Understanding the multiplication in celluar chain complex of $\mathbb{R}P^n$
If $n$ is even the sequence goes:
$... 0 \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow^2 \mathbb{Z} \rightarrow^0 \mathbb{Z} \rightarrow^2 \mathbb{Z} \rightarrow^0 ... \rightarrow^0 \mathbb{Z} \rightarrow 0$
And for $n$ odd the sequence goes:
$... 0 \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow^0 \mathbb{Z} \rightarrow^2 \mathbb{Z} \rightarrow^0 \mathbb{Z} \rightarrow^2 ... \rightarrow^2 \mathbb{Z} \rightarrow 0$
Why does the map alternate between multiplication by two and zero?
I understand why it is so for the map from the $2$ skeleton to the $1$ skeleton, you are wrapping the boundary of a disc $D^2$ twice around a circle $S^1$. In my mind, every map would be multiplication by $2$, but obviously this is not true.
Would somebody care to explain? Thanks!