Tail-swapping is a sign-reversing involution on the intersecting systems

The set of lattice vertices lying on at least two paths of $\mathbf{P}$ is finite and nonempty. Choose its lexicographically least vertex $x$, and then choose the lexicographically least pair of indices $i<j<r$ such that both $P_i$ and $P_j$ pass through $x$.

Write $P_i={\alpha }_i{\beta }_i$ and $P_j={\alpha }_j{\beta }_j$, where ${\alpha }_i$ and ${\alpha }_j$ end at $x$ and ${\beta }_i$ and ${\beta }_j$ start at $x$. Define a new $r$-tuple ${\mathbf{P}}^{\mathrm{\prime }}$ by replacing $P_i$ with ${\alpha }_i{\beta }_j$, replacing $P_j$ with ${\alpha }_j{\beta }_i$, and leaving every other path unchanged; and put ${\pi }^{\mathrm{\prime }}:=\pi \circ (ij)$.

The new tuple ${\mathbf{P}}^{\mathrm{\prime }}$ is a ${\pi }^{\mathrm{\prime }}$-system: the swapped paths are still monotone because each is a concatenation of monotone segments meeting at the same lattice point $x$, and their endpoints are $A_i\to E_{\pi (j)}$ and $A_j\to E_{\pi (i)}$ respectively, while every other endpoint is unchanged.

Since ${\pi }^{\mathrm{\prime }}=\pi \circ (ij)$ and $(ij)$ is a transposition, [L1] gives $\mathrm{sgn}({\pi }^{\mathrm{\prime }})=-\mathrm{sgn}(\pi )$.

The same choices $x,i,j$ are recovered from $({\pi }^{\mathrm{\prime }},{\mathbf{P}}^{\mathrm{\prime }})$. At every lattice vertex, swapping the two tails preserves the number of paths passing through that vertex: it only exchanges the labels $i$ and $j$ after $x$. Thus the set of vertices lying on at least two paths, and hence its lexicographically least member $x$, is unchanged. The paths with indices $i$ and $j$ still both pass through $x$, and the set of indices of paths passing through $x$ is unchanged, so the least pair there is again $i,j$. Applying the construction again swaps the same tails back.

Steps 2.1 and 3.1 show that the construction defines a map $\tau$ from $\mathcal{I}(A,E)$ to itself and that $\tau (\tau (\pi ,\mathbf{P}))=(\pi ,\mathbf{P})$ for every $(\pi ,\mathbf{P})$; step 2.2 gives the sign change. So $\tau$ is the required sign-reversing involution.