Let $X$ be a space, $k=k_1+\dotsb+k_r$ and let $G:=\mathfrak{S}_{k_1}\times\dotsb\times \mathfrak{S}_{k_r}$ act freely on the right on $X$. Fix a commutative ring $R$ and another space $Y$.

Then the singular complex $S_\bullet(X)$ (over $R$) is a right $RG$-module. On the other hand, $S_\bullet(Y)^{\otimes k}$ is a left $RG$-module by permuting the factors.

Starting with a simplex $\sigma$ in $X/G$, we can choose a lift $\tilde{\sigma}$ in $X$. Consider now the following map

$$S_q(X/G)\times \prod_{i=1}^r S_{p_i}(Y) \to (S(X)\otimes_G S(Y)^{\otimes d})_{q+\sum k_ip_i},$$ $$(\sigma,a_1,\dotsc,a_r)\mapsto \tilde{\sigma}\otimes_G(a_1^{\otimes k_1}\otimes\dotsb\otimes a_r^{\otimes k_r}).$$

If $R=\mathbb{Z}/2$ and $k=k_1=2$, then it is clear to me that this map sends tuples of cycles to cycles and tuples of cycles and boundaries to boundaries and thus, induce a map in homology. I am wondering if in general we get an induced map

$$H_q(X/G)\times \prod_{i=1}^r H_{p_i}(Y) \to H_{q+\sum k_ip_i}(S(X)\otimes_G S(Y)^{\otimes d})$$

(Edit: I think it is clear if $\# G=k_1!\dotsm k_r!$ is invertible in $R$. However, I am especially interested in the case $R=\mathbb{Z}/2$)

  • $\begingroup$ what is $\mathfrak{S}_{k_1}$ for you? Symmetric group in $k_1$ elemens? $\endgroup$ – Praphulla Koushik Feb 24 at 10:21
  • $\begingroup$ Yes, exactly, the $k_i$th permutation group. $\endgroup$ – FKranhold Feb 24 at 10:23

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