Cartan Formula for Steenrod square on cocycles Let $x_n,y_n,\cdots$ be cocycles in $Z^n(X,\mathbb{Z}_2)$ (not cohomology classes in $H^n(X,\mathbb{Z}_2)$).
Let $Sq^k(x)\equiv x_n \cup_{n-k} x_n$ be the Steenrod square (This definition is valid for cocycles).
Is the following expression valid for cocycles
$$
Sq^k(x_n\cup y_m) =\sum_{i=0}^k  Sq^{i}(x_n)\cup Sq^{k-i}(y_m)
$$
The above Cartan Formula is true for cohomology classes. I like to ask if it is even true for cocycles?
 A: This is not necessarily true. The first warning one should give is that there are multiple definitions of "$\cup_{i}$" on cocycles, and different ones satisfy different identities. However, this is unlikely to hold in general.
One of the few examples that can be calculated model-independently is the following.
If $n=m=1$ and $k=2$ the left-hand side is $x\cup y \cup x \cup y$, while the right-hand side is $x \cup x \cup y \cup y$, which are almost always distinct cochains using the Alexander-Whitney product (this assumes you define $Sq^k x = 0 $ for $k > |x|$).
One of the simpler example that I can find for the explicit versions of the cup-$i$ operations defined by McClure and Smith in "Multivariable cochains and little n-cubes" is when $k=0$. I'm finding that the left-hand side usually has an extremely large number of cross-terms which don't ultimately contribute. In their notation, for example, when $n=m=1$, the right-hand side (in their notation) is
$$
\langle 131242 \rangle(x, y, x, y)
$$
while the left-hand side is
$$
\langle 131234 \rangle(x, y, x, y) +
\langle 123234 \rangle(x, y, x, y) +
\langle 134124 \rangle(x, y, x, y) +
\langle 123424 \rangle(x, y, x, y)
$$
and the fact that all of these are equivalent to $x \cup y$ is non-obvious.
