Proof Question: Determinant of Block Lower Triangular Matrix

Question

Consider the matrix $B=\left[\begin{array}{c|c}A&0\\ \hline C&D\end{array}\right]$ where $A$ is a matrix of size $k\times k$ and $D$ is a matrix of size $n\times n$. I am trying to prove that $\det(B)=\det(A)\cdot \det(D)$ like user "Zilin J." did here.

---

Here is the area where I get stuck in the proof.

$\begin{align*} \det B &= \sum_{\sigma\in S_{n+k}}\operatorname{sgn}\sigma\prod_{i=1}^{n+k}b[i, \sigma(i)]\\ &= \text{haven't got here yet..}\\ \end{align*}$

-I understand the proof where it states if $\sigma(i) = j$, $i\le k$, and $j > k$, then the corresponding summand $\prod_i b[i,\sigma(i)]$ is $0$. Thus, referring to the determinant above, it is clear we could get rid of those elements in $S_{n+k}$ where that occurs in the lower limit of our sum above.

-But the yellow part is where I get stuck. My question is how are we able to deduce the yellow part below? Else, we consider $\sigma(i)\neq j$ or $i>k$ or $j\leq k$. In other words, if $\sigma(i)=j$, then $i>k$ or $j\leq k$. [Note I'm using $P\implies Q \equiv \lnot P\lor Q$ to get that other words part.]

Any permutation $\sigma\in S_{n+k}$ for which no such $i$ and $j$ exist can be uniquely "decomposed" into two permutations, $\pi$ and $\tau$, where $\pi\in S_k$ and $\tau\in S_n$ such that $\sigma(i) = \pi(i)$ for $i \le k$ and $\sigma(k+i) = k+\tau(i)$ for $i \le n$.

Answer 1

Let $\sigma\in S_{k+n}$ be a permutation such that for all $i\le k$, it is not the case that $\sigma(i)>k$, i.e. it takes $\sigma(i)\le k$.
That just defines $\pi\in S_k$, as $\{1,\dots,k\}$ is invariant under $\sigma$, so that we can set $\pi:=\sigma\restriction_{\{1,\dots,k\}}$.
And, consequently, as $\sigma$ is a one-to-one mapping on $\{1,2,\dots,k+n\}$, it must map all the remaining elements to a remaining element, i.e. $\sigma$ also leaves $\{k+1,\dots,k+n\}$ invariant, thus defining $\tau$.
Observe that $\sigma=\pi\tau=\tau\pi$.