Proof We start by showing that $(\widehat L,d_2)$ is complete. Let $(\alpha _n)_n$ be a Cauchy sequence in $(\widehat L,d_2)$ . Then

$$ \begin{align*}\forall \, M\in \mathbb{N}\;\; \exists\, N\in \mathbb{N} \;\; \forall \, m,n>N \;\; (|\alpha_n\wedge \alpha_m|>M).\end{align*} $$

Thus, for all $k\in \mathbb {N}$ , $\alpha _n^{[k]}$ stabilizes when $n\to \infty $ and $\lim _{n\to \infty }\alpha _n^{[k]}$ is a prefix of $\lim _{n\to \infty }\alpha _n^{[k+1]}$ . Thus, there is a unique word $\beta $ in $\tilde A^\infty $ such that $\beta ^{[k]}=\lim _{n\to \infty }\alpha _n^{[k]}$ and so $\beta =\lim _{n\to \infty } \alpha _n$ . Since L is prefix-closed, then, for all $k\in \mathbb {N}$ , $\beta ^{[k]}=\lim _{n\to \infty }\alpha _n^{[k]}\in L$ , and so $\beta \in \widehat L$ . Hence $(\widehat L,d_2)$ is complete. Notice that L is clearly dense in $\widehat L$ , since for all $N\in \mathbb {N}$ and $\alpha \in \widehat L$ , $\alpha ^{[N]}\in L$ and $|\alpha \wedge \alpha ^{[N]}|=N$ , and so $(\widehat L,d_2)$ is the completion of $(L,d_2)$ .

We will now prove that $(\widehat L,d_2)$ is a totally bounded metric space and that suffices, as it is complete. Since $\tilde A$ is finite, for a given $n\in \mathbb {N}$ , there are finitely many words in $\tilde A^*$ of length n and so, finitely many of such words in L. Let $\varepsilon>0$ and take $k\in \mathbb {N}$ such that $2^{-k}<\varepsilon $ . Let $V=\{x\in L\,\big \lvert \, |x|=k\}$ , which is finite. It is clear that

$$ \begin{align*}\widehat L=\bigcup_{x\in V}B(x,\varepsilon),\end{align*} $$

since for every $\alpha \in \widehat L$ , we have that $\alpha \in B(\alpha ^{[k]};\varepsilon )$ .

Proof We prove this by induction on $d_A(u\pi ,v\pi )$ . If $d_A(u\pi ,v\pi )\leq 1$ , we are done. So, assume that the result holds for all $u,v\in L$ with $|u\wedge v|=0$ such that $d_A(u\pi ,v\pi )\leq k$ for some integer $k>1$ and let $w,w'\in L$ be words such that $|w\wedge w'|=0$ and $d_A(w\pi ,w'\pi )=k+1$ . Take a geodesic path $\xi =x_1\ldots x_{k+1}$ connecting $w\pi $ and $w'\pi $ and let $\zeta =\overline {(w\xi ^{[k]})\pi }\in L$ . Now, if $\zeta $ does not share a prefix neither with w nor with $w'$ , then $|w|+|w'|\leq |w|+|\zeta |+|\zeta |+|w'|\leq kN+N=Nd_A(w\pi ,w'\pi )$ . Suppose w.l.o.g. that $\zeta $ shares a prefix with w (and so, not with $w'$ ). Then let $\alpha =\zeta \wedge w$ and put $w=\alpha w"$ and $\zeta =\alpha \zeta '$ . Since L is factorial, then $\zeta '$ and $w"$ are words in L such that $|w"\wedge \zeta '|=0$ and $d_A(w"\pi ,\zeta '\pi )=k.$ By hypothesis, we have that $|w"|+|\zeta '|=|w|-|\alpha |+|\zeta |-|\alpha |\leq Nk$ . Also, $|w'|+|\zeta |\leq N$ . Since $|\alpha |\leq |\zeta |$ , we have that $|w|+|w'|\leq |w'|+|\zeta | +|w|-|\alpha |+|\zeta |-|\alpha |\leq N(k+1).$

Proof Assume 1. Then, $\widehat {\pi _L}$ is a continuous bijection between compact spaces and so it must be a homeomorphism. Hence, it has a continuous inverse $\widehat {\pi _L}^{-1}$ . This is a continuous map between compact spaces, and so it is uniformly continuous and its restriction to G, $\pi _L^{-1}$ , is also uniformly continuous. Thus, there is some integer $N_L>0$ such that $|u\wedge v|=0\Rightarrow (u\pi |v\pi )\leq N_L$ for all $u,v\in L$ . Take words $u,v\in L$ , with $d_A(u\pi ,v\pi )\leq 1$ and $|u\wedge v|=0$ . Then

$$ \begin{align*}(u\pi|v\pi)=\frac 12 (|u|+|v|-d_A(u\pi,v\pi))\leq N_L\end{align*} $$

