Uniqueness of Determinants in Hoffman’s Linear Algebra 
The sequence $(\sigma (1),…,\sigma (n))$ can be obtained from the sequence $(1,2,…,n)$ by a finite number of interchanges of pairs of elements. For example, if $\sigma (1)\neq 1$, we can transpose $1$ and $\sigma (1)$, obtaining $(\sigma (1),..., 1,…)$. Proceeding in this way we shall arrive at the sequence $(\sigma (1),…,\sigma(n))$ after $n$ or less such interchanges of pairs.
For a given permutation $\sigma$, when we pass from $(1,2,...,n)$ to $(\sigma (1),\sigma (2),…,\sigma (n))$ by interchanging pairs, the number of interchanges is always even or always odd. This basic combinatorial fact can be proved without any reference to determinants; however, we should like to point out how it follows from the existence of a determinant function on $n\times n$ matrices.
Let us take $K$ to be the ring of integers. Let $D$ be a determinant function on $n\times n$ matrices over $K$. Let $\sigma$ be a permutation of degree $n$, and suppose we pass from $(1,…,n)$ to $(\sigma (1),…,\sigma (n))$ by $m$ interchanges of pairs $(i, j)$, $i\neq j$. As we showed $(-1)^m=D(e_{\sigma (1)},…,e_{\sigma (n)})$ that is, the number $(-1)^m$ must be the value of $D$ on the matrix with rows $e_{\sigma (1)},…,e_{\sigma (n)}$. If $D(e_{\sigma (1)},…,e_{\sigma (n)})=1$, then $m$ must be even. If $D(e_{\sigma (1)},…,e_{\sigma (n)})=-1$, them $m$ must be odd.

Hoffman’s claim we can get $(\sigma (1),…,\sigma (n))$ from $(1,2,…,n)$ by a finite number of interchanges of pairs of elements. He gave one algorithm to do that. Que: How to rigorously show that algorithm works? IMO algorithm is bit vague, I mean, define $\sigma :J_4\to J_4$ such that $\sigma (1)=2$, $\sigma (2)=4$, $\sigma (3)=1$ and $\sigma (4)=3$. Since $\sigma(1)\neq 1$, $(1,2,3,4)\to (2,1,3,4)$. Since $\sigma(2)\neq 2$, I think we can’t do $(2,1,3,4)\to (4,1,3,2)$. Let $\mu:J_4\to J_4$ such that $\mu=(2,1,3,4)$. Since $\sigma (2)\neq \mu (2)$, $(2,1,3,4)\to (2,4,3,1)$. So algorithm is not straightforward it requires proof. Que: How to really prove following claim: sequence $(\sigma (1),…,\sigma (n))$ can be obtained from the sequence $(1,2,…,n)$ by a finite number of interchanges of pairs of elements.
“……… number of interchanges is always even or always odd” that’s an interesting result. I don’t find Hoffman’s proof using determinant function satisfactory. I think proof is circular. In section 5.2 Hoffman showed existence of determinant function. Let $D_1,D_2:M_n(K)\to K$ be determinant function. Which make sense because we haven’t yet proved uniqueness of determinant function. What if $D_1 (e_{\sigma (1)},…,e_{\sigma (n)})=1$ and $D_2 (e_{\sigma (1)},…,e_{\sigma (n)})=-1$? Since we took arbitrary determinant map $D$, we have $m=\pm 1$. Which is absurd. Now one might argue we get $D(e_{\sigma (1)},…,e_{\sigma (n)})$ either $1$ or $-1$, using definition of alternative. So there is only one value $1$ or $-1$, but that depends on sequence/map $\sigma$, precisely in how many interchanges of pairs we get $(\sigma(1),…,\sigma(n))$ from $(1,…,n)$. But if we know that info/data we know what $m$ is. If we know value of $m$, we don’t need to find $D(e_{\sigma (1)},…,e_{\sigma (n)})$. That’s why I said proof is circular.
There is a simple proof of “…… number of interchanges is always even or always odd”. Proof: Assume towards contradiction, $\exists m,n\in \Bbb{N}\cup \{0\}$ such that $m$ is even and $n$ is odd. Then $D(e_{\sigma (1)},…,e_{\sigma (n)})=(-1)^m=(-1)^n=1=-1$. Thus we reach contradiction.
Que: Why Hoffman’s took $K=(\Bbb{Z},+,\cdot)$ instead of an arbitrary commutative ring?

This video took completely different approach. Defined parity of a permutation $\sigma$ as $(-1)^{\text{# of inversion in $\sigma$}}$. We call $\sigma$ even (or odd), if $\text{# of inversion in $\sigma$}$ is even (or odd).
 A: *

*Induct on $n$. Given a permutation $\sigma$, put $\tau = (\sigma(n), n)$ (with $\tau = 1$ if $\sigma(1) = 1$). Then $\tau \sigma(n) = n$ by construction, so we can consider $\sigma \tau$ as an element of $S_{n-1}$. In other words, embed in $S_{n-1}$ in $S_n$ as the group of permutations fixing $n$. Thus we can write $\sigma \tau = \tau_1 \cdots \tau_k$ for some transpositions (i.e., interchanges of elements) $\tau_1, \dots, \tau_k$ in $S_{n-1}\subset S_n$; that is, $\sigma = \tau_1 \cdots \tau_k \, \tau$.



*Defining the sign $\operatorname{sgn}\sigma$ of a permutation $\sigma\in S_n$ via determinants is odd, since the construction is usually in the opposite direction (even in more abstract treatments that consider the determinant as, e.g., a map on the wedge product of a space). The well-defintion of $\operatorname{sgn}$ does follow from the uniqueness of a homomorphism $GL_n(k) \to k^\times$ for a field $k$ with $\operatorname{char} k\not = 2$ as suggested, but I don't know what Hoffman did to prove the latter.



*The natural embedding of $\mathbb{Z}_2 = \{\pm 1\}$ in $\mathbb{Z}$ is presumably his motivation, but I'd be surprised if taking $K = \mathbb{Q}$ (so that you're working over a field) or $K = \mathbb{C}$ (so that you also have algebraic closure) didn't simplify things. (I also have to admit that using $K$ to denote a ring that isn't a field bothers me a bit, but that's probably just me.)

