Cellular homology of projective space $\mathbb{R}P^n$ For the projective space the cell decomposition is $e_0 \cup \dots \cup e_n$ and the attaching map is $a_1 a_1 \dots  a_k a_k$ for the $k$-th cell.
So for $k \leq n$ I thought that this means that the boundary map between the cellular chain groups is multiplication by two because the boundary map of the $k$-th cell coincides with the attaching map. So I thought $im \partial_k = \langle a_1 a_1 + \dots + a_k a_k \rangle = 2 \langle a_1 + \dots + a_k \rangle \cong 2 \mathbb{Z}$.

Why is this wrong? In the example on p. 144 in Hatcher the chain complex looks like this 

$$ \dots  \xrightarrow{0}&#160;\mathbb{Z} \xrightarrow{2}&#160;\mathbb{Z} \xrightarrow{0}&#160;\mathbb{Z}\xrightarrow{2}&#160;\mathbb{Z} \dots$$

How can the boundary map be the $0$ map? Many thanks for your help.
you can use of local degree and the fact degree of reflection is $(-1)^{n+1}$ .or if you use $I^n$, $\partial_k=&lt;a_1-(-1)^{k+1}a_1+...&gt;$ because degree of reflection is $(-1)^{n+1}$
you can use of local degree and the fact degree of reflection is $(-1)^{n+1}$ .or if you use $I^n$, $\partial_k=&lt;a_1-(-1)^{k+1}a_1+...&gt;$ because degree of reflection is $(-1)^{n+1}$