Write an Artin group as an HNN-extension

Question

Assume that $A_\Gamma$ is an Artin group and $\chi:A_\Gamma\to(\mathbb{Z},+)$ is a group homomorphism of the following form. $\Gamma=\Gamma_1\cup\Gamma_2$ with $\Gamma_1\cap\Gamma_2=\emptyset,A_{\Gamma_1}=\mathbb{Z}^{|\Gamma_1|},A_{\Gamma_2}=\mathbb{Z}^{|\Gamma_2|}$, all the edges between $\Gamma_1$ and $\Gamma_2$ are labelled with an even number $>2,\chi(v)=1~\forall~v\in\Gamma_1$ and $\chi(v)=-1~\forall~v\in\Gamma_2$. i.e. $A_\Gamma$ has the following presentation: $$A_\Gamma=\left\langle\begin{array}{l|cl} & vw=wv~\forall~u,v\in\Gamma_1 \\ v\in\Gamma&vw=wv~\forall~u,v\in\Gamma_2\\ & (vw)^l=(wv)^l~\forall~v\in\Gamma_1~w\in\Gamma_2\text{ connected with }l(u ,v)=2l\\ \end{array}\right\rangle$$ where $l( u ,v)$ denoted the label of the edge between $u$ and $v$.

I want to wirte $A_\Gamma$ as an $HNN$-extension $H\ast_{L,v}$ where both $L$ and $H$ lie in $\ker(\chi)$. I know this can be done due to Theorem B3.1 of Strebel's sigma invariants notes (https://arxiv.org/pdf/1204.0214.pdf), but I want to compute the explicit presentation for both $L$ and $H$. This is my attempt:

First of all, let us fix $v\in\Gamma_1$ and define $w_0:=v^{-1}w~\forall~w\in\Gamma_1\setminus\lbrace v\rbrace$ and $w_0=vw~\forall~w\in\Gamma_2$, we aim to rewrite the relations of $A_\Gamma$ in terms of the $w_0$ and conjugations by $v$.

• Let $w\in\Gamma_1\setminus\lbrace v\rbrace$, then: $$vw=wv\Leftrightarrow v^2w_0=vw_0v\Leftrightarrow vw_0=w_0v\Leftrightarrow w_0^v=w_0$$
• Let $u,w\in\Gamma_1\setminus\lbrace v\rbrace$ then: $$uw=wu\Leftrightarrow vu_0vw_0=vw_0vu_0\Leftrightarrow u_0w_0=w_0u_0$$
• Let $u,w\in\Gamma_2$ then: $$uw=wu\Leftrightarrow v^{-1}u_0v^{-1}w_0=v^{-1}w_0v^{-1}u_0\Leftrightarrow u_0v^{-1}w_0=w_0v^{-1}u_0\Leftrightarrow$$$$\Leftrightarrow w_0^{-1}u_0=v^{-1}u_0w_0^{-1}v\Leftrightarrow u_0w_0^{-1}=(w_0^{-1}u_0)^v$$
• If $w\in\Gamma_2$ is connected to $v$ via a edge of label $2l$ then: $$(vw)^l=(wv)^l\Leftrightarrow v(vw)^l=v(wv)^l\Leftrightarrow vw_0^l=w_0^lv\Leftrightarrow(w_0^l)^v=w_0^l$$
• If $u\in\Gamma_1\setminus\lbrace v\rbrace$ and $w\in\Gamma_2$ are connected with an edge of label $2l$, then: $$(uw)^l=(wu)^l\Leftrightarrow (vu_0v^{-1}w_0)^l=(v^{-1}w_0vu_0)^l\Leftrightarrow (u_0w_0)^l=v^{-1}(w_0u_0)^lv\Leftrightarrow (w_0u_0)^l=\left( (u_0w_0)^l\right)^v$$

Hence, the presentation of $A_\Gamma$ can be rewritten as follows: $$\left\langle \begin{array}{l|cl} & w_0^v=w_0~\forall~w\in\Gamma_1\setminus \lbrace v\rbrace\\ & u_0w_0=w_0u_0~\forall~u,v\in\Gamma_1 \\ v,w_0~\forall~w\in\Gamma\setminus\lbrace v\rbrace& (w_0^{-1}u_0)^v=u_0w_0^{-1}~\forall~u,w\in\Gamma_2 \\ & (w_0^l)^v=w_0^l~\forall~w\in\Gamma_2\text{ with }l(\lbrace u ,v\rbrace)=2l\\ & \left((u_0w_0)^l\right)^v=(w_0u_0)^l~\forall~u\in\Gamma_1\setminus\lbrace v\rbrace,w\in\Gamma_2\text{ with }l(\lbrace u,w\rbrace)=2l \end{array}\right\rangle$$ From this, I thought that: $$H=\left\langle w_0~\forall~w\in\Gamma\setminus\lbrace v\rbrace \right| u_0w_0=w_0u_0~\forall~u,w\in\Gamma_1\setminus\lbrace v\rbrace\rangle$$ $$L=\left\langle \begin{array}{l } w_0~\forall~w\in\Gamma_1\setminus\lbrace v\rbrace\\ w_0^{-1}u_0~\forall~u,w\in\Gamma_2\\ w_0^l~\forall~w\in\Gamma_2\text{ with }l(\lbrace u ,v\rbrace)=2l\\ (u_0w_0)^l~\forall~u\in\Gamma_1\setminus\lbrace v\rbrace\text{ and }w\in\Gamma_2\text{ with }l(\lbrace u,v \rbrace)=2l\\ \end{array} \right\rangle$$ But there is no way to define an isomorphism $\phi:L\to L$ such that: $$A_\Gamma\cong\langle H,v\mid \phi(g)^v=g~\forall~g\in L\rangle=H\ast_{L,v}$$

I guess that I have to work in a different way with the relations to get an expression that will have an obvious decomposition as an HNN-extension, but I don't know how to do that. Also, I don't know how to make the brackets biggers for those presentations, so I'll be happy to get an answer for that too.

Thank's for your help.