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I was trying to answer by myself at this question. This is what I've done:

Let $e^\alpha_n$ be an $n$-cell of $X$ (I use the notations in the previous question). We have the attaching map $\chi^\alpha_n:\partial e^\alpha_n\cong S^{n-1}\rightarrow X_{n-1}$ where $X_{n-1}$ is the $(n-1)$-skeleton. So we have the maps

$\chi^{\alpha\beta}_n:S^{n-1}\rightarrow X_{n-1}\rightarrow\frac{X_{n-1}}{X_{n-1}-e^\beta_{n-1}}\cong S^{n-1}$. The differential on the cellular chain complex is defined by

$d(e^\alpha_n)=\sum_\beta\mathrm{deg}(\chi^{\alpha\beta}_{n})e^\beta_{n-1}$.

If $g\in G$ let's define $g\alpha$ as follows: $ge_n^\alpha=e^{g\alpha}_n$.

A Little computation show that the differential is a map of $G$-modules if and only if $\mathrm{deg}(\chi^{(g\alpha)(g\beta)}_n)=\mathrm{deg}(\chi^{\alpha\beta}_n)$.

Now we have the diagram \begin{matrix} H(S^{n-1}) &\rightarrow& H(S^{n-1})\\\\ \downarrow &&\downarrow\\\\ H(S^{n-1})&\rightarrow&H(S^{n-1}) \end{matrix} where the first orizzontal arrow is induced by $\chi^{\alpha\beta}_n$, the second orizzontal arrow by $\chi^{(g\alpha)(g\beta)}_n$ and the vertical by the multiplication by $g$. What I have to show is that the image of $1$ is the same in the two orizzontal arrow. So any of you could help me in showing this, please?


I asked this question also on mathoverflow

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    A continuous map $f:X\to Y$ is cellular if it preserves skeleta, i.e., $f(X_n)\subset Y_n$. A cellular action is one where the action map $G\times X \to X$ is cellular (note:discrete groups are CW complexes). More generally, we can define group objects and group actions in any category, so it may be helpful to think of this as applying general definitions to the category of CW-complexes. N.B. The cellular approximation theorem makes cell maps a reasonable class of morphisms for homotopy theoretic purposes.2011-09-25

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