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?

• A totally reflexive comment on any question about signs in algebraic topology: have you read Tyler Lawson's "In which I try to get the signs right for once"? Jan 6 '20 at 18:03
• Ah, thatʼs cool! Thank you! It also solves my confusion: I forgot to use the sign rule $(f\otimes g)(a\otimes b) = (-1)^{|g|\cdot |a|} \cdot f(a)\otimes g(b)$. Jan 6 '20 at 21:13
• When in doubt, invoke the Koszul rule of signs: if one symbol passes through another, multiply by $-1$ raised to the product of their degrees. Jan 10 '20 at 15:12

• Yes, I think that everything should be fine. I was just confused because I missed the sign rule for $(f\otimes g)(a\otimes b)$. Jan 20 '20 at 22:58