Does a permutation determined by set of Inversions?

Question

For $\sigma\in \mathfrak{S}_n$, symmetric group on $\{1,2\dots,n\}$ we denote the set of inversions $\{(i,j) | 1\leq i< j\leq n\text{ and }\sigma(i)> \sigma(j)\}$ by $ \mathcal{I}(\sigma)$. I believe that if $ \mathcal{I}(\sigma) = \mathcal{I}(\tau)$ for any permutations $\sigma,\tau\in\mathfrak{S}_n$, then $\sigma=\tau$. I don’t know how to prove this.

Answer 1

Thanks @Mike Earnest, I figured that $|\{(i,j)\in \mathcal{I}(\sigma)| i<j\}|- |\{(a,i)\in \mathcal{I} (\sigma) | a<i\}|+i=\sigma(i)$. Hence set of inversions determines the permutation.