Cup product formula. Here is the section of a paper I was reading named "Note on Cup-Products" by I.M. James:


*A formula for cup-products. The cohomology theory in what follows has coefficients in the ring of integers. Consider a $CW-$complex $X,$ and a pair of elements $a \in H^{p}(X), b \in H^{{q}}(X),$ such that $a \smile b=0,$ where $p, q \geqq 2 .$ Suppose that $H^{p+q-1}(X)$ is finite. Then I define an integer-valued function, $h,$ on $\pi_{p+q-1}(X)$ as follows. Let $\lambda \in \pi_{p+q-1}(X) .$ Let $X^{*}$ denote the complex which is obtained by attaching an oriented $(p+q)$ -cell to $X$ by a map of homotopy class $\lambda$. Let $c$ denote the cohomology class which is carried by the cell. There are unique elements $a^{\prime} \in H^{p}\left(X^{*}\right), b^{\prime} \in H^{q}\left(X^{*}\right),$ which map into $a, b$ respectively, under the injection. Since $a \cup b=0$ in $X,$ there is an integer $m$ such that $a^{\prime} \cup b^{\prime}=m c .$ I define $h(\lambda)=m .$ We prove:

THEOREM (4.1). The function h constitutes a homomorphism of $\pi_{p+q-1}(X)$ into the group of integers.
We prove (4.1) in the following form:
LEMMA (4.2). Let $\lambda_{1}, \lambda_{2}, \lambda_{3} \in \pi_{p+q-1}(X)$ be elements such that $\lambda_{1}+\lambda_{2}$
$+\lambda_{3}=0 .$ Then
$$
h\left(\lambda_{1}\right)+h\left(\lambda_{2}\right)+h\left(\lambda_{3}\right)=0
$$
In what follows, let $t$ be an indexing integer which takes values $1,2,3 .$ Let $X_{t}$ denote the complex which is obtained by attaching a $(p+q)$ -cell $e_{t}$ to $X$ by a map of homotopy class $\lambda_{t},$ so that
$$
X_{2} \cap X_{3}=X_{1} \cap X_{3}=X_{1} \cap X_{2}=X.
$$
Let $X^{\prime}=X_{1} \cup X_{2} \cup X_{3} .$ Consider the injections:
$$
H^{r}\left(X^{\prime}\right) \xrightarrow{j_{t}} H^{r}\left(X_{t}\right) \xrightarrow{i_{t}}{{H}}^{r}(X).
$$
Let $a^{\prime}, b^{\prime}$ denote the cohomology classes of $X^{\prime}$ such that $i_{t} j_{t}\left(a^{\prime}\right)=a,$ $i_{t} j_{t}\left(b^{\prime}\right)=b,$ and let $c_{t}$ denote the class which is carried by $e_{t} .$ Since $a \cup b=0$ in $X,$ there are integers $m_{t}$ such that
$$
a^{\prime} \cup b^{\prime}=m_{1} c_{1}+m_{2} c_{2}+m_{3} c_{3}
$$
However, by the naturality of the cup-product,
$$
\begin{aligned}
j_{t}\left(a^{\prime}\right) \cup j_{t}\left(b^{\prime}\right) &=j_{t}\left(a^{\prime} \cup b^{\prime}\right) \\
&=m_{1} j_{t}\left(c_{1}\right)+m_{2} j_{t}\left(c_{2}\right)+m_{3} j_{t}\left(c_{3}\right) \\
&=m_{t} j_{t}\left(c_{t}\right) .
\end{aligned}
$$
Hence $m_{t}=h\left(\lambda_{t}\right),$ by the definition of $h .$ Since $\lambda_{1}+\lambda_{2}+\lambda_{3}=0,$ by hypothesis, there is a map $f: S^{p+q} \rightarrow X^{\prime}$ which maps $S^{p+q}$ onto each of the three cells $e_{t}$ with degree $1 .$ Let $f^{*}: H^{r}\left(X^{\prime}\right) \rightarrow H^{r}\left(S^{p+q}\right)$ denote the homomorphism induced by $f,$ which is trivial unless $r=p+q .$ Then $f^{*}\left(c_{t}\right)=c,$ the cohomology class of $S^{p+q} .$ Hence
$$
\begin{aligned}
\left(m_{1}+m_{2}+m_{3}\right) c &=f^{*}\left(m_{1} c_{1}+m_{2} c_{2}+m_{3} c_{3}\right) \\
&=f^{*}\left(a^{\prime} \cup b^{\prime}\right) \\
&=f^{*}\left(a^{\prime}\right) \cup f^{*}\left(b^{\prime}\right),
\end{aligned}
$$
which is zero. Therefore $m_{1}+m_{2}+m_{3}=0,$ since $H^{p+q}\left(S^{p+q}\right)$ is freely generated by $c .$ Since $m_{t}=h\left(\lambda_{t}\right),$ this proves $(4.2) .$ The passage from (4.2) to (4.1) is elementary, and will be omitted.
It is worth mentioning that $h(\lambda)$ can equally well be defined to be the functional cup-product [12] of $a$ and $b$ with respect to $\lambda.$

