 Solution

First suppose $(W,S)$ is a Coxeter system. The usual exchange condition implies that multiplication by a simple generator changes length by exactly one. Let $w=t_1\cdots t_n$ be reduced and suppose both $s_{i}w$ and $w{s}_j$ have length $n+1$. The length of $s_{i}w{s}_j$ is therefore either $n+2$ or $n$. In the latter case, apply exchange to the reduced word $s_i{t}_1\cdots t_n$ followed by $s_j$. If exchange deleted one of the $t_k$, multiplying the resulting equality on the left by $s_i$ would express the length-$n+1$ element $w{s}_j$ using only $n-1$ generators. Hence exchange must delete the initial $s_i$, giving $s_{i}w{s}_j=w$. This is exactly the folding condition.

Conversely, suppose the folding condition holds, and let $W_M$ be the abstract Coxeter group with generators $S$ and matrix $(m_{ij})$. The defining relations hold in $W$, so there is a surjective homomorphismIt remains to prove injectivity. Take any word in the kernel. If its image word in $W$ is not reduced, choose its shortest nonreduced prefix $ut$, where $u$ is reduced and $t$ is its last generator. Part b gives $ut=v$, where $v$ is obtained by deleting one letter from $u$. Hence $u=vt$, and $u$ and $vt$ are two reduced expressions for the same element. By the assumed braid-equivalence theorem they are related by braid moves. Those moves are defining relations in $W_M$, after which the end of the prefix becomes $vtt$ and $t^2=1$ shortens the original word by two.

Repeating this process turns the kernel word, using only Coxeter relations, into a word that is reduced in $W$. Since its image is the identity, that reduced word is empty. The original word is therefore already the identity in $W_M$, so $\ker \pi =1$. Hence $W\cong W_M$ is the Coxeter group with generators $S$.

Solved by gpt-5.6-sol high.