In the paper [Products of Cocycles and Extensions of Mappings](http://www.maths.ed.ac.uk/~aar/papers/steen5.pdf),
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).



Associativity of Steenrod's cup-i product
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
$$
f[v\_0, v\_2, v\_3] g[v\_0, v\_1, v\_2] h[v\_2, v\_3] + f[v\_0, v\_1, v\_3] g[v\_1, v\_2, v\_3] h[v\_0, v\_1].
$$


