Transversality is generic

Question

Let $M$ and $N$ be submanifolds of $\mathbb R^n$. I am trying to prove that for almost every $x\in \mathbb R^n$, $M+x$ and $N$ intersect transversely. Intuitively, transversality is a "generic" condition, so if we slightly change $x$, we can make the intersection transverse. However, I do not know how to make this intuition precise. Can anyone explain?

Answer 1

Let's define a map which is critical exactly when $M+x$ is not transverse to $N$. Then you can throw Sard's theorem at the problem and conclude that the non-transverse points have measure zero.

Define $f:M\times\mathbb{R}^n\to\mathbb{R}^n$ by $(m,x)\mapsto m + x$. Define $B = f^{-1}(N)$. Now let $g$ be the projection of $B$ onto the $\mathbb{R}^n$ coordinate.

The idea here is to fix $N$ and let $f$ capture the motion of $M$ through $\mathbb{R}^n$. The preimage of $N$ under $f$, what we're calling $B$, we can think of as a family of sets $(M+x)\cap N\subset M$ parametrized by $\mathbb{R}^n$. You want to verify that for almost every $x$, $(M+x)\pitchfork N$, i.e., at each point $p$ of intersection $(M+x)\cap N$, $T_pM + T_pN = T_p\mathbb{R}^n$.

I'll let you take it from here, unless you'd like me to finish the problem. The gist of what's left is relating the surjectivity of $dg$ to the transveraslity of $M$ and $N$.