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.



> 
> 1. Is $\Sigma\_2$ a smooth submanifold?
> 2. Is $\phi$ a smooth or at least a continuous map? In other words how bad is the action?
> 3. How does this generalize to triple points and $n$-points? I.e., can we have any kind of group of $n$ elements acting?
> 
> 
> 



Is identification of double points of an immersion smooth?
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**](http://dx.doi.org/10.1090/memo/0250), Mem. Am. Math. Soc. 250, 60 p. (1981). [ZBL0493.57012](https://zbmath.org/?q=an:0493.57012)


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 $\{x\_2,\ldots, x\_r\}$.


