Proof. As recalled in Remark 3.1, we have an inclusion $\mathbf {QD}_{v_0,\mu _{\mathcal E}}(\Gamma _{\mathcal E})\subset \mathbf {QD}_{v_0,\mu }(\Gamma )$ . Hence, we can assume $\mathcal E=\emptyset $ . We will proceed by induction on the number of edges of $\Gamma $ . If $\Gamma $ has only one edge, the result is clear. Otherwise, fix an edge $e\in E(\Gamma )$ and consider the contraction $\iota \colon \Gamma \to \Gamma /\{e\}$ of e. Recall that the map $\iota _*\colon \mathbf {QD}_{v_0,\mu }(\Gamma )\to \mathbf {QD}_{\iota (v_0),\iota _*(\mu )}(\Gamma /\{e\})$ taking $(\mathcal E,D)$ to $\iota _*(\mathcal E,D)$ is surjective and closed (see [Reference Abreu and PaciniAP20, Proposition 4.11]).

First of all, we assume that $\iota _*(\emptyset ,D)=\iota _*(\emptyset ,D')$ . This means that $D(v)=D'(v)$ for every vertex $v\in V(\Gamma )$ not incident to e. If e is a loop, then $D=D'$ , and we have nothing to prove. Otherwise, let s and t be the end-vertices of e and assume that $D(t)\ge D'(t)$ . Set $n:=D(t)-D'(t)=D'(s)-D(s)$ and define the divisors $D_i$ on $\Gamma ^{\{e\}}$ for $i=1,\dots ,n$ and $D^{\prime }_i$ on $\Gamma $ for $i=0,\dots ,n$ taking a vertex v to

$$ \begin{align*} & D_i(v)= \begin{cases} D(v) & \text{ if } v\not\in\{s,t\}\\ 1 & \text{ if } v=v_e \\ D(v)-i = D'(v)+ n -i& \text{ if } v=t\\ D(v)+i-1 & \text{ if } v=s \end{cases} \end{align*} $$

and

$$ \begin{align*} D^{\prime}_i(v)= \begin{cases} D(v) & \text{ if } v\not\in\{s,t\}\\ D(v)-i =D'(v)+n-i& \text{ if } v=t\\ D(v)+i & \text{ if } v=s. \end{cases} \end{align*} $$

Let $e_1:=e_2:=\dots :=e_n:=e$ . Note that $(\emptyset ,D_i^{\prime })$ and $(\{e\},D_i)$ are $(v_0,\mu )$ -quasistable because both $D_i(V)$ and $D^{\prime }_i(V)$ are greater than or equal to either $D(V)$ or $D'(V)$ , for every $V\subset V(\Gamma )\subset V(\Gamma ^{\{e\}})$ . We see that $(\emptyset ,D)$ and $(\emptyset , D')$ are upper-connected in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ by means of the edges $e_1,\dots ,e_n$ and the divisors $D_1,\dots , D_n,D^{\prime }_0,\dots ,D^{\prime }_n$ .

Now we consider the general case. By the induction hypothesis, $\iota _*(\emptyset ,D)$ and $\iota _*(\emptyset , D')$ are upper-connected in $\mathbf {QD}_{\iota (v_0),\iota _*(\mu )}(\Gamma /\{e\})$ by means of edges $e_1,\ldots , e_n$ of $\Gamma /\{e\}$ , and divisors $D_{e,i}\in \mathbf {QD}_{\iota (v_0),\iota _*(\mu )}(\Gamma /\{e\},\{e_i\})$ for $i=1,\dots ,n$ and $D^{\prime }_{e,i}\in \mathbf {QD}_{\iota (v_0),\iota _*(\mu )}(\Gamma /\{e\})$ for $i=0,\dots ,n$ . Since $\iota _*$ is surjective, there are divisors $D_i\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\{e_i\})$ for $i=1,\dots ,n$ , such that $\iota _*(\{e_i\},D_i)=(\{e_i\},D_{e,i})$ . By Remark 2.1 and the fact that $\iota _*$ is closed, we have that there are $(v_0,\mu )$ -quasistable divisors $D_i^{\prime }$ and $D_i^{\prime \prime }$ on $\Gamma $ such that

$$ \begin{align*} (\emptyset, D_i^{\prime})\leq (\{e_i\},D_i), \;\;\;\;& (\emptyset, D_i^{\prime\prime})\leq (\{e_i\},D_i),\\ \iota_*(\emptyset, D_i^{\prime})=(\emptyset, D_{e,i}'),\;\; \;\;& \iota_*(\emptyset, D_i^{\prime\prime}) = (\emptyset, D_{e,i-1}'). \end{align*} $$

This means that $\iota _*(\emptyset , D)=\iota _*(\emptyset , D_1^{\prime \prime })$ , $\iota _*(\emptyset , D_i^{\prime })=\iota _*(\emptyset , D_{i+1}")$ and $\iota _*(\emptyset , D')=\iota _*(\emptyset , D_{n}^{\prime })$ . By the previous case, we have that the pairs $((\emptyset , D),(\emptyset , D_1^{\prime \prime }))$ , $((\emptyset , D_i^{\prime \prime }), (\emptyset , D_{i+1}"))$ and $((\emptyset , D'),(\emptyset ,D_n^{\prime }))$ are pairs of upper-connected pseudo-divisors in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ . Since $(\emptyset , D_i^{\prime })\leq (\{e_i\},D_i)$ and $(\emptyset , D_i^{\prime \prime })\leq (\{e_i\},D_i)$ , it follows that $(\emptyset , D_i^{\prime \prime })$ and $(\emptyset , D_i^{\prime })$ are upper-connected in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ . This proves that $(\emptyset ,D)$ and $(\emptyset , D')$ are upper-connected in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ , concluding the proof.

