K{\rightarrow}\Lambda\cup\{\infty\}$, a subgroup $H\subset SL_2(K)$ acts freely without inversions on the corresponding Bruhat–Tits tree $BT$ if and only if the translation length of any element $h\in H\setminus\{1\}$ is non-zero. But the translation length of a matrix $m\in SL_2(K)$ can be computed in terms of the valuation of its trace by the formula $l_{BT}(m)=\max\{0,-2v(\operatorname{Tr}(m))\}$, so the freeness (without inversion) of the action translates into $v(\operatorname{Tr}(h))<0$ for all $h\in H\setminus\{1\}$ [@Chi_book Lemma 4.3.5 p.148]. Therefore, since for all $i$ and all $g_i\in \Gamma_i\setminus\{1\}$, $\operatorname{Tr}(j_i(g_i))$ has negative valuation, all the elements $g^*\in \Gamma^*\setminus\{1\}$ satisfy $v^*(\operatorname{Tr}(j^*(g^*)))<0$, which means that $\Gamma^*$ acts freely without inversion on $BT^*$.

Finally, there remains to check that the marked group $(\Gamma,S)$ embeds into $\Gamma^*$ (see for instance [@CG_compactifying]). We use the notation $S=(s_1,\dots,s_n)$ and $S_i=(s_1^{(i)},\dots,s_n^{(i)})$. Consider the family $S^*=(s_1^*,\dots,s_n^*)$ of elements of $\Gamma^*$ defined by $s_1^*=(s_1^{(i)})_{i\in{{\mathbb{N}}}}, \dots,s_n^*=(s_n^{(i)})_{i\in{{\mathbb{N}}}}$. The definition of the convergence of marked groups says that if an $S$–word represents the trivial element (resp. a non-trivial element) in $\Gamma$, then for $i$ sufficiently large, the corresponding $S_i$–word is trivial (resp. non-trivial) in $\Gamma_i$. Since $\omega$ is non-principal, this implies that the corresponding $S^*$–word is trivial (resp. non-trivial). This means that the map sending $(s_1,\dots,s_n)$ to $(s_1^*,\dots,s_n^*)$ extends to 