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!

"Higher cup-1 product of coboundaries is also a coboundary?"math.stackexchange.com/q/3159473/141334 $\endgroup$ – annie heart Mar 23 at 16:45