Proof. We construct a map $\mathbf {QD}_{v_0}(\Gamma )\to \mathbf {QD}_{v_1}(\Gamma )$ that takes a pseudo-divisor $(\mathcal E,D)$ in $\mathbf {QD}_{v_0}(\Gamma )$ to $(\mathcal E,D+v_0-v_1)$ . This map is well-defined, indeed. Fix $(\mathcal E,D)$ a $v_0$ -quasistable pseudo-divisor and set $D'=D+v_0-v_1$ . For every $V\subset V(\Gamma ^{\mathcal E})$ , we have that $\beta _{\Gamma ,D}(V)$ is an integer number by Equation (7). Moreover,

$$\begin{align*}\beta_{\Gamma,D'}(V)=\begin{cases} \beta_{\Gamma,D}(V) + 1 & \text{ if }v_0\in V, v_1\notin V,\\ \beta_{\Gamma,D}(V) - 1 & \text{ if }v_0\notin V, v_1\in V,\\ \beta_{\Gamma,D}(V) & \text{ otherwise}. \end{cases} \end{align*}$$

It follows that $(\mathcal E, D-v_0+v_1)$ is $v_1$ -quasistable.

The fact that the map is a morphism of posets is clear, and it has a natural inverse that takes $(\mathcal E,D')$ to $(\mathcal E, D'-v_0+v_1)$ ; hence, it is an isomorphism.

Proof. Recall that we write $\mathbf P=\{\alpha ,\beta ,\gamma ,\delta \}$ (see Figure 1). We set

$$\begin{align*}(\mathcal E_1,D_1):=g(\alpha), \;\;(\mathcal E_2,D_2):=g(\beta), \; \;(\mathcal E_3,D_3):=g(\gamma), \; \; (\mathcal E_4,D_4):=g(\delta). \end{align*}$$

By definition of specialization, we have $\mathcal E_3\cup \mathcal E_4\subset \mathcal E_1\cap \mathcal E_2$ , with $|\mathcal E_1|=|\mathcal E_2|=|\mathcal E_3|+1=|\mathcal E_4|+1$ . Hence, we have three cases:

1. 1. either $\mathcal E_3=\mathcal E_4$ and $\mathcal E_1\neq \mathcal E_2$ ,

2. 2. or $\mathcal E_3\neq \mathcal E_4$ and $\mathcal E_1=\mathcal E_2$ ,

3. 3. or $\mathcal E_3=\mathcal E_4$ and $\mathcal E_1=\mathcal E_2$ .

We begin with Case (1). In this case, we define $\mathcal E:=\mathcal E_3=\mathcal E_4$ , which means that $\mathcal E_1=\mathcal E\cup \{e_1\}$ and $\mathcal E_2=\mathcal E\cup \{e_2\}$ for some distinct edges $e_1,e_2\in E(\Gamma )$ . We have that $(\mathcal E,D_3)$ and $(\mathcal E,D_4)$ must be different (since g is injective), and hence, they are the two pseudo-divisors on $\Gamma $ of type $(\mathcal E,D')$ to which both $(\mathcal E\cup \{e_1\},D_1)$ and $(\mathcal E\cup \{e_2\}, D_2)$ specialize described in Remark 2.4. In particular, neither $e_1$ nor $e_2$ is a loop; otherwise, there will be only one of these specializations.

Let us prove that $e_1$ and $e_2$ are parallel edges. Assume, by contradiction, that there exists a vertex v incident to $e_1$ and not to $e_2$ . Then, by Remark 2.4, it follows that $D_3(v)=D_2(v)$ and $D_4(v)=D_2(v)$ and also, without loss of generality, that $D_3(v)=D_1(v)$ and $D_4(v)=D_1(v)+1$ , giving rise to a contradiction. This proves that $e_1$ and $e_2$ are incident to same pair of vertices, meaning that they are parallel.

Denote by $s,t$ the end-vertices of $e_1$ and $e_2$ . Again by Remark 2.4 and up to switch $D_3$ with $D_4$ , we have that either $D_3=D_1-v_{e_1}+s=D_2-v_{e_2}+s$ or $D_3=D_1-v_{e_1}+t=D_2-v_{e_2}+s$ . We can rule out the second possibility as follows. If $D_3=D_1-v_{e_1}+t=D_2-v_{e_2}+s$ , then $D_4=D_1-v_{e_1}+s=D_2-v_{e_2}+t$ ; hence, $D_3(t)=D_1(t)+1=D_2(t)$ and $D_4(t)=D_1(t)=D_2(t)+1$ , which is a contradiction. It follows that $D_3=D_1-v_{e_1}+s=D_2-v_{e_2}+s$ , giving the poset described in item (1) of the statement with $D:=D_1-v_{e_1}=D_2-v_{e_2}$ .

We move to Case (2). In this case, we define $\mathcal E:=\mathcal E_3\cap \mathcal E_4$ , and hence, $|\mathcal E_3|=|\mathcal E_4|=|\mathcal E|+1$ . This means that $\mathcal E_3=\mathcal E\cup \{e_1\}$ , $\mathcal E_4=\mathcal E\cup \{e_2\}$ , $\mathcal E_1=\mathcal E_2=\mathcal E\cup \{e_1,e_2\}$ for some distinct edges $e_1,e_2\in E(\Gamma )$ .

Let us prove that $e_1$ and $e_2$ are parallel edges. Let $V_0$ be the set of vertices incident to both $e_1$ and $e_2$ . Assume, by contradiction, that $|V_0|\le 1$ . Let v be a vertex not incident to $e_2$ . Since $(\mathcal E\cup \{e_1\},D_3)$ is an elementary specialization of both $(\mathcal E\cup \{e_1,e_2\},D_1)$ and $(\mathcal E\cup \{e_1,e_2\},D_2)$ , by Remark 2.4, we have that $D_1(v)=D_3(v)=D_2(v)$ . We can argue similarly for any vertex not incident to $e_1$ . We deduce that $D_1(v)=D_2(v)$ for every vertex $v\not \in V_0$ . Since $D_1$ and $D_2$ have the same degree and since $|V_0|\le 1$ , we have that $D_1=D_2$ , which is a contradiction. This proves that $|V_0|=2$ (i.e., $e_1$ and $e_2$ are parallel edges).

Let s and t be the end-vertices of $e_1$ and $e_2$ . Then, for $i=3,4$ , we have four cases:

1. (i) either $D_i(s)=D_1(s)+1 = D_2(s)+1$ ,

2. (ii) or $D_i(s) = D_1(s) = D_2(s)+1$ ,

3. (iii) or $D_i(s) = D_1(s)+1 = D_2(s)$ ,

4. (iv) or $D_i(s) = D_1(s) = D_2(s)$ .

