$\forall s,t \in S \ \lcm(\varphi(s),\varphi(t)) = \varphi(\lcm(s,t))$

The following results on such morphisms are basically due to J. Crisp, who proved them in [@crisp] in the case of finite-type Artin groups.

Let $U,V \in M$. If $\varphi(U) | \varphi(V)$ then $U | V$.

By induction on $\ell(V)$. Since $\forall s \in S \ \ell(\varphi(s)) \geq 1 = \ell(s)$, we have $\ell(\varphi(U)) \geq \ell(U)$. Since $\varphi(U) | \varphi(V)$, we have $\ell(\varphi(U)) \leq \ell(\varphi(V))$ hence $\ell(U) \leq \ell(\varphi(V))$. Hence $\varphi(V) = 1$ implies $\ell(U) = 0$ and $U = 1$, which settles the case $\ell(V) = 0$.

We thus assume $\ell(V) \geq 1$. The case $U=1$ being clear, we can assume $U \neq 1$. Then there exists $s,t \in S$ with $s | U$ and $t|V$. It follows that $\varphi(t) | \varphi(V)$ and $\varphi(s) | \varphi(U) | \varphi(V)$, hence $\lcm(\varphi(s),\varphi(t)) | \varphi(V)$.

Now $\lcm(s,t) = t m$ for some $m \in M$ and $V = t V'$ for some $V' \in M$, hence $\varphi(t) \varphi(m) | \varphi(V) = \varphi(t) \varphi(V')$ and this implies $\varphi(m) | \varphi(V')$ by cancellability in $M$. Since $\ell(V') < \ell(V)$, from the induction assumption follows that $m | V'$ hence $tm | V$ that is $\lcm(s,t) | 