Cup product formula.

Question

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

4. 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.

Answer 1

1. 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 $m \in \mathbb{Z}$.

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

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

4. 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.

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

6. 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 $X_t$.

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

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