Compute $S_n(X;G)$ with $X=\mathbb{R}P^2$ and $G=\mathbb{Z}/2\mathbb{Z}$ Consider $X=\mathbb{R}P^2$ and $G=\mathbb{Z}/2\mathbb{Z}$, I want to compute the cellular chain complex $S_{\ast}(X;G)$.
An element in $S_n(X;G)$ is of the form $\sum_{i=1}^n g_i\alpha_i$ with $g_i\in G$ and $\alpha_i:\Delta^n\to X$.
The result should be $$0\to G\to G\to G\to 0,$$ where every map is the zero map.
So how do we get this groups and the maps?
 A: $\mathbb R\mathrm P^2$ has a CW structure of three cells: $e_0,e_1,e_2$, where $e_i$ has dimension $i$.  The attachment map of $e_1$ maps both ends to $e_0$.  The attachment map of $e_2$ maps the boundary circle to the $1$-skeleton by a degree-two map.
The cellular chain complex has groups $S_n(\mathbb R\mathrm P^2;\mathbb{Z}/2\mathbb{Z})$ generated by $e_n$, for $n=0,1,2$.  The other groups are trivial.
To compute the boundary maps in general, for the boundary map $\partial_n:S_n(\mathbb R\mathrm P^2;\mathbb{Z}/2\mathbb{Z})\to S_{n-1}(\mathbb R\mathrm P^2;\mathbb{Z}/2\mathbb{Z})$ with $n>1$, we make a matrix $(a_{ij})_{ij}$ recording the degree of the boundary map from the $j$th $n$-cell to the quotient of the $(n-1)$-skeleton collapsing everything but the $i$th $(n-1)$-cell.  The $n=1$ case is the same as for singular homology.  In the case of $\mathbb R\mathrm P^2$:


*

*Since the attachment map of $e_1$ maps both $S^0$ points to the same point, the boundary map is $0$, so the matrix is $[0]$.

*Since the attachment map of $e_2$ maps the boundary $S^1$ to the $1$-skeleton (an $S^1$) by a degree-2 map, the matrix is $[2]$, which with coefficients in $\mathbb{Z}/2\mathbb{Z}$ is $[0]$.


Therefore, the cellular chain complex is $$0\to\mathbb{Z}/2\mathbb{Z}\xrightarrow{0}\mathbb{Z}/2\mathbb{Z}\xrightarrow{0}\mathbb{Z}/2\mathbb{Z}\to 0$$
