Proof that $\det(A)=\det(A^T)$ using permutations. I'm reading a proof for the identity $\det(A) = \det(A^T)$ and I'm trying to udnerstand why the following rows are equivalent:
$$\eqalign{
  & \det ({A}) = \sum\limits_{\pi  \in {S_n}} {{\mathop{\rm sgn}} (\pi ) \cdot {a_{\pi (1),1}}} ...{a_{\pi (n),n}}  \cr 
  & \det ({A^T}) = \sum\limits_{\pi  \in {S_n}} {{\mathop{\rm sgn}} ({\pi ^{ - 1}}) \cdot {a_{1,{\pi ^{ - 1}}(1)}}} ...{a_{n,{\pi ^{ - 1}}(n)}} \cr} $$
Intuitively, I can understand that each term appears in both of the summations.
How to understand it algebraically?
 A: $f : \sigma \in S_n \rightarrow f(\sigma) =\sigma^{-1} \in S_n$ is a bijection (you can trivially check it is its own inverse), and this is a general fact that for a finite sum $\sum_{i \in A}u_i = \sum_{j \in B}u_{f(j)}$ if $f$ is a bijection from $B$ to $A$ (actually it also works for some infinite sums, such as sums with only positive terms). 
As you said, it just states that in both sums each term appears exactly once and since addition is commutative, order doesn't matter.
A: $\pi \in S_n \Rightarrow \pi^{-1} \in S_n$ and vice versa, so your (finite!) sums are equal.
A: Here
$\begin{align*} \det(A)=&\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i, \sigma(i)}\\
=&\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma^{-1})\prod_{\alpha\in\{(i,\sigma(i))| i\in \{ 1, 2, ..., n\}\}}a_\alpha\\
=&\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma^{-1})\prod_{\alpha\in\{(\sigma^{-1}(i),i| i\in \{ 1, 2, ..., n\}\}}a_\alpha \text{ by the Lemma below}\\
=&\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma^{-1})\prod_{i=1}^n a_{\sigma^{-1}(i),i} \\
=&\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{\sigma(i),i} \\
=& \det(A^T).
\end{align*}$
Lemma: Let $\sigma$ be arbitrary in $S_n$. Then, $\{(i,\sigma(i))| i\in \{ 1, 2, ..., n\}\}=\{(\sigma^{-1}(i), i)| i\in \{ 1, 2, ..., n\}\}$.
Proof: Let $x\in LHS$. Thus, $x=(i, \sigma(i))$ for some $i\in \{1, 2, ..., n\}$. Thus, $\sigma^{-1}(m)=i$ for some $m\in \{1, 2, ..., n\}$. Note $(\sigma^{-1}(m),m)\in RHS$. So, it suffices to show $m=\sigma(i)$. We know $m=\sigma(i)$ as $\sigma(\sigma^{-1}(m))=\sigma(i)\implies m=\sigma(i)$. An analogous argument shows the other way.
