Aluffi's proof that \det(AB)=\det(A)\det(B) for commutative rings 1) How does \mathbb{Z} being initial

Step 1
If ${\displaystyle A=\left(a_{ij}\right)}$ and ${\displaystyle B=\left(b_{ij}\right)}$ are in ${\displaystyle M_n\left(R\right)}$, where R is a commutative unitary ring, then there is a homorphism
${\displaystyle \varphi \left(n\right):\mathbb{Z}[x_{11},\cdots ,x_{\cap },y_{11},\cdots ,y_{\cap }]\to R}$
such that
${\displaystyle \varphi \left(n\right)=n\times 1_R}$
for
${\displaystyle n\in \mathbb{Z},\varphi \left(x_{ij}\right)=a_{ij}}$
and
${\displaystyle \varphi \left(y_{ij}\right)=b_{ij}}$
If one proves that ${\displaystyle det\left(XY\right)=detXdetY}$ for the matrices
${\displaystyle X=\left(x_{ij}\right)}$,
respectively
${\displaystyle Y=\left(y_{ij}\right)}$,
then apply ${\displaystyle \varphi }$ to this relation and get
${\displaystyle det\left(AB\right)=detAdetB}$
But you know that
${\displaystyle det\left(XY\right)=detXdetY}$
holds for matrices X, Y whose entries are in a field. This is why the book suggests to look at these matrices in the field of fractions of the ring
${\displaystyle \mathbb{Z}[x_{11},\cdots ,x_{\cap },y_{11},\cdots ,y_{\cap }]}$

Step 1
Prove that
${\displaystyle det\left(AB\right)=det(A_{det\left(B\right)}}$ for square matrices A, B over any integral domain.
First question:
Here is the main point: Consider the ring
${\displaystyle A=\mathbb{Z}[x_1,\cdots ,x_n]=\mathbb{Z}\left[x\right]}$.
Suppose you have two polynomials ${\displaystyle f\left(x\right),g\left(x\right)\in A}$ such that ${\displaystyle f\left(x\right)=g\left(x\right)}$.
Then, for any ring R, we can interpret f(x) and g(x) as polynomials in ${\displaystyle R\left[x\right]}$. Moreover, f(x) and g(x) continue to be equal as polynomials in ${\displaystyle R\left[x\right]}$. Thus, by considering the evaluation map, you can substitute ${\displaystyle x_1,\cdots ,x_n}$ as elements from the ring R and you have an identity.
For example, we have the polynomial equality
${\displaystyle (x_1-x_2)(x_1+x_2)=x_1^2-x_2^2}$
in ${\displaystyle \mathbb{Z}[x_1,x_2]}$ this leads to the fact that
${\displaystyle (a-b)(a+b)=a^2-b^2}$
in any (commutative and unital) ring.
Step 2
The formal proof of this is obtained by using the fact that given elements ${\displaystyle a_1,\cdots ,a_n\in R}$, there is a well-defined (unique) homomorphism ${\displaystyle Z\left[x\right]\to R}$ such that ${\displaystyle x_i\mapsto a_i}$. (The initial-ity of ${\displaystyle \mathbb{Z}}$ in Ring determines the map on the level of the constants uniquenely.)
To apply it to your case, note that the ${\displaystyle det\left(AB\right)}$ and ${\displaystyle det\left(A\right)det\left(B\right)}$ can be viewed as polynomials in ${\displaystyle \mathbb{Z}[x,y]}$ (when you set all the entries of A and B to be interdeterminates) and you have shown that they are equal there. Thus, they are equal over any ring.

Step 1 A Bourbaki-proof: You may use the exterior product to prove the formula for any two matrices A,B (of the same rank) over any commutative unital ring: Let $E:=R\{e_1,\cdots ,e_n\}$ be a free R-module of rank n and let $A,B\in En{d}_R(E)$ be $n\times n$ -matrices with coefficients in R (R any commutative unital ring). It follows ${\wedge }^{n}A,{\wedge }^{n}B\in En{d}_R({\wedge }^{n}E)$ are R-linear endomorphisms of the rank one free R-module ${\wedge }^{n}E\cong R{e}_1\wedge \cdots \wedge e_n:=Re$ It has the property that ${\wedge }^{n}A(ue)=det(A)ue$ is multiplication with the element $det(A)\in R$. From this it follows since ${\wedge }^n$ is a functor (see the below link) that $det(AB):={\wedge }^n(A\circ B)={\wedge }^{n}A\circ {\wedge }^{n}B:=det(A)det(B)$