Associativity of Steenrod's cup-i product In the paper Products of Cocycles and Extensions of Mappings,
Steenrod introduced the cup-i product (and Steenrod square). I would like to ask if Steenrod's cup-i product associative or not? The paper did not discuss this property (it seems).
 A: The cup-i products are not associative for $i > 0$.
For example, Steenrod's cup-1 product has the following description (taken mod 2 for expedience). For cocycles $f$ of degree $p$ and $g$ of degree $q$, and a simplex of degree $p+q-1$ with vertices $[v_0, v_1, \dots, v_{p+q-1}]$, the cup-1 product is defined by
$$
(f \cup_1 g) [v_0 \dots v_{p+q-1}] = \sum_{i=0}^{p-1} f[v_0, \dots, v_i, v_{q+i}, v_{p+q-1}] \cdot g[v_i, \dots, v_{q+i}].
$$
In other words, you sum up over all ways to apply $g$ to a "middle" portion of the simplex.
If $f$ and $g$ have degree 2 and $h$ has degree 1, we find
$$
\begin{align*}
((f \cup_1 g) \cup_1 h) [v_0, v_1, v_2, v_3] =& (f[v_0, v_2, v_3] g[v_0, v_1, v_2] + f[v_0, v_1, v_3] g[v_1, v_2, v_3]) \\&\cdot (h[v_0,v_1]+ h[v_1, v_2] + h[v_2, v_3]) \\
(f \cup_1 (g \cup_1 h)) [v_0, v_1, v_2, v_3] =& f[v_0, v_2, v_3] g[v_0, v_1, v_2] \cdot (h[v_0, v_1] + h[v_1, v_2]) \\&+ f[v_0, v_1, v_3] g[v_1, v_2, v_3] \cdot (h[v_1, v_2] + h[v_2, v_3])
\end{align*}
$$
and the difference between the two is
$$
f[v_0, v_2, v_3] g[v_0, v_1, v_2] h[v_2, v_3] + f[v_0, v_1, v_3] g[v_1, v_2, v_3] h[v_0, v_1].
$$