and so $|u|+|v|\leq 2N_L+d_A(u\pi ,v\pi )\leq 2N_L+1.$

Now, we will prove the converse. Assume that G admits a set of generators A and a factorial language of unique representatives $L\subseteq \mathrm {Geo}_A(G)$ such that given words $u,v\in L$ with $d_A(u\pi ,v\pi )\leq 1$ and $|u\wedge v|=0$ , we have that $|u|+|v|$ is bounded by some constant $C>0$ . Let $g,h\in G$ and put $u=|\bar g\wedge \bar h|$ . Consider the L-geodesics $\alpha ,\beta ,\gamma $ connecting $u\pi $ to g, $u\pi $ to h and g to h, respectively. Since G is hyperbolic, then $[[\alpha ,\beta ,\gamma ]]$ is $\delta $ -thin. Put $q_0=g$ , $q_n=h$ and $q_i=g(\gamma ^{[i]}\pi )$ , for $i\in \{1,\ldots ,n-1\}$ . Since $q_0\in \alpha $ and $q_n\in \beta $ , there are some $j\in \{0,\ldots , n-1\}$ , $p_1\in \alpha $ and $p_2\in \beta $ such that $d_A(q_j, p_1)\leq \delta $ and $d_A(q_{j+1},p_2)\leq \delta $ . Now,

$$ \begin{align*} (g|h)=&\frac 12(d_A(1,g)+d_A(1,h)-d_A(g,h))\\ =&\frac 12 (|u|+d_A(u\pi,p_1)+d_A(p_1,g)+|u|+d_A(u\pi,p_2)\\ &+d_A(p_2,h)-d_A(g,q_j)-1-d_A(q_{j+1},h))\\ =&|u| +\frac 12 (d_A(u\pi,p_1)+d_A(u\pi,p_2))+\frac 12 (d_A(p_1,g)-d_A(g,q_j))\\ &+\frac 12 (d_A(p_2,h)-d_A(h,q_{j+1}))-\frac 12. \end{align*} $$

We have that $\frac 12 (d_A(p_1,g)-d_A(g,q_j))\leq \frac \delta 2$ , because

$$ \begin{align*}d_A(p_1,g)\leq d_A(p_1,q_j)+d_A(q_j,g)\leq \delta +d_A(q_j,g).\end{align*} $$

Similarly, $\frac 12 (d_A(p_2,h)-d_A(h,q_{j+1}))\leq \frac \delta 2$ . So,

$$ \begin{align*}(g|h)\leq |\bar g\wedge \bar h|+\delta-\frac 12+\frac 12 (d_A(u\pi,p_1)+d_A(u\pi,p_2)).\end{align*} $$

But now, consider the subpath $\alpha '\subseteq \alpha $ from $u\pi $ to $p_1$ and the subpath $\beta '\subseteq \beta $ from $u\pi $ to $p_2$ . They are labeled by words in L, since L is factorial, sharing no prefix and ending at distance at most $2\delta +1$ since

$$ \begin{align*}d_A(p_1,p_2)\leq d_A(p_1,q_j)+d_A(q_j,q_{j+1})+d_A(q_{j+1},p_2)\leq 2\delta+1.\end{align*} $$

By Lemma 3.1, we have that $|\alpha '|+|\beta '|\leq (2\delta +1)C$ , and so $\frac 12 (d_A(u\pi ,p_1)+d_A(u\pi ,p_2))\leq \frac {(2\delta +1)C}{2}$ and

$$ \begin{align*}(g|h)\leq |\bar g\wedge \bar h|+\delta-\frac 12+\frac {(2\delta+1)C}{2}.\end{align*} $$

This implies that $\pi _L^{-1}$ is uniformly continuous since

$$ \begin{align*}\forall \,M\in \mathbb{N}\;\;\exists N\in \mathbb{N} \left((g|h)>N\Rightarrow |\bar g\wedge \bar h|> M\right)\!.\end{align*} $$

Therefore, it extends to some continuous mapping $\widehat {\pi _L^{-1}}$ . Hence, $\widehat {\pi _L}\widehat {\pi _L^{-1}}$ is a continuous extension of the identity to the completion. Since that extension is unique, it must be the identity on $\hat L$ . Similarly, $\widehat {\pi _L^{-1}}\widehat {\pi _L}$ must be the identity on $\hat G$ and so $\widehat {\pi _L} \widehat {\pi _L^{-1}}$ are mutually inverse continuous mappings, thus homeomorphisms.

Proof Let $L\subseteq \tilde A^*$ be a prefix-closed language such that $\pi |_L$ is bijective and G has bounded triangles for L with constant N. Since L is prefix closed, the undirected graph T (obtained by identifying inverse edges) with vertex set $V=L$ having an edge (labeled by $a\in A$ ) between x and y if and only if $y=xa$ is a tree, i.e., between any pair of vertices, there is one and only one reduced path in T. We have that T is a subgraph of the Cayley graph $\Gamma =\Gamma _X(G)$ . Hence, for all $u,v\in L$ , we have that $d_{\Gamma} (u,v)\leq d_T(u,v)$ . Now, let $u,v\in L$ and put $w=u\wedge v$ , $u=wu'$ and $v=wv'$ . We have that $d_T(u,v)= |u'|+|v'|\leq Nd_{\Gamma} (u\pi ,v\pi )$ . Therefore, $\Gamma $ is quasi-isometric to a tree, which by [Reference Antolín1, Theorem 4.7(B2)], implies that G is virtually free.

If G is virtually free, then, by Proposition 1.1, there is a finite set of generators A and a factorial language $L\subseteq \tilde A^*$ such that $\widehat \pi _L$ is bijective; which, by Theorem 3.2, implies that there is a finite set of generators A and a factorial (hence, prefix-closed) language of unique representatives $L\subseteq \tilde A^*$ such that G has bounded triangles for L.