With the same argument used above, Cases (i) and (iv) would imply that $D_1=D_2$ , which is a contradiction. In Case (ii), we have that $D_i=D_1-v_{e_{5-i}}+t=D_2-v_{e_{5-i}}+s$ , which means that $D_1+t=D_2+s$ . Similarly, in Case (iii), we have that $D_1+s=D_2+t$ . So the same case must hold for both $i=3$ and $i=4$ . This means that $D_3(v)=D_4(v)$ for every $v\in V(\Gamma )$ , giving the poset described in item (2) of the statement with $D:=D_3-v_{e_1}=D_4-v_{e_2}$ .

Finally, we consider Case (3). In this case, we define $\mathcal E:=\mathcal E_3=\mathcal E_4$ , which means that $\mathcal E_1=\mathcal E_2=\mathcal E\cup \{e\}$ for some edge $e\in E(\Gamma )$ . Since g is injective, we have that $(\mathcal E,D_3)$ and $(\mathcal E,D_4)$ are different. Hence, they are the two pseudo-divisor of type $(\mathcal E,D')$ to which both $(\mathcal E\cup \{e\},D_1)$ and $(\mathcal E\cup \{e\},D_2)$ specialize, described in Remark 2.4. This implies that $D_1=D_2$ , which is a contradiction with the fact that $g(\alpha )\neq g(\beta )$ .

Proof. The rank of the elements $\alpha $ and $\beta $ in $\mathbf P$ is 1, and g is a morphism of ranked posets. Then $g(\alpha )$ and $g(\beta )$ have rank 1 in $\mathbf {PD}(\Gamma )$ , and hence, $g(\mathbf P)$ is the poset described in item (1) of Proposition 4.2 with $\mathcal E=\emptyset $ .

Proof. Since g is a morphism of ranked posets and the rank of $\alpha _1$ is 2, we have $g(\alpha _1)=(\{e_1,e_2\},D)$ for some edges $e_1$ and $e_2$ of $\Gamma $ and a divisor D on $\mathbf {PD}(\Gamma ,\{e_1,e_2\})$ . Let $s_1$ and $t_1$ (respectively, $s_2$ and $t_2$ ) be the (possibly coincident) end-vertices of $e_1$ (respectively, of $e_2$ ). By Remark 2.4, there are at most 4 pseudo-divisors $(\mathcal E',D')$ of rank 1 (i.e., with $|\mathcal E'|=1$ ) such that $(\mathcal E',D')<(\{e_1,e_2\},D)$ : they are the pseudo-divisors of the set

(9) $$ \begin{align} \left\{ (\{e_1\}, D-v_{e_2}+s_2), \; (\{e_1\}, D-v_{e_2}+t_2), \; (\{e_2\}, D-v_{e_1}+s_1), \; (\{e_2\}, D-v_{e_1}+t_1)\right\}. \end{align} $$

Since g is an injective morphism of ranked poset, the set in Equation (9) is equal to $\{g(\beta _1), g(\beta _2), g(\beta _3), g(\beta _4)\}$ . In particular, neither $e_1$ nor $e_2$ are loops (i.e., $s_1\neq t_1$ and $s_2\neq t_2$ 0.

The induced subposets $\{\beta _1,\beta _2,\gamma _1,\gamma _2\}$ and $\{\beta _3,\beta _4,\gamma _2,\gamma _3\}$ of $\mathbf R$ are isomorphic to the poset $\mathbf P$ . By Corollary 4.3, we see that $e_1$ and $e_2$ are parallel edges of $\Gamma $ , and, without loss of generality, we have that $s:=s_1=s_2$ , $t:=t_1=t_2$ , and

$$ \begin{align*} g(\beta_1)=(\{e_1\}, D-v_{e_2}+s), \; \; & g(\beta_2)=(\{e_2\}, D-v_{e_1}+s),\\ g(\beta_3)=(\{e_1\}, D-v_{e_2}+t), \; \; & g(\beta_4)=(\{e_2\}, D-v_{e_1}+t). \end{align*} $$

Finally, there are exactly 3 pseudo-divisors of rank 0 (i.e., of type $(\emptyset , D")$ ) that are smaller than at least one pseudo-divisor in the set $\mathcal U=\{g(\beta _1),g(\beta _2),g(\beta _3),g(\beta _4)\}$ . By Remark 2.4, they are

$$ \begin{align*} (\emptyset, D^{\prime\prime}_1) := &(\emptyset, D-v_{e_1}-v_{e_2}+2s),\\ (\emptyset, D^{\prime\prime}_2) := &(\emptyset, D-v_{e_1}-v_{e_2}+s+t),\\ (\emptyset, D^{\prime\prime}_3) := & (\emptyset, D-v_{e_1}-v_{e_2}+2t). \end{align*} $$

The first and the third are smaller than exactly two of the pseudo-divisors in the set $\mathcal U$ , while the second is smaller that every pseudo-divisor in $\mathcal U$ . Thus, we have $g(\gamma _1)=(\emptyset , D^{\prime \prime }_1)$ , $g(\gamma _2)=(\emptyset , D^{\prime \prime }_2)$ , and $g(\gamma _3)=(\emptyset , D^{\prime \prime }_3)$ . This finishes the proof.

