Every immersion can be deformed to have only transverse self-intersections

Question

I asked this question some time ago in MSE, but obtained no answer. Maybe this is the right place to post it.

Let $f : M^n \to \overline{M}^{n+k}$ be an immersion between smooth manifolds. Is it true that there exists a smooth map $F : M \times [0,1] \to \overline{M}$ such that the following conditions hold?

1. $F_0 = f$;
2. $F_t : M \to \overline{M}$ is an immersion for every $t \in [0,1]$;
3. $F_1(M)$ has only transverse self-intersections.

(Here, $F_t(p) = F(p,t)$ for every $(p,t) \in M \times [0,1]$.)

If this does not hold in this full generality, is it true for hypersurfaces ($k=1$)? For $\overline{M} = \mathbb{R}^{n+k}$?

Answer 1

The statement follows easily from Sard's lemma.

Suppose $k=\dim N-\dim N$. Cover $M$ by a countable collection of charts $s_i\:U_i\to M$ such that $f\circ s_i$ extends to embeddings $U_i\times \mathbb{R}^k\to N$. Applying Srad's lemma you can remove bad self-intersection from each chart. Combining these perturbations you get the result. One has to be bit careful not to make new bad self-intersection, but is can be easily achieved by making smaller-and-smaller perturbations in each chart.

The proof is very similar to Thom transversality theorem, and I am sure it is well known to someone.