This is a theorem about fibre bundles, stated in page 21, Theorem 3.6
Let $\xi= (E,B,p)$ be an $n$-vector bundle. Let $F_0$ be the fibre $F$ without its nonzero element, and $E_0$ be the total space with out the zero section.
The group $H^i(E,E_0; \Bbb Z_2)$ is zero for $i<n$ and $H^n(E,E_0; \Bbb Z_2)$ contains a unique class $u$ such that for each fibre $F= \pi^{-1}(b)$ the restriction, $$ u|(F,F_0) \in H^n(F,F_0; \Bbb Z_2)$$ is the unique nonzero class in $H^n(F,F_0; \Bbb Z_2)$. Furthermore, the correspondence $x \mapsto x \cup u $ defines an isomoprhism $$ \cdot \cup u : H^k(E,\Bbb Z_2) \rightarrow H^{k+n}(E,E_0; \Bbb Z_2)$$
My question is, how exactly is the last map (highlighted) defined?
My intuitive attempt: By definition an element in $H^{k+n}(E,E_0; \Bbb Z_2)$ is represented by some cocycle $u:C_{n}(E,E_0) \rightarrow \Bbb Z_2$.
This again lifts to an element, $u':C_{n}(E) \rightarrow \Bbb Z_2$. Hence we can define $$ x \cup u: C_{k+n}(E) \rightarrow \Bbb Z_2$$ So it suffices to check this vanishes on $C_{k+n}(E_0)$, which does as $u$ vanishes on $C_n(E_0)$. So we have an induced map $x \cup u:C_{k+n}(E,E_0) \rightarrow \Bbb Z_2$.
Now we then have a series of check: that this operation gives an element in cohomology class, and is well defined on cohomology class.
I think my argument (even if it is right) is extremely inefficient, is there a neat way to see this?