Proof. Set $D_1:=D-v_e+s$ and $D_2:=D-v_e+t$ and recall Remark 2.4. For $v\in V(\Gamma )$ , we have

$$ \begin{align*} &D_1(v)=\begin{cases} D(v)&\text{ if }v\neq s,\\ D(v) + 1&\text{ if }v=s, \end{cases} & D_2(v)= \begin{cases} D(v)&\text{ if }v\neq t,\\ D(v) + 1&\text{ if }v=t. \end{cases} \end{align*} $$

Let $s_0$ and $t_0$ be the end-vertices of $e_0$ . We can assume without loss of generality that

$$\begin{align*}D'(v)=\begin{cases} D_1(v)& \text{ if }v\neq s_0,\\ D_1(v) + 1&\text{ if }v= s_0. \end{cases} \end{align*}$$

Assume by contradiction that

$$\begin{align*}D'(v)=\begin{cases} D_2(v)&\text{ if }v\neq s_0,\\ D_2(v) + 1&\text{ if }v= s_0. \end{cases} \end{align*}$$

Hence, we would have that $D_1=D_2$ , a contradiction. Then we have that $s_0\neq t_0$ and

$$\begin{align*}D'(v)=\begin{cases} D_2(v)&\text{ if }v\neq t_0,\\ D_2(v)+1&\text{ if }v= t_0. \end{cases} \end{align*}$$

If $t_0\not \in \{s,t\}$ , then $D'(t_0)=D_2(t_0)+1 = D(t_0)+1$ and $D'(t_0)=D_1(t_0)=D(t_0)$ , which is a contradiction. So we have that $t_0\in \{s,t\}$ , and analogously, we have that $s_0\in \{s,t\}$ . This proves that $\{s_0,t_0\}=\{s,t\}$ , and hence, the edges $e,e_0$ are parallel edges of $\Gamma $ .

Question. Are there examples of triples $(\Gamma , v_0,\mu )$ and $(\Gamma ',v_0^{\prime },\mu ')$ for which $\mathbf {QD}_{v_0,\mu }(\Gamma )$ and $\mathbf {QD}_{v_0^{\prime },\mu '}(\Gamma ')$ are isomorphic but it is not true that the biconnected components of $\Gamma /\operatorname {\mathrm {Br}}(\Gamma )$ and $\Gamma '/\operatorname {\mathrm {Br}}(\Gamma ')$ are of the same number and isomorphic under some bijection?

Proof. For (1), we notice that the following map is an isomorphism of posets:

$$ \begin{align*} \mathbf{QD}_{v_0,\mu}(\Gamma)&\to \mathbf{QD}_{v_0,\mu'}(\Gamma)\\ (\mathcal E,D)&\mapsto (\mathcal E, D-D_0). \end{align*} $$

For (2), it is enough to notice that $\mu _{\operatorname {\mathrm {can}}}(\Gamma _0)=\mu _{\operatorname {\mathrm {can}}}(\Gamma )-D_0$ , where $D_0=\sum _{v\in V(\Gamma )}w_{\Gamma }(v)v$ .

Proof. Let $e\neq e_1$ be an edge of $\Gamma $ parallel to $e_1$ and $e_2$ . Using that $D\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\{e_1,e_2\})$ , it is easy to see that $D-v_{e_2}+v_e$ is in $\mathbf {QD}_{v_0,\mu }(\Gamma , \{e,e_1\})$ . We also have

$$ \begin{align*} (\{e,e_1\},D-v_{e_2}+v_e)&\geq (\{e_1\},D-v_{e_2}+s),\\ (\{e,e_1\},D-v_{e_2}+v_e)&\geq(\{e_1\},D - v_{e_2}+t). \end{align*} $$

That means that the number of quasistable pseudo-divisors $(\mathcal E,\widetilde {D})$ on $\Gamma $ with $|\mathcal E|=2$ such that

$$\begin{align*}(\mathcal E,\widetilde{D})\geq (\{e_1\},D-v_{e_2}+s) \;\;\text{ and } \;\; (\mathcal E,\widetilde{D}) \geq (\{e_1\},D-v_{e_2}+t) \end{align*}$$

is at least the number of edges parallel to $e_1$ and different from $e_1$ .

However, let us see that the number of pseudo-divisors $(\mathcal E',\widetilde {D}')$ on $\Gamma '$ with $|\mathcal E'|=2$ such that

$$\begin{align*}(\mathcal E',\widetilde{D}')\geq (\{e_1^{\prime}\},D'-v_{e_2^{\prime}}+s') \;\text{ and } \; (\mathcal E',\widetilde{D}')\geq (\{e_1^{\prime}\},D' -v_{e_2^{\prime}}+t') \end{align*}$$

is exactly one. Indeed, we have that $(\{e_1^{\prime },e_2^{\prime }\},D')$ satisfy this condition. Moreover, any other such pseudo-divisor must satisfy that $\mathcal E'=\{e_1^{\prime },e_2^{\prime }\}$ .

Assume that we have another such pseudo-divisor $(\{e_1^{\prime },e_2^{\prime }\}, \widetilde {D}')$ . The poset

$$\begin{align*}\{ (\{e_1^{\prime},e_2^{\prime}\}, \widetilde{D}'),\; (\{e_1^{\prime},e_2^{\prime}\}, D'), ( \{e_1^{\prime}\}, D'-v_{e_2^{\prime}}+ s'), \; (\{e_2^{\prime}\}, D'-v_{e_1^{\prime}}+t') \} \end{align*}$$

is a copy of the poset $\mathbf P$ inside $\mathbf {QD}_{v_0^{\prime },\mu '}(\Gamma ')$ . Proposition 4.2 characterizes such copies of $\mathbf P$ inside $\mathbf {QD}_{v^{\prime }_0,\mu '}(\Gamma ')$ , and it is clear that we are in the case of Item (2). Hence, $D'-v_{e_2^{\prime }}+s'-v_{e_1^{\prime }}=D'-v_{e_1^{\prime }}+t'-v_{e_2^{\prime }}$ , which is a contradiction since $s'\ne t'$ .

Since f is an isomorphism of posets, we deduce that there exists exactly one edge parallel to $e_1$ , which must be $e_2$ . Since $(\{e_1,e_2\},D)$ is a quasistable pseudodivisor, by Remark 3.1, we have that the removal of $e_1,e_2$ does not disconnect the graph; hence $e_1,e_2$ is a special pair of $\Gamma $ . Arguing similarly for $f^{-1}$ , we have that $e_1^{\prime },e_2^{\prime }$ is a special pair of $\Gamma '$ .

Proof. The result is clear if $D_1=D_2$ , so we can assume that $D_1\neq D_2$ . By Proposition 3.3, it is sufficient to prove the result when there exists a pseudo-divisor $(\{e,e_0\}, D)$ that specializes to both $(\{e\},D_1)$ and $(\{e\}, D_2)$ . By Remark 2.4, the edge $e_0$ is not a loop. and, denoting by s and t the end-vertices of $e_0$ , we have $D_1=D-v_{e_0}+s$ and $D_2=D-v_{e_0}+t$ . Set M $D_i^{\prime }:=\delta _{\Gamma ^{\prime }}(f(\{e\},D_i))$ for $i=1,2$ .