first part


Second part


Third part

I have some questions about it:
1- Why " since $a \smile b = 0$ in $X,$ there is an integer $m$ such that $a' \smile b' = mc$" where $c$ is the cohomology class which is carried by the oriented $(p + q)$-cell. is there a proof for this?
2- Why the section is named " a formula for cup-products", where is the formula?
3- I need an example of spaces which has infinite $H^{p+q-1}(X)$?
4- why the author defined $h(\lambda) = m$ like this? is there any thing in the literature that say that in a situation similar to what we have we should define our function like that? (I got some help that this is a generalization of hopf invariant)
5- Is there a book that contains a proof of naturality of cup product?
6- when the author said "Hence $m_t = h(\lambda_t)$", I do not understand how the previous equality leads to this, what is the relation between $h$ and $j_t$?
7- How is $f^*(c_t) = c$?
8- Why $f^{*}\left(a^{\prime}\right) \cup f^{*}\left(b^{\prime}\right)$ is zero?
Maybe I am overthinking things, frustrated from not understanding and hence beginning to ask trivial questions. Forgive me please if I did that.
Any help will be greatly appreciated.
 A: *

*By Mayer-Vietoris (or excision) applied to $X^* = X \cup_{S^{p+q-1}, \lambda} D^{p+q-1}$, there is an exact sequence $$\cdots \to \mathbb{Z} \{c\} \cong H^{p+q-1} S^{p+q-1} \to H^{p+q} X^* \to H^{p+q} X \to \cdots.$$  The element $a' \cup b' \in H^{p+q} X^*$ is sent to $a \cup b \in H^{p+q} X$, which is $0$ by assumption, so by exactness $a' \cup b'$ can be identified with an element coming from $H^{p+q-1} S^{p+q-1}$ which has the form $mc$ for some $m \in \mathbb{Z}$.


*Perhaps $a' \cup b' = mc$ is the formula for cup products referred to.


*Just take an infinite wedge of $(p+q-1)$-spheres.


*Asking for a "reason" why an author decides to do something depends on the background context and their goals.  As you point out, this is a generalization of the Hopf invariant, which should be reason enough to study it.


*Any good algebraic topology book will do this, for example Hatcher, Proposition 3.10.


*This is the definition of $h$, applied to $j_t(a')$ and $j_t(b')$, which are lifts of $a$ and $b$ in the cohomology of $X$ to the cohomology of $X_t$.


*By construction $f$ maps $S^{p+q-1}$ (which represents $c$) onto $e_t$ (which represents $c_t$) with degree $1$.  So $f^* c_t = 1 \cdot c = c$.


*$f^*(a') \smile f^*(b') = 0$ because $f^*(a') = 0$ and $f^*(b') = 0$, which follows from the fact that $H^* S^{p+q}$ has no cohomology in those degrees.
