Let $R[G]$ be a group ring. Then, I have two questions about the chain complex, $C_n(\overset{\sim}{X}; \mathbb{Z}[\pi_1(X)] )$ (where$R=\mathbb{Z}, G=\pi_1(X)$, and $\overset{\sim}{X}$ is a universal cover of $X$). For example, let $M=S^1$ and $\overset{\sim}{M} $ be a universal cover of $M$.(of course, $\overset{\sim}{M} = \mathbb{R} $) Then, I want to think about the following chain complex, $C_1(\overset{\sim}{M}) \overset{\partial}\longrightarrow C_0(\overset{\sim}{M}) $. I based on the following clip.
The first question, my thought is conflicted in the clip. consider $C_0(\overset{\sim}{M})$
My thought : The chain $C_0(\overset{\sim}{M})$ is a $~\underline{left}~$ $\mathbb{Z} [\pi_1(M)]$-module, we can consider
$$ \mathbb{Z} [\pi_1(X)] \times C_0(\overset{\sim}{M}) \to C_0(\overset{\sim}{M}) , (\sum m_i\gamma_{i} ,\overset{\sim}{x}) \mapsto \left ( \sum m_i\gamma_{i} \right )(\overset{\sim}{x}):= \sum m_i\gamma_{i} \overset{\sim}{x} ..........(★) $$ where $m_i \in \mathbb{Z}, \gamma_{i} \in \pi_1(M)$. I understand (★) as a deck transformation. Indeed, $\mathbb{Z}[\pi_1(M)]=\mathbb{Z}[t,t^{-1}]$ This means that the map $(1\cdot \gamma_{i})( \overset{\sim}{x})=\gamma_{i}\overset{\sim}{x} $ means that the vertex $\overset{\sim}{x}$ moves up one floor along the positive orientation like the below figure, so the scalar $m_i \in \mathbb{Z}$ determines both the orientation the vertex moves and how many floor goes up and down. (Of course, such idea is also valid in $C_1(\overset{\sim}{M})$, i.e, $\mathbb{Z} [\pi_1(X)] \times C_1(\overset{\sim}{M}) \to C_1(\overset{\sim}{M}) $). Am I on the right track?
Lecture : However, according to the clip, the lecturer mentioned that the following chain complex $C_*(\overset{\sim}{M})$ is a $~\underline{right}~$ $\mathbb{Z} [\pi_1(M)]$-module. And then he explained that $C_1(\overset{\sim}{M})=\mathbb{Z}[t,t^{-1}](=\mathbb{Z} [\pi_1(X)] )$, But, such statement is not plausible when considering my thought, the definition of $\mathbb{Z}[\pi_1(M)]$-module. How justify the lecturer's idea? And why the following chain complex should be $~\underline{right}~$ $\mathbb{Z} [\pi_1(M)]$-module?(Actually, I found the clue in page 328 in Hatcher's textbook
a $~\underline{left}~$ $\mathbb{Z} [\pi_1(M)]$-module is regarded as $~\underline{right}~$ $\mathbb{Z} [\pi_1(M)]$-module by setting $\alpha \cdot \gamma := \gamma^{-1} \cdot \alpha$....
, but I cannot see why such setting is reasonable.
The second question is how to define the boundary map. According to the lecture, the lecturer said that the boundary map $\partial=t-1$,
$$C_1(\overset{\sim}{M}) \overset{\partial ~\overset{?}{=}~ t-1}\longrightarrow C_0(\overset{\sim}{M}) $$
But, I cannot surely understand why the boundary operator becomes $(t-1)$. To begin with, the natrual way is the definition of the boundary. Thus, when computing difference between two points, (if $a \in C_1(\overset{\sim}{M}))$, $ \partial a = \gamma \overset{\sim}{x}- \overset{\sim}{x}=\overset{\sim}{x}(\gamma -1)$, here $(\gamma -1)$ comes up. However, it is obviously strange to say $\partial :=t-1$ based on the following result.