Assume that $e_1^{\prime }\neq e_2^{\prime }$ . For $i=1,2$ , we have that

$$\begin{align*}f(\{e,e_0\},D)\geq f(\{e\},D_i)=(\{e_i^{\prime}\},D_i^{\prime}), \end{align*}$$

which implies that $f(\{e,e_0\},D)=(\{e_1^{\prime },e_2^{\prime }\},D')$ , for some divisor $D'\in \mathbf {QD}_{v^{\prime }_0,\mu '}(\Gamma ',\{e^{\prime }_1,e^{\prime }_2\})$ .

Again by Remark 2.4, there is an end-vertex $s_i^{\prime }$ of $e_i^{\prime }$ for $i=1,2$ , such that $D_1^{\prime }=D'-v_{e_2^{\prime }}+s_2^{\prime }$ and $D_2'=D'-v_{e_1^{\prime }}+s_1^{\prime }$ . If we set $\widetilde {D}' := D'-v_{e_1^{\prime }}-v_{e_2^{\prime }}+s_1^{\prime }+s_2^{\prime }$ , then $(\emptyset , \widetilde {D}')\leq (\{e_i^{\prime }\},D_i^{\prime })$ for $i=1,2$ . Set $(\emptyset ,\widetilde {D}):=f^{-1}(\emptyset , \widetilde {D}')$ . Therefore, for $i=1,2$ , we have

$$\begin{align*}(\emptyset, \widetilde{D})\leq (\{e\},D_i)\leq (\{e,e_0\},D). \end{align*}$$

We see that $(\{e,e_0\},D) $ satisfies the hypotheses of Lemma 4.6; hence, the edges e and $e_0$ of $\Gamma $ are parallel.

Now consider the image of the poset $\mathbf R_{e,e_0}(D)\subset \mathbf {QD}_{v_0,\mu }(\Gamma )$ via the isomorphism f. By Proposition 4.5, we have $f(\mathbf R_{e,e_0}(D))=\mathbf R_{e_1^{\prime },e_2^{\prime }}(D')$ , and the edges $e^{\prime }_1,e^{\prime }_2$ are parallel, with

$$\begin{align*}f(\{e\},D-v_{e_0}+s)&=(\{e_1^{\prime}\},D'-v_{e_2^{\prime}}+s_2^{\prime})\\f(\{e\},D-v_{e_0}+t)&=(\{e_2^{\prime}\},D'-v_{e_1^{\prime}}+s_1^{\prime}). \end{align*}$$

By contradiction, assume that $s':=s_1^{\prime }=s_2^{\prime }$ , and let $t'$ be the other end-vertex of $e^{\prime }_1$ and $e^{\prime }_2$ . Then the two pseudo-divisors $(\emptyset , D'-v_{e_1^{\prime }}-v_{e_2^{\prime }}+2s')$ and $(\emptyset , D'-v_{e_1^{\prime }}-v_{e_2^{\prime }}+s'+t')$ of $\mathbf {R}_{e_1^{\prime },e_2^{\prime }}(D')$ would be smaller than both $(\{e_1^{\prime }\},D_1^{\prime })$ and $(\{e_2^{\prime }\},D_2^{\prime })$ . On the other hand, there is only one element of $\mathbf {R}_{e,e_0}(D)$ smaller then $(\{e\},D_1)$ and $(\{e\},D_2)$ , which is $(\emptyset , D-v_{e}-v_{e_0}+s+t)$ . We have a contradiction, which proves that $s_1^{\prime }\neq s_2^{\prime }$ . So the end-vertices of $e^{\prime }_1$ and $e^{\prime }_2$ are $s':=s^{\prime }_2$ and $t':=s^{\prime }_1$ . We see that the hypotheses of Lemma 5.6 are satisfied; hence, the pairs $e,e_0$ and $e_1^{\prime },e_2^{\prime }$ are special pairs of $\Gamma $ and $\Gamma '$ , respectively, and we are done.

Proof. The result readily follows from Proposition 5.7.

Proof. By Corollary 5.8, we can define a bijection $f_E\colon \operatorname {\mathrm {ND}}(\Gamma )\to \operatorname {\mathrm {ND}}(\Gamma ')$ as follows.

1. 1. For each edge $e\in \operatorname {\mathrm {ND}}(\Gamma )$ that does not belong to a special pair of $\Gamma $ , we set $f_E(e)$ to be the unique edge of $\Gamma '$ satisfying $\{f_E(e)\} =\epsilon _{\Gamma ^{\prime }}(f(\{e\},D))$ for every divisor $D\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\{e\})$ . (Notice that $\mathbf {QD}_{v_0,\mu }(\Gamma ,\{e\})$ is not empty since e is not a bridge of $\Gamma $ ).

2. 2. For each special pair $\{e_1,e_2\}$ of $\Gamma $ , we let $f_E(e_1), f_E(e_2)$ be the edges of $\Gamma '$ such that

   $$\begin{align*}\{f_E(e_1),f_E(e_2)\} = \epsilon_{\Gamma^{\prime}}(f(\{e_1,e_2\},D)), \end{align*}$$
   for every divisor $D\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\{e_1,e_2\})$ . (Here, there is a choice to be made: a different choice would switch the values $f_E(e_1)$ and $f_E(e_2)$ .) Notice that $\{f_E(e_1),f_E(e_2)\}$ is a special pair, and hence, it is contained in $\operatorname {\mathrm {ND}}(\Gamma )$ .

