Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
If the group acts cellularly (that is, permuting cells and their corresponding attachment maps) the differential is always a map of $G$-modules... –  Mariano Suárez-Alvarez Sep 20 '11 at 4:19
    
yes, but I'm trying to prove this without knowing that –  Berry Sep 22 '11 at 3:57
    
But that is part of the definition of a cellular action, Berry. –  Mariano Suárez-Alvarez Sep 22 '11 at 5:17
    
Could you give me a precise definition of cellular action, please? –  Berry Sep 24 '11 at 22:09
    
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. –  Aaron Sep 25 '11 at 23:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.