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@LSpice no, $[n]^2 = \{\{x,y\}: x\neq y\}$, so $[n]^2$ has ${n\choose 2} = n(n-1)/2$ elements, and indeed, any $P\in [n]^2$ is a subset of $[n]$, so for $a\in S_n$, the concept of $a^{-1}(P)$ is meaningful.
I don't think so, @IlyaBogdanov, because we take the union over some edges $e\in E$, and this union gives a subset of $V$ -> which we require to equal $V$. Does that make sense?
