Coxeter elements of tree type are conjugate by source and sink firings

(Firings connect tree orientations.) Every orientation of a tree is acyclic because a directed cycle would be an undirected cycle. Prove firing connectivity by induction on the number of vertices. For one vertex there is only one orientation; for two vertices a single firing of either endpoint reverses the only edge. For a tree with at least three vertices, an endpoint $v$ of a longest simple path is a leaf: a neighbour outside the path would extend it, and a second neighbour on the path would create a cycle. Let $u$ be its neighbour and put $T^{\mathrm{\prime }}=T\setminus \{v\}$. This is a smaller tree, since a path between remaining vertices cannot use a leaf internally and deleting a vertex creates no cycle. By induction, a finite firing sequence changes $\omega {\mathrm{\mid }}_{T^{\mathrm{\prime }}}$ to ${\omega }^{\mathrm{\prime }}{\mathrm{\mid }}_{T^{\mathrm{\prime }}}$. Lift each firing at $x\mathrm{\ne }u$ directly, since its incident edges are unchanged. Immediately before a firing at $u$, its incident arrows in $T^{\mathrm{\prime }}$ all point in one direction; if the edge $uv$ points the other way, fire the leaf $v$ first, which is always legal and flips only $uv$. Now fire $u$. The restriction to $T^{\mathrm{\prime }}$ follows the inductive sequence. At the end, if $uv$ has the wrong direction, fire $v$ once more. This reaches ${\omega }^{\mathrm{\prime }}$ and proves (1).

(Orderings encode acyclic orientations.) An ordering gives no directed cycle because the position strictly increases along each oriented edge. Conversely, in a finite acyclic orientation there is a source: otherwise repeatedly following an incoming edge would revisit a vertex and give a directed cycle. Remove a source and repeat to obtain an ordering realizing every edge direction. If two such orderings realize the same orientation, they are linear extensions of the partial order generated by its directed paths. To connect the extensions, move the first vertex of one extension left through the preceding vertices of the other; each crossed vertex is incomparable with it, and induction repeats this on the remaining vertices. Incomparable vertices cannot be joined by an edge, so their generators commute by [F4]. Thus all these orderings give the same product, proving the orientation-to-element assertion.

(One firing is conjugation.) If $s$ is a source, choose a realizing ordering that starts with $s$ and write $c=sz$. After firing $s$, the ordering with $s$ moved to the end realizes the new orientation, so its product is $zs=s(sz)s^{-1}$ by $s^2=1$. If $s$ is a sink, choose a realizing ordering ending in $s$, write $c=zs$, and move $s$ to the beginning after firing; the new product is $sz=s(zs)s^{-1}$. This also covers an isolated vertex: it is both source and sink, firing changes no edge, and it commutes with all other generators. Hence each firing conjugates by its simple reflection.

(Componentwise conjugacy.) If $S=\mathrm{\varnothing }$, both products are $1$. Otherwise suppose every connected component of $\mathrm{\Gamma }$ is a tree; finite type guarantees this by [F2]. Restrict the orderings defining $c,c^{\mathrm{\prime }}$ to each component. By step 1.1 a finite firing sequence connects the resulting orientations, and by step 2.1 each firing conjugates the corresponding component product by a simple reflection. Thus each component pair is conjugate by some $w_i\in W_{S_i}$. The component decomposition in [F6] combines these into $w=(w_i)\in W$ with $c^{\mathrm{\prime }}=wc{w}^{-1}$. If $\mathrm{\Gamma }$ is connected, this conjugator is a product of simple reflections, as each firing uses one. This proves (3).