Let us prove that for every $(\mathcal E,D)\in \mathbf {QD}_{v_0,\mu }(\Gamma )$ , there is a unique divisor $D'\in \mathbf {QD}_{v_0^{\prime },\mu '}(\Gamma ', f_E(\mathcal E))$ such that $f(\mathcal E,D)\sim (f_E(\mathcal E), D')$ . By Remark 5.10, it suffices to prove that $\epsilon _{\Gamma ^{\prime }}(f(\mathcal E,D))$ and $f_E(\mathcal E)$ are equivalent subsets of $E(\Gamma ')$ . Set $\mathcal E':=\epsilon _{\Gamma ^{\prime }}(f(\mathcal E,D))$ . For each subset $\mathcal E_0\subset \mathcal E$ , there exists a divisor $D_0\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\mathcal E_0)$ such that $(\mathcal E_0,D_0)\leq (\mathcal E,D)$ (see Remark 2.4). Moreover, we have $f(\mathcal E_0,D_0)\leq f(\mathcal E,D)$ , and hence, $\epsilon _{\Gamma ^{\prime }}(f(\mathcal E_0,D_0))\subset \epsilon _{\Gamma ^{\prime }}(f(\mathcal E,D))=\mathcal E'$ . Thus, the following conditions hold:

1. 1. If an edge $e\in \mathcal E$ does not belong to a special pair, then $f_E(e)\in \mathcal E'$ .

2. 2. If $\{e_1,e_2\}\subset \mathcal E$ is a special pair, then $f_E(\{e_1,e_2\})\subset \mathcal E'$ .

3. 3. If an edge $e_1\in \mathcal E$ belongs to a special pair $\{e_1,e_2\}$ with $e_2$ not in $\mathcal E$ , then either $f_E(e_1)\in \mathcal E'$ or $f_E(e_2)\in \mathcal E'$ , but $\{f_E(e_1),f_E(e_2)\}\not \subset \mathcal E'$ . Moreover, $\{f_E(e_1),f_E(e_2)\}$ is a special pair of $\Gamma '$ .

This concludes the proof that $\epsilon _{\Gamma ^{\prime }}(f(\mathcal E,D))$ and $f_E(\mathcal E)$ are equivalent.

Next, we prove that $f_E$ is a weakly cyclic equivalence. By Remark 2.3, it is enough to prove that $f_E$ and $f_E^{-1}$ take maximally nondisconnecting subsets to maximally nondisconnecting subsets. Let $\mathcal E\subset E(\Gamma )$ be a maximally nondisconnecting subset. Then there exists exactly one divisor $D\in \mathbf {QD}_{v_0,\mu }(\Gamma ,\mathcal E)$ (see Remark 3.1). We also have that $(\mathcal E, D)$ is maximal in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ . We set $(\mathcal E',D'):=f(\mathcal E,D)$ . Then $(\mathcal E',D')$ is maximal in $\mathbf {QD}_{v_0^{\prime },\mu '}(\Gamma ')$ , which implies that $\mathcal E'$ is a maximally nondisconnecting subset (see Remark 3.1). Since $f_E(\mathcal E)$ and $\mathcal E'$ are equivalent, we have that $f_E(\mathcal E)$ is also a maximally nondisconnecting subset. The number of spanning trees of $\Gamma $ and $\Gamma '$ is equal to the number of maximal elements of $\mathbf {QD}_{v_0,\mu }(\Gamma )$ and $\mathbf {QD}_{v^{\prime }_0,\mu '}(\Gamma ')$ , respectively; hence, they are equal because f is an isomorphism. Since the number of spanning trees of $\Gamma $ and $\Gamma '$ are the same, it follows that $f_E^{-1}$ also takes maximally nondisconnecting subsets to maximally nondisconnecting subsets. This concludes the proof that $f_E$ is a weakly cyclic equivalence.

Now we will prove that f and $f^{-1}$ take equivalent pseudo-divisors to equivalent pseudo-divisors. We proceed by induction on the rank of a pseudo-divisor. Let $(\mathcal E_1,D_1)$ and $(\mathcal E_2,D_2)$ be two equivalent pseudo-divisors of rank k on $\Gamma $ . If $k=0$ – that is, if $|\mathcal E_1|=|\mathcal E_2|=0$ – then $D_1=D_2$ , and hence, $f(\mathcal E_1,D_1)=f(\mathcal E_2,D_2)$ , and we are done. The same reasoning holds for $f^{-1}$ .

By the induction hypothesis, f and $f^{-1}$ send equivalent pseudo-divisors of rank strictly less than k to equivalent pseudo-divisors. We will prove the induction step only for f. The reasoning for $f^{-1}$ is similar. It is enough to prove the result for $\mathcal E_1=\mathcal E\cup \{e_1\}$ and $\mathcal E_2=\mathcal E\cup \{e_2\}$ , for some $\mathcal E\subset E(\Gamma )$ and for some special pair $\{e_1,e_2\}$ of $\Gamma $ such that $\mathcal E\cap \{e_1,e_2\}=\emptyset $ . Let s and t be the end-vertices of $e_1$ and $e_2$ . Define

$$ \begin{align*} D_s :=& D_1 - v_{e_1}+s=D_2-v_{e_2}+s\\ D_t :=& D_1 - v_{e_1}+t=D_2-v_{e_2}+t. \end{align*} $$

Notice that we have

(10) $$ \begin{align} D_s(s)=D_t(s)+1 \quad\text{ and } \quad D_s(t)=D_t(t)-1. \end{align} $$

By Remark 2.4, we have that

$$ \begin{align*} (\mathcal E,D_s)\leq (\mathcal E_1,D_1), \; \; & (\mathcal E,D_s)\le(\mathcal E_2,D_2),\\ (\mathcal E,D_t)\leq (\mathcal E_1,D_1), \; \; & (\mathcal E,D_t)\leq (\mathcal E_2,D_2).\end{align*} $$

In particular, the set $\{(\mathcal E_1,D_1),(\mathcal E_2,D_2),(\mathcal E,D_s),(\mathcal E,D_t)\}$ is a poset isomorphic to the poset $\mathbf P$ in Definition 4.1. Therefore, the image of this set via f must be one of the images described in Proposition 4.2. Set $(\mathcal E_1^{\prime },D_1^{\prime })=f(\mathcal E_1,D_1)$ and $(\mathcal E_2^{\prime },D_2^{\prime })=f(\mathcal E_2,D_2)$ .

