero matrix, then construct the $f'$ such that $df'_{x} = df_{x} + A \neq 0$ $ \forall x \in \mathbb{R}^{n}$ and $|f-f'| < \epsilon$ on $K$, but I have no idea how to go about finding that A. Ensuring $F_{A} \pitchfork \{0\}$ doesn't seem to do this from my understanding.
I can give some other approaches to the problem I have tried if necessary, but I just need something to connect the hint to the problem.
By the Transversality Theorem, if $F$ is transverse to a submanifold $Z$ (in this case, G&P tell you to choose $Z=\{0\}\subset M(n)$), then for almost all $A$, the map $F_A$ will be transverse to $Z$. Since $n>1$, dimensions tell you that the only way this can happen is for the image of $F_A$ to miss $0$ entirely. (If $f\colon X \to Y$ and $\dim X<\dim Y$, you can only have a regular value by default.)
Now use compactness of $K$ to choose the matrix $A$ small enough (close enough to the zero matrix) so that $\sup_{x\in K} |Ax|<\epsilon$, and set $f'(x)=f(x)+Ax$.