We have a vector bundle $p:E\rightarrow B$ of rank $n$, zero section $\xi:B\rightarrow E$ and denote $E-0 := E-\text{im}( \xi)$. For $x\in B$ let $F_x=p^{-1}(x)$ and for $V$ - nbhd of $x$ we put $E_V=p^{-1}(V)$.
Take generator $U_V\in H^n(E_V,E_V-0)\cong\mathbb{Z}$ and concider such a map $$j^*_x : H^*(E_V,E_V-0)\rightarrow H^*(F_x,F_x-0),$$
which is the induced map of inclusion $j_x :(F_x,F_x-0)\hookrightarrow(E_V,E_V-0).$
I would like to rise $U_V$ to generators in $H^n(F_x,F_x-0)$.
If $V$ is contractible, it implies that $j^*_x$ is an isomorphism and we are done. Otherwise, using Künneth Theorem we get
$$H^*(E_V,E_V-0)\cong H^{*-i}(V) \quad \text{for}\quad *\geq n, $$
and therefore we can take a generator $U_V\in H^n(E_V,E_V-0)$ corresponds to $1\in H^0(V)$ up to above isomorphism. On the intuition level I see more or less that $j^*_x(U_V)$ should be a generator in $H^n(F_x,F_x-0)$ but how to write it in more formal way?
I see that - on the intuition level - in the following way:
We take a generator $1\in H^0(V)$ and due to above isomorphism get the generator $U_V\in H^n(E_V,E_V-0)$. Clearly $F_x\subset E_V$ and this allows us to construct the embedding $j_x$.
Intuitively, since induced map $j^*_x$ takes elements from a bigger set to the smaller one, having generator on whole $H^*(E_V,E_V-0)$ we would obtain it on a fiberwise level ($F_x$). But then, it would seem that correspondence $1\sim U_V$ is only for guaranteeing $U_V$ being the generator.
Indexes in cohomology groups make me confused also; $1$ belongs to $0-$th group and $U_V$ into the $n-$th one.