In the first section of J. P. May’s *General algebraic approach to Steenrod operations*, May defines for $\pi\subseteq\Sigma_r$ an integer $q\in\mathbb{Z}$ and a commutative ring $\Lambda$, the $\Lambda\pi$-module $\Lambda(q)=\Lambda$ with sign action $\sigma\lambda = (-1)^{qs(\sigma)}\cdot \lambda$ where $(-1)^{s(\sigma)}$ is the sign of $\sigma$.

For a $\Lambda$-chain complex $K$ we consider $K^{\otimes r}(q)= K^{\otimes r}\otimes \Lambda(q)$ with the diagonal action. I assume that this translates to the explicit sign rule for the transposition $\sigma_{i,i+1}$ in $K^{\otimes r}(q)$ $$\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r) = (-1)^{q+|a_i|\cdot |a_{i+1}|}\cdot (a_1\otimes\dotsb \otimes a_{i+1}\otimes a_i\otimes\dotsb\otimes a_r).$$ Am I correct with this? He later considers cycles $a,b\in K_q$ and $c\in K_{q+1}$ with $dc=a-b$. Now let $I$ be the cellular chain complex of the intervall, i.e. $I_1=\Lambda\langle e\rangle$ and $I_0=\Lambda \langle e_0,e_1\rangle$, and $de=e_1-e_0$. We consider the chain map of degree $q$ $$f:I\to K, e\mapsto (-1)^qc, e_1\mapsto a, e_2\mapsto b.$$ This satisfies $dfe=(-1)^qfde$ and hence is a chain map. May now claims that $f^{\otimes r}:I^{\otimes r}\to K^{\otimes r}(q)$ is $R\pi$-equivariant. I don’t see why: We have $$f^{\otimes r}(\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r)) = (-1)^{|a_i|\cdot |a_{i+1}|} f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r),$$ whereas on the other side, we have $$\sigma_{i,i+1}\cdot f^{\otimes r}(a_1\otimes\dotsb\otimes a_r)=(-1)^{q+(|a_i|+q)\cdot(|a_{i+1}|+q)}\cdot f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r).$$ There are equal iff $q\cdot (|a_i|+|a_{i+1}|)$ is even, but this is not necessarily the case. Concretely, if $q=1$ and $\sigma=\sigma_{1,2}$, then $$f^{\otimes 2}(\sigma\cdot e\otimes e_1) = f^{\otimes 2}(e_1\otimes e)=a\otimes (-c)\ne a\otimes c=\sigma\cdot (-c)\otimes a=\sigma\cdot f^{\otimes 2}(e\otimes e_1).$$ What am I missing?