By contradiction, assume that we are in the situation described in item (2) of Proposition 4.2. This implies that there exist parallel edges $\{e_1^{\prime },e_2^{\prime }\}$ , a subset $\mathcal E'\subset E(\Gamma ')\setminus \{e_1^{\prime },e_2^{\prime }\}$ and a divisor $D'$ on $\Gamma ^{{\prime }\mathcal E'}$ such that $f(\mathcal E,D_s) = (\mathcal E'\cup \{e_1^{\prime }\},D'+v_{e_1^{\prime }})$ and $f(\mathcal E,D_t)=(\mathcal E'\cup \{e_2^{\prime }\},D'+v_{e_2^{\prime }})$ . By induction hypothesis (recall that $|\mathcal E|=k-1$ ), we have that $\mathcal E'\cup \{e_1^{\prime }\}$ and $\mathcal E'\cup \{e_2^{\prime } \}$ are both equivalent to $f_E(\mathcal E)$ ; hence, $\mathcal E'\cup \{e_1^{\prime }\}$ and $\mathcal E'\cup \{e_2^{\prime }\}$ are equivalent subsets of $E(\Gamma ')$ . We deduce that $\{e_1^{\prime },e_2^{\prime }\}$ is a special pair, and hence, $(\mathcal E'\cup \{e_1^{\prime }\},D'+v_{e_1^{\prime }})$ and $(\mathcal E'\cup \{e_2^{\prime }\},D'+v_{e_2^{\prime }})$ are equivalent pseudo-divisors of $\Gamma '$ . However, the ranks of $(\mathcal E'\cup \{e_1^{\prime }\},D'+v_{e_1^{\prime }})$ and $ (\mathcal E'\cup \{e_2^{\prime }\},D'+v_{e_2^{\prime }})$ are equal to $k-1$ , so by the induction hypothesis, we get that $f^{-1}(\mathcal E'\cup \{e_1^{\prime }\},D'+v_{e_1^{\prime }})=(\mathcal E,D_s)$ and $f^{-1}(\mathcal E'\cup \{e_2^{\prime }\},D'+v_{e_2^{\prime }})=(\mathcal E,D_t)$ are equivalent pseudo-divisors of $\Gamma $ . This implies that $D_s(v)=D_t(v)$ for every $v\in V(\Gamma )$ , which contradicts Equation (10).

We deduce that we are in the situation described in item (1) of Proposition 4.2. Then there exist parallel edges $e_1^{\prime },e_2^{\prime }$ of $\Gamma '$ , a subset $\mathcal E'\subset E(\Gamma ')\setminus \{e^{\prime }_1,e^{\prime }_2\}$ of $\Gamma '$ and a divisor $D'\in \Gamma ^{{\prime }\mathcal E'}$ such that $(\mathcal E_1',D^{\prime }_1)=(\mathcal E'\cup \{e_1^{\prime }\},D'+v_{e^{\prime }_1})$ and $(\mathcal E_2^{\prime },D^{\prime }_2)=(\mathcal E'\cup \{e_2^{\prime }\},D'+v_{e^{\prime }_2})$ . Hence, $D^{\prime }_1(v)=D^{\prime }_2(v)$ for every $v\in V(\Gamma )$ , and so $D^{\prime }_1$ and $D^{\prime }_2$ satisfy Condition (1) of Definition 5.9. Moreover,

1. 1. the subsets $\mathcal E'\cup \{e_1^{\prime }\}$ and $f_E(\mathcal E\cup \{e_1\})$ of $E(\Gamma ')$ are equivalent, by construction.

2. 2. the subsets $\mathcal E'\cup \{e_2^{\prime }\}$ and $f_E(\mathcal E\cup \{e_2\})$ of $E(\Gamma )$ are equivalent, by construction.

3. 3. the subsets $f_E(\mathcal E\cup \{e_1\})$ and $f_E(\mathcal E\cup \{e_2\})$ of $E(\Gamma ')$ are equivalent since $f_E$ sends special pairs to special pairs.

This implies that $\mathcal E^{\prime }_1=\mathcal E'\cup \{e^{\prime }_1\}$ and $\mathcal E^{\prime }_2=\mathcal E'\cup \{e_2'\}$ are equivalent, and hence, $f(\mathcal E_1,D_1)\sim f(\mathcal E_2,D_2)$ , concluding the proof.

Proof. Let us prove that $h_f$ is a bijection. We begin by proving that $h_f$ is injective. Assume that $h_f(\mathcal E_1,D_1)=h_f(\mathcal E_2,D_2)$ for some pseudo-divisors $(\mathcal E_1,D_1)$ and $(\mathcal E_2,D_2)$ on $\Gamma $ . Since $\epsilon _{\Gamma '}(h_f(\mathcal E_i,D_i))= f_E(\mathcal E_i)$ , we have that $f_E(\mathcal E_1)=f_E(\mathcal E_2)$ , which implies that $\mathcal E_1=\mathcal E_2=:\mathcal E$ (recall that $f_E$ is a bijection; see Proposition 5.12). Writing $(\mathcal E_i^{\prime },D_i^{\prime }):=f(\mathcal E,D_i)$ for $i=1,2$ , we have that

$$\begin{align*}(\mathcal E_1^{\prime},D_1^{\prime})\sim h_f(\mathcal E,D_1) = h_f(\mathcal E,D_2) \sim (\mathcal E_2^{\prime},D_2^{\prime}); \end{align*}$$

hence, $(\mathcal E_1^{\prime },D_1^{\prime })\sim (\mathcal E_2^{\prime },D_2^{\prime })$ . By Proposition 5.12, we have that

$$\begin{align*}(\mathcal E,D_1) = f^{-1}(\mathcal E_1^{\prime},D_1^{\prime})\sim f^{-1}(\mathcal E_2^{\prime},D_2^{\prime}) = (\mathcal E,D_2), \end{align*}$$

and hence, $D_1=D_2$ by Remark 5.10. This finishes the proof of the injectivity of $h_f$ . Since $\mathbf {QD}_{v_0,\mu }(\Gamma )$ and $\mathbf {QD}_{v^{\prime }_0,\mu '}(\Gamma ')$ are finite sets of the same cardinality, it follows that $h_f$ is bijective.

