Use of Steenrod's higher cup product and the graded-commutativity In Steenrod's   ``Products of Cocycles and Extensions of Mappings (1947),'' which derives  [Theorem 5.1]

$$
\delta(u\cup_{i} v)=(-1)^{p+q-i}u\cup_{i-1}v+(-1)^{pq+p+q}v\cup_{i-1}u+\delta u\cup_{i}v+(-1)^pu\cup_{i}\delta v
$$
where $u\in C^p$, $v\in C^q$ are cochains.

Take $u\in C^2$, $v\in C^3$.
Suppose $u=\delta w\in B^2 \subset C^2$ is a coboundary and $v\in C^3$ is a cochain. We have the following result the cup-1 result:
$$
\delta(\delta w\cup_{1} v)=+(\delta w)\cup v+(-1)v\cup (\delta w)+\delta (\delta w)\cup_{1}v+(\delta w)\cup_{1}\delta v
$$
This means that
$$
\delta(\delta w\cup_{1} v)=+(\delta w)\cup v+(-1)v\cup (\delta w)+ (\delta w)\cup_{1}\delta v
$$
In other words, whether the commutativity of $(\delta w)\cup v-v\cup (\delta w)$ boils down to
$$
(\delta w)\cup v-v\cup (\delta w)=-  (\delta w)\cup_{1}\delta v +\delta(\delta w\cup_{1} v) \tag{1}
$$
My questions:

(I) Whether a 2-coboundary $u=(\delta w)$ and a 3-cohain $v$ are commutative in the normal cup product up to a cobounary term $\delta(...)$? What is the property of graded-commutativity (up to (-1) power of combinations of dimensions) of $(\delta w)\cup v-v\cup (\delta w)$?
(II) Namely,
is the right hand side in eq (1), this term $(\delta w)\cup_{1}\delta v$ is also a coboundary? If so
$$
(\delta w)\cup_{1}\delta v =\delta \alpha?
$$
What is $\alpha$?

See a related issue:
"Higher cup-1 product of coboundaries is also a coboundary?" https://math.stackexchange.com/q/3159473/141334
Thank you!!! <3 Please help/comment/advice/give Refs!

*

*Ref: N. E. Steenrod, Products of cocycles and extensions of mappings, Annals of Mathematics 48 290–320 (1947)
 A: For $(\delta w) \cup v - v \cup (\delta w)$ to be a coboundary, it would need to also be a cocycle: so we would have to have
$$
0 = \delta((\delta w) \cup v - v \cup (\delta w)) = (\delta w) \cup (\delta v) - (\delta v) \cup (\delta w)
$$
Let $X$ be the standard 6-simplex $[v_0,v_1,\dots,v_6]$. Define the following cochains on $X$:
$$
\begin{align*}
v(\sigma) &=
\begin{cases}
1 &\text{if }\sigma = [v_3,v_4,v_5,v_6]\\
0 &\text{otherwise}
\end{cases}
\\
w(\sigma) &=
\begin{cases}
1 &\text{if }\sigma = [v_0,v_1]\\
0 &\text{otherwise}
\end{cases}
\end{align*}
$$
Then we have
$$
\begin{align*}
[(\delta w) \cup (\delta v) - (\delta v) \cup (\delta w)]([v_0,v_1,\dots,v_6]) =\ &
(\delta w)([v_0,v_1,v_2]) \cdot (\delta v)([v_2,v_3,v_4,v_5,v_6]) \\&- 
(\delta v)([v_0,v_1,v_2,v_3,v_4]) \cdot (\delta w)([v_4,v_5,v_6])
\\=\ & w(\partial[v_0,v_1,v_2]) \cdot v(-\partial[v_2,v_3,v_4,v_5,v_6])
\\&- v(-\partial[v_0,v_1,v_2,v_3,v_4]) \cdot w(\partial[v_4,v_5,v_6])
\\=\ & \dots
\\=\ & -1
\end{align*}
$$
Therefore, $(\delta w) \cup v - v \cup (\delta w)$ is not a cocycle.
