Right action on universal-cover chains

Let $X$ be a nonempty connected CW complex with basepoint $x$, let $p:\tilde{X}\to X$ be a universal cover (Universal covering spaces, Every nonempty path-connected locally path-connected semilocally simply connected space has a universal cover), put $\pi ={\pi }_1(X,x)$, and let $R$ be a commutative unital ring. Identify $\pi$ with the deck group by the no-reversal isomorphism of For a path-connected locally path-connected semilocally simply connected base, the deck group of a universal cover is isomorphic to the fundamental group. Write $gc$ for the induced left deck action on a singular or cellular chain.

The right group-ring action on universal-cover chains is $c\cdot g:=g^{-1}c(g\in \pi ),$ extended additively and $R$-linearly to [[def-group-ring|$R[\pi ]$]]. It is a right action because $(c\cdot g)\cdot h=h^{-1}(g^{-1}c)=(gh{)}^{-1}c=c\cdot (gh),c\cdot 1=c\mathrm{.}$ Here the multiplication $[g][h]=[gh]$ and the unit $[1]$ are those constructed in The group ring $R[G]$ is a unital $R$-algebra with basis $G$, and each $g\in G$ is a unit of $R[G]$; bilinearity extends the displayed group action uniquely to all finite formal sums in $R[\pi ]$. Every deck map is cellular for the lifted CW structure and commutes with the singular and cellular boundaries. Hence $\mathrm{\partial }(c\cdot g)=(\mathrm{\partial }c)\cdot g,$ and $C_{\ast }^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(\tilde{X};R)$ and $C_{\ast }^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(\tilde{X};R)$ are chain complexes of right $R[\pi ]$-modules.

Choosing one lift and one orientation of every cell of $X$ makes each cellular chain group a free right $R[\pi ]$-module on those lifted oriented cells. This is only a basis choice, not part of the chain complex. Empty chain degrees and the zero ring give zero modules. The inversion in the definition is forced by the library's first-loop-first deck convention; omitting it would reverse the balanced tensor formulas below.