Is identification of double points of an immersion smooth? Let $f:M^m\to N^n$ be a generic map between smooth manifolds $n>m$. Depending on the pair $(m,n)$ generic maps will have a singular set of double points $\Sigma_2\subset M$.
Let $\phi:\Sigma_2\to \Sigma_2$ be the map of sets that  sends $x\in \Sigma_2$ to the other point $y\in \Sigma_2$ such that $f(x)=f(y)$.
$\phi$ can be thought as a $\mathbb{Z}/2$-action  on $\Sigma_2$.
For the sake of concreteness one can think of a generic immersion of a $M^3\to N^4$, then the set of double points is generically of dimension 2.


*

*Is $\Sigma_2$ a smooth submanifold?

*Is $\phi$ a smooth or at least a continuous map? In other words how bad is the action?

*How does this generalize to triple points and $n$-points? I.e., can we have any kind of group of $n$ elements acting?


 A: I think the answer to the first 2 questions is yes. Most of the details are in the thesis of Ralph Herbert:
Herbert, Ralph J., Multiple points of immersed manifolds, Mem. Am. Math. Soc. 250, 60 p. (1981). ZBL0493.57012
The important construction here is the $r$-tuple point manifold $\Delta_r(f)$ of the immersion $f:M^m\looparrowright N^n$, which for $r\ge2$ is defined as follows. Let $F(M,r)\subseteq M^{(r)}$ be the ordered configuration space of $r$-tuples $(x_1,\ldots , x_r)$ of points of $M$ such that $x_i\neq x_j$ whenever $i\neq j$; it is an open submanifold of $M^{(r)}$. Now consider the restriction of the $r$-th Cartesian power of $f$ to this configuration space, which will by abuse of notation will be denoted $f^{(r)}:F(M,r) \looparrowright N^{(r)}$. Generically, $f^{(r)}$ is transverse to the thin diagonal $d_r(N)= \{(n,\ldots , n)\}\subseteq N^{(r)}$ (see Herbert; also Gollubitsky and Guillemin, Stable mappings and their singularities, Chapter III, Corollary 3.3). Then $$\Delta_r(f) :=(f^{(r)})^{-1}(d_r(N))= \{(x_1,\ldots , x_r)\in F(M,r) \mid f(x_i)=f(x_j)\mbox{ for all }1\le i,j\le n\} $$
is a submanifold of $F(M,r)$ of codimension $rn-n$, therefore of dimension $rm-(rn-n)=n-r(n-m)$. If $M$ is a closed manifold, then so is $\Delta_r(f)$ (compactness is not obvious, it uses the fact that $f$ is locally an embedding). Note that $\Delta_r(f)$ carries smooth free actions of the symmetric group $\mathfrak{S}_r$ and the symmetric group $\mathfrak{S}_{r-1}$ which permute the last $r$ and $r-1$ coordinates, respectively. Let $M_r(f):=\Delta_r(f)/\mathfrak{S}_{r-1}$.
Now consider the map
$$
\mu_r(f): M_r(f)\to M,\qquad (x_1,[x_2,\ldots , x_r])\mapsto x_1
$$
given by projection onto the first co-ordinate. This $\mu_r(f)$ can be shown to be an immersion. Its image is the set of $x\in M$ such that $|f^{-1}f(x)|\ge r$.
Now restrict to the case $r=2$. The immersion $\mu_2(f):M_2(f)\looparrowright M$ is not an embedding when $f$ has triple points or higher. But if we remove the points of the domain where $\mu_2(f)$ fails to be injective, we get an injective immersion onto you $\Sigma_2$ (the "pure" double points), which in fact is an embedding. (A perhaps more convincing argument using general position which works for all $\Sigma_r$ is given on page 25 of Herbert.)
Thus $\Sigma_2\subseteq M$ is an embedded submanifold, and the involution $\phi:\Sigma_2\to \Sigma_2$ is smooth as it is conjugate to the restriction of invoution on $M_2(f)=\Delta_2(f)$ which permutes the factors.
For triple points and higher, I don't see any group action on $\Sigma_r$, since when $f^{-1}f(x)=\{x,x_2,\ldots , x_r\}$ there is no natural way to order the set $\{x_2,\ldots, x_r\}$.
