I'm working on a problem about Artin groups, and to simplify this problem I want to take a quotient that allow us to go to an easier Artin group, but I'm not sure if the quotient is well defined. This is the statement of my problem:


Let $A\_\Gamma$ be an Artin group, where $\Gamma$ is a complete graph, $f:A\_\Gamma\to\mathbb{Z}$ a group homomorphism and $a\in V(\Gamma)$ such that $f(a)=1$. The artin group is generated by the vertices of the graph, and there are two types of generators: the $v\in V(\Gamma)$ such that $f(v)=0$ and the $v\in V(\Gamma)$ such that $f(v)\neq0$. My objective is to define a normal subgroup $N\trianglelefteq A\_\Gamma$ such that $A\_\Gamma/N\cong A\_{\Gamma'}$ where $\Gamma'\subset\Gamma$ is the induced subgraph by the $v\in V(\Gamma)$ such that $f(v)=0$ and $a$, i.e. we want a new Artin group given by colapsing all vertices with non-zero image to $a$. To do so, I wanted to define $N$ as the normal subgroup generated by the set:
$$\lbrace v^{-1}a^{f(v)} \mid v\in V(\Gamma),f(v)\neq0\rbrace$$
(Remark: I $N\leq\ker(f)$ if possible, that is why I need to define those ''weird'' generators instead of $v^{-1}a$)
It is obvious that in the quotient $A\_\Gamma/N$ all the $v\in V(\Gamma)$ with $f(v)\neq 0$ are identified with some power of $a$, which allow us to get rid of those generators. However, I can't see if this quotient is indeed the Artin group I want or sth different.


Any hint will be thanked.


Edit: Inspired by @MoisheKohan I was able to see that I can reduce the problem to the right-angled case by first taking a quotient with the normal subgroup generated by the commutators of the generators. In this situation, since $\Gamma$ is a complete graph $A\_\Gamma\cong\mathbb{Z}^n$, where $n$ is the number of generators of $\mathbb{Z}$, so the problem should be easier.



Quotient of an Artin group is an Artin group
(Edit: Missed that the graph should be complete, now it is.)


If I'm understanding all this correctly, the answer is "no". Take three vertices, $a$, $b$, and $c$. Take an edge labeled 4 from $a$ to $b$, an edge labeled 4 from $a$ to $c$, and an edge labeled 2 from $b$ to $c$. Take $f$ to be $f(a)=f(b)=1$, $f(c)=0$. So now you're modding out that $a$ and $b$ get identified. In this quotient, $a$ and $c$ suddenly commute, which is not what the induced subgraph $\Gamma'$ wants (it's $a$ and $c$ with an edge labeled 4). If you want to phrase it in terms of quotient maps instead of normal subgroups, the problem is that the quotient map isn't well defined: you have that $b$ and $c$ commute, but their images are $a$ and $c$, which don't commute.


