This is the mirror of [previous post](https://math.stackexchange.com/questions/32757/invariance-of-walls-self-intersection-under-the-regular-homotopy).


For an immersion $f\colon N^n\to M^{2n}$ with fixed lift $\tilde{f}\colon N\to \tilde{M}$ and $N,M,\tilde{M}$ are oriented, first we define a unordered double point set $S\_2[f]=\lbrace(x\_1,x\_2)\in N\times N | x\_1\neq x\_2, f(x\_1)=f(x\_2)\rbrace/(x\_1,x\_2)\sim(x\_2,x\_1)$.


Second, for each $x=[(x\_1,x\_2)]\in S\_2[f]$, we can think an equivariant intersection number $I[x]=a(x)w(x)\in \lbrace \pm g|g\in\pi\_1(M)\rbrace\subset \mathbb{Z}[\pi\_1(M)]$ as follows:


$a(x)\in \pi\_1(M)$ satisfies $a(x)\circ\tilde{f}(x\_1)=\tilde{f}(x\_2)$, here we are regarding $a(x)\colon \tilde{M}\to \tilde{M}$.


$w(x)$ is defined to be $1$ or $-1$ if the isomorphism between oriented vector spaces $(d(a(x)\tilde{f})\_{x\_1},d\tilde{f}\_{x\_2})\colon T\_{x\_1}N\oplus T\_{x\_2}N\to T\_{y}\tilde{M}$, where $y=a(x)\circ\tilde{f}(x\_1)=\tilde{f}(x\_2)$. 


Eventually, we define the Wall's self intersection $\mu(f)=\sum\_{x\in S\_2[g]}I[x]\in \mathbb{Z}[\pi\_1(M)]/\lbrace\sum a\_g(g-(-1)^ng^{-1})\rbrace$ as in [Wall's Surgery on compact manifolds Section 5](http://www.maths.ed.ac.uk/~aar/books/scm.pdf) or [Andrew Ranicki's book page 259](http://www.maths.ed.ac.uk/~aar/books/surgery.pdf).


Here, the the equivalence relation $\sum a\_g(g-(-1)^ng^{-1})$ is added to eliminate the ambiguity in the choice of representative element in $S\_2[f]$, namely $[(x\_1,x\_2)]=[(x\_2,x\_1)]\in S\_2[f]$. 


Now, For $t\in [0,1]$, let $f\_t\colon N^n\to M^{2n}$ be an immersion (or regular homotopy by definition) and the regular homotopy $\tilde{f\_t}\colon N\to \tilde{M}$ is also given.


I want to prove that in this situation $\mu(f\_0)=\mu(f\_1)$. i.e. the self-intersection number is an invariant under the regular homotopy.


Here is the argument given in Ranicki's book page 259.


Let $f\colon N\times[0,1]\to M\times [0,1]$ be the trace of $f\_t$. i.e. $f(x,t)=(f\_t(x),t)$. 


Then, $f\times f^{-1}(\Delta(M\times[0,1])-\Delta(N\times[0,1])$ is 1-dimensional manifolds. 


i.e. The unordered double point set of $f\_1$ is obtained from the unordered double point set of $f\_0$ by a sequence of "birth" and "death" of cancelling pair consisting two points in unordered double point set. Hence, such cancelling pair contributes to the self intersection number by $\underline{a-(-1)^nw(a)a=0}\in Q\_{(-1)^n}(\mathbb{Z}[\pi\_1(M)])$. Hence, $\mu(f\_0)=\mu(f\_1)$. 


I want to prove rigorously that the "birth" or "death" of each cancelling pair contributes to self-intersection number as much as the underlined part. I think that I should compare the orientation of tangent space of $N$ at some double points and tangent space of $M$ at the image of them. But I can't fill in the details.


Are there anybody who can help me?



Regular homotopy invariance of Wall's self-intersection form.
The double points of a generic immersion $f:P^p \to Q^q$ of closed manifolds without triple points constitute a closed $(2p-q)$-dimensional manifold $S\_2[f]$ (defined as in the question). This is also true for a generic immersion $(f,\partial f):(P,\partial P) \to (Q,\partial Q)$ of manifolds with boundary: in this case the double points constitute a $(2p-q)$-dimensional manifold with boundary $(S\_2[f],S\_2[\partial f])$. In particular, this is the case for the trace immersion of the regular homotopy
$$(f,\partial f):(P,\partial P)=N^n \times (I,\partial I) \to (Q,\partial Q)=M^{2n}\times (I,\partial I)$$ for $p=n+1$, $q=2n+1$, with $(S\_2[f],S\_2[\partial f])$ a 1-dimensional manifold with boundary. Following Wall work in the universal cover $\tilde{M}$ : the count of the double points of $\partial f$ is
$$\mu(\partial f)=\delta\mu(f)-(-)^n\overline{\delta\mu(f)}\in Z[\pi\_1(M)]$$ with $\delta\mu(f)\in Z[\pi\_1(M)]$ the count of the double point arcs of $f$. The double point circles of $f$ can be ignored here.


