The ordinary Thom isomorphism says $H^{\*+n}(E,E\_{0}) \simeq H^{\*}(X)$, where $E$ is a vector bundle over $X$ and $E\_{0}$ is $E$ minus the zero section. Now assume that $S$ is a non vanishing section for the vector bundle $E$. In each fiber $E\_{x}$ we remove two points $0\_{x}$ and $S(x)$. Then we put $E\_{0,1}$for the union of all 2-points punctured fibers.


**Motivating by the ordinary Thom isomorphism, my question is**



> 
> What should be a relevant right side of the following equality(equivalency):
> 
> 
> 


\begin{equation}
H^{\*+n}(E,E\_{0,1}) \simeq \;?
\end{equation}


What should be a generalized Thom class?


Does this right side depend on choosing a particular non vanishing section $S$?


It is obvious that we can generalize the main question to multi- point punctured fibers. That is, assume we have m sections $S\_{1},\ldots ,S\_{m}$ such that we have m distinct vector $S\_{1}(x),\ldots,S\_{m}(x)$. We remove these m ponts from each fiber $E\_{x}$we denote the resulting total space by $E\_{1,2,\ldots m}$. We search for a relevant right side for:


\begin{equation}
H^{\*+n}(E,E\_{1,2, \ldots, m})=?
\end{equation}



A generalized Thom Isomorphism
You get two copies of $H^\*(X)$. In general ($k$ pairwise disjoint sections), by excision it's just like disjoint union of $k$ copies of the original bundle, hence $k$ copies of $H^\*(X)$. (An extra observation is that, for one section, the result does not depend on its choice, as any section is homotopic to $0$.)


There's no generalized Thom class: the usual one would do. You can regard it as a class in $H^n(E,E\setminus D)$, where $D$ is a disk bundle containing all the sections. This class restricts to $H^n(E,E\_{\text{whatever}})$ in your notation.


