TITLE: Adem relations of Steenrod square without modding out the coboundaries
QUESTION [8 upvotes]: In the paper Products of Cocycles and Extensions of Mappings,
Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$:
$$ 
Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \  x_n \in C^n(M^d;\mathbb{Z}_2)
$$
Note that the above definition works for any $\mathbb{Z}_2$-valued cochain $x_n$.
My question is that if the Adem relations
$$
Sq^i Sq^j = \sum_{k=0}^{[i/2]} {j-k-1 \choose i-2k} Sq^{i+j-k} Sq^k
$$
are still valid when acting on cochains (or cocycles) without modding out the coboundaries?
(All the references that I can find state Adem relations up to the coboundaries.)

REPLY [5 votes]: Theorem 3.1 in Peter May's "A General Algebraic Approach to Steenrod Operations" gives a better behaved definition for $Sq^k(x)$ for (co)chains $x$ which are not assumed to be cocycles.   It probably makes sense to think about these questions (Adem relations or Cartan formula on the cocycle level) using them rather than just the cup-(n-k) formula.    I find it highly implausible that either of these (Adem relations or Cartan formula) should hold 'on the nose', no matter what definition extending the usual one on (co)cycles one gives.