Let $E$ be a commutative monoid with identity $e$, $S\subseteq E$, and let $S'$ be the submonoid of E generated by $S$.
For every invertible element $x$ of E, $x^{*}$ denotes the inverse of $x$.
Let $R(x,y)$ denote the following relation:
$$\text{there exist }a,b\in E\text{ and }p,q,s\in S'\text{ such that }x=(a,p), y=(b,q)\text{ and }aqs=bps;$$
this is an equivalence relation on the product monoid $E\ \times\ S'$, compatible with its law. Denote the quotient monoid $E\times S'/R$ by $E_{S}$.
For $a\in E$ and $p\in S'$, let $a/p$ denote the equivalence class of $(a,p)$ modulo R. Then, mapping
$$\varepsilon: E\longrightarrow E_{S},\ a\mapsto a/e,$$
is a homomorphism.
For every $a\in E$ and $p\in S'$, we have $a/p=(a/e)(e/p)$: i.e., $a/p=\varepsilon(a)\varepsilon(p)^{*}.$
Therefore, $\varepsilon(E) \cup\varepsilon(S)^{*}$ generates $E_{S}$, where $\varepsilon(S)^{*}$ denotes the set of invertible elements of $\varepsilon(S)$.
Now, we have two special cases:
(i) $E_{E}$, the group of fractions/differences of $E$, generated by $\varepsilon(E) \cup\varepsilon(E)^{*}$;
(ii) $E_{S}$, where $S$ is the set of cancellable elements of $E$. In this case, $\varepsilon$ is injective.
Now apply the above to the commutative, additive monoid $\mathbb{N}$ of natural numbers; we get
$$\mathbb{Z}:=\mathbb{N}\times\mathbb{N}\ /R;$$
since this construction satisfies (i) and (ii), it is a group and the associated $\varepsilon$ is injective.
I have seen people define multiplication on $\mathbb{Z}$ as a new law of composition on the quotient $\mathbb{N}\times\mathbb{N}\ /R$. ---But consider the following result:
Lemma 2. Let $E$ be a monoid and $x\in E$.
(1) There exists a unique homomorphism f of $\mathbb{N}$ into E with $f(1)=x$ and $$f(n)=\top^{n}x\text{, for all n in }\mathbb{N},$$ where $\top$ denotes the law of E.
(2) If $x$ is invertible, there exists a unique homomorphism g of $\mathbb{Z}$ into $E$ such that $g(1)=x$ and $g$ coincides with $f$ on $\mathbb{N}$.
(Bourbaki Algebra Chapter 1, § 2, no. 6)
Bourbaki defines integer multiplication by applying Lemma 2 to the case where $\mathbb{Z}=E$:
...for every $m\in\mathbb{Z}$ there exists an endomorphism of $\mathbb{Z}$ characterized by $f_{m}(1)=m$. If $m\in\mathbb{N}$, the mapping $n\mapsto mn$ is an endomorphism of the magma $\mathbb{N}$; hence $f_m(n)=mn$ for all $m,n$ in $\mathbb{N}$.
Okay, let's apply Lemma 2 to an element $m$ of $\mathbb{Z}$. We have a homomorphism $g_{m}:\mathbb{N}\rightarrow\mathbb{Z},n\mapsto nm$, which we can uniquely extend to an endomorphism of $\mathbb{Z}$
$$\bar{g}_{m}:a-p\mapsto g_{m}(a)-g_{m}(p)$$
such that $g_{m}=\bar{g}_{m}\circ\varepsilon$ and
$$\bar{g}_{m}(1-0)=(\bar{g}_{m}\circ\varepsilon)(1)=g_{m}(1)=m;$$
therefore, $\bar{g}_{m}$ satisfies Lemma 2.
(Since we are considering the additive group $\mathbb{Z}$, I have changed $a/p=\varepsilon(a)\varepsilon(p)^{*}$ to $a-p=\varepsilon(a)+(-\varepsilon(p))$.)
I am not yet really comfortable, technically that is, with the notion of "identification" (in the sense of (ii)), and so I am not really sure if I like letting $m$ denote a generic element of $\mathbb{Z}$, since it really is a set of equivalence classes; $m$ is equal to either $\varepsilon (n)$ or $-\varepsilon(k)$ for some $n,k\in\mathbb{N}$.
For example, how (I know that one can) is one to verify that the above notion of multiplication satisfies commutativity? Take $x,y\in\mathbb{Z}$; then presumably one has to show that $\bar{g}_{x}(y)=\bar{g}_{y}(x)$---but what should this proof look like, if one does not want to make use of the "profitable" identification $\mathbb{N}\simeq\varepsilon (\mathbb{N})$?
This is seemingly a rather confused post. I guess I would like someone to clarify two things for me:
(a) How should one utilize identifications, of the sort associated with injective homomorphisms (see (ii)), in proofs?
(b) Is there a more clear way to express this whole business of uniquely extending a monoid homomorphism to a larger domain while preserving its properties?
Thank you in advance.