# Singular homology: Lifting simplices gives map in homology

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 $$R=\mathbb{Z}/2$$)

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