Proof Transverse Submanifolds

Question

Let $f:M\rightarrow\mathbb{R}^p$ differentiable and $N$ submanifold of $\mathbb{R}^p$. Show that for all $\epsilon>0$ exist $v\in\mathbb{R}^p$ whit $||v||<\epsilon$ such that $f(x)=f(x)+v$ is transverse to $N.$

Answer 1

The comments provide the proof already, but since I don't think this is a homework problem (a standard proof can be found in Guillemin & Pollack), I'll just spell everything out.

Define a map $F: M\times \mathbb{R}^p \to \mathbb{R}^p$ by $F(x,v) = f(x) + v$. One easily checks that this is smooth, and moreover, at any $(x,v)\in M\times \mathbb{R}^p$ one computes that the differential is: $$[dF_{(x,v)}] =\begin{bmatrix} \begin{bmatrix} \ & \ & \ \\ \ & df_x & \\\ \ & \ & \ \\ \end{bmatrix} & \begin{bmatrix} 1 & \ & 0 \\ \ & \ddots & \\\ 0 & \ & 1\\ \end{bmatrix} \end{bmatrix} $$ So, the map $F$ is submersion at every point, which implies that $F$ is transversal to $N$. Now, applying the transversality theorem, one concludes that for almost all $v_0\in \mathbb{R}^p$, the map $f_{v_0}(x) := F(x, v_0)$ is transversal to $N$. In particular, since any open ball around the origin $B_\epsilon(0)$ has a positive measure, there exists $v\in B_\epsilon(0)$ such that $f_{v_0}$ is transversal to $N$, as desired.