R-orientations of Manifolds Let $M$ be an $n$-manifold and $R$ a commutative ring with $1$.  Then for every $x\in M$ the relative homology group $H_{n}(M,M-{x};R)$ is isomorphic to $R$ and an $R$-orientation of $M$ is defined to be a function that assigns to each $x\in M$ a unit in $H_{n}(M,M-{x};R)$ subject to a local consistency condition.  This definition is where I am struggling.  This isomorphism is merely a group isomorphism, so how does it make sense to talk about a unit in $H_{n}(M,M-{x};R)$?
There is no canonical group isomorphism $H_{n}(M,M-{x};R)\rightarrow R$, and even if there was this doesn't seem helpful since homology groups aren't rings.  It seems like we are being invited to use what ever ismorphism $H_{n}(M,M-{x};R)\rightarrow R$ in order to associate the homology groups $H_{n}(M,M-{x};R)$ with $R$.  This would be fine if it's the case that if $\alpha$ and $\beta$ are any two isomorphisms $H_{n}(M,M-{x};R)\rightarrow R$ that there exists a ring isomorphism $\phi:R\rightarrow R$ such that $\phi\circ \beta=\alpha\circ id_{H_n}$.  And also if $B$ is a ball in $\mathbb{R}^{n}$ containging $x$ that the isomorphism $H_{n}(M,M-{B};R)\rightarrow H_{n}(M,M-{x};R)$ must send units to units.  But are these things true?
 A: No, it is not merely a group isomorphism. It is a module isomorphism.
Let $R$ be a ring, and consider $R$ as a module over itself. Suppose $\varphi: R \to R$ is a module homomorphism. Then $\varphi(r) = \varphi(1 \cdot r) = \varphi(1) \cdot r$ for all $r \in R$, by the fact that this is a module homomorphism.
So every module endomorphism of $R$ is of the form "multiplication by an element of $R$". Write $\varphi_a: R \to R$ for the map $\varphi_a(r) = ar$.
Then $\varphi_a \varphi_b = \varphi_{ab}$. It follows that $\varphi_a$ is invertible as a module homomorphism if and only if $a$ is invertible as an element of $R$. As products of units are units, it follows that any module isomorphism $\varphi: R \to R$ sends units to units.

Note that "units in $R$" are definable module-theoretically, which provides an alternative proof.
An element $r \in R$ is a unit if and only if the submodule generated by $r$ --- also known as the ideal $(r)$ --- is the entire module $R$. A unit of a ring is the same as a generator of $R$, considered as a module over itself (that is, we say $m \in M$ is a generator of a module if $mR = M$; notice that if $M = R$ considered as a module over itself, the condition $rR = R$ is precisely the condition that $r$ is a unit). As isomorphisms send module generators to module generators, it follows that a module isomorphism $R \to R$ sends units to units.