Let us prove that $h_f$ is a morphism of ranked posets. It is clear that $h_f$ preserves the rank of pseudo-divisors. Assume that $(\mathcal E_1,D_1)\geq (\mathcal E_2,D_2)$ in $\mathbf {QD}_{v_0,\mu }(\Gamma )$ . In particular, $\mathcal E_2\subset \mathcal E_1$ . We have that

$$ \begin{align*} &f(\mathcal E_1,D_1)\geq f(\mathcal E_2,D_2), \\ & h_f(\mathcal E_i,D_i)\sim f(\mathcal E_i,D_i),\\ & \epsilon_{\Gamma^{\prime}}(h_f(\mathcal E_2,D_2))=f_E(\mathcal E_2)\subset f_E(\mathcal E_1)=\epsilon_{\Gamma^{\prime}}(h_f(\mathcal E_1,D_1)). \end{align*} $$

Thus, $h_f(\mathcal E_1,D_1)\geq h_f(\mathcal E_2,D_2)$ by Remark 5.11, concluding the proof that $h_f$ is a morphism of ranked posets.

Using the same reasoning, we also have that $h_{f^{-1}}\colon \mathbf {QD}_{v^{\prime }_0,\mu '}(\Gamma ')\to \mathbf {QD}_{v_0,\mu }(\Gamma )$ is a morphism of ranked posets. It remains to prove that $h_{f^{-1}}$ is the inverse of $h_f$ . Fix $(\mathcal E',D')=h_f(\mathcal E,D)$ . We have the following equivalences:

1. 1. $h_{f^{-1}}(\mathcal E',D')\sim f^{-1}(\mathcal E',D')$ , by the definition of $h_{f^{-1}}$ .

2. 2. $f^{-1}(\mathcal E',D')\sim f^{-1}(f(\mathcal E,D))=(\mathcal E,D)$ because $(\mathcal E',D'`)\sim f(\mathcal E,D)$ by the definition of $h_f$ and because $f^{-1}$ takes equivalent divisors to equivalent divisors (see Proposition 5.12).

Therefore, $h_{f^{-1}}(\mathcal E',D')\sim (\mathcal E,D)$ . By the definition of $h_f$ and $h_{f^{-1}}$ , we have $\mathcal E'=f_E(\mathcal E)$ and $\epsilon _{\Gamma }(h_{f^{-1}}(\mathcal E',D'))=f_E^{-1}(\mathcal E')$ . It follows that

$$\begin{align*}\epsilon_{\Gamma}(h_{f^{-1}}(\mathcal E',D'))=f_E^{-1}(\mathcal E')=\mathcal E=\epsilon_{\Gamma}(\mathcal E,D). \end{align*}$$

Hence, $h_{f^{-1}}(\mathcal E',D')= (\mathcal E,D)$ by Remark 5.10. This finishes the proof.

Proof. By Remark 3.1, the poset $\mathbf {QD}_{v_0}(\Gamma )$ is a singleton. So it is enough to prove that the divisor D given by the formula in the statement is $v_0$ -quasistable. By Equation (7), we have $\beta _{\Gamma ,D}(V) = D(V)- g_V+1$ for every hemisphere $V\subset V(\Gamma )$ . Since $\Gamma $ is a tree, we have $g_V = 0$ for every hemisphere $V\subset V(\Gamma )$ . We also have

$$\begin{align*}D(V)=\begin{cases} 0& \text{ if } v_0\notin V,\\ -1& \text{ if } v_0\in V.\\ \end{cases} \end{align*}$$

It follows that

$$\begin{align*}\beta_{\Gamma,D}(V) = \begin{cases} 0 & \text{ if } v_0\in V, \\ 1 & \text{ if } v_0\notin V. \end{cases} \end{align*}$$

This proves that D is $v_0$ -quasistable.

Proof. By Remark 3.1 we can assume that $\mathcal E_1=\emptyset $ and $\Gamma \setminus \{v_1\}$ is a tree.

For $e\in E(v_1)$ , we set $S_e:=E(v)\setminus \{e\}$ . By Lemma 5.15, there exists a unique divisor $D_{S_e}\in \mathbf {QD}_{v_0}(\Gamma ,S_e)$ , and we have $D_{S_e}(v_0)=-1$ and $D_{S_e}(u)=0$ for every $u\in V(\Gamma )\setminus \{v_0\}$ . Note that $S_e$ is a maximally nondisconnecting subset of $\Gamma $ , and, vice versa, any maximally nondisconnecting subset of $\Gamma $ is of the form $S_e$ for some $e\in E(v_1)$ . In particular, by Remark 3.1, we have that $\{(S_e,D_{S_e})\}_{e\in E(v_1)}$ is the set of all maximal elements of $\mathbf {QD}_{v_0}(\Gamma )$ .

Set $D_1:=D_{S_e}-\sum _{\widetilde {e}\in S_e} v_{\widetilde {e}} + (\operatorname {\mathrm {val}}(v)-1)v$ . We have that $(\emptyset , D_1)$ is a specialization of $(S_e,D_{S_e})$ , and since $(S_e,D_{S_e})$ is $v_0$ -quasistable, it follows that $(\emptyset , D_1)$ is $v_0$ -quasistable as well (see Remark 3.1). This proves the existence of $D_1$ . Note that $D_1$ is independent of the choice of $e\in E(v_1)$ .

However, if $\widetilde {D}_1$ is another such divisor, then $(\emptyset ,\widetilde {D}_1)$ is smaller than a maximal element $(S_e,D_{S_e})$ for some $e\in E(v)$ . By Lemma 5.15 and Equation 13, we can write $\widetilde {D}_1(v_1)=D_{S_e}(v_1)+\operatorname {\mathrm {val}}(v_1)-1$ . Since $|S_e|=\operatorname {\mathrm {val}}(v_1)-1$ , the only possible way for $(S_e,D_{S_e})$ to specialize to $(\emptyset ,\widetilde {D}_1)$ is if $\widetilde {D}_1=D_{S_e}-\sum _{\widetilde {e}\in S_e} v_{\widetilde {e}} + (\operatorname {\mathrm {val}}(v)-1)v$ . This means that $\widetilde {D}_1=D_1$ and finishes the proof of the first statement.

Fix $S\subsetneqq E(v_1)$ . There exists $e\in E(v_1)$ such that $S\subset S_e$ . The divisor $D_S:=D_{S_e}-\sum _{\widetilde {e}\in S_e\setminus S}v_{\widetilde {e}}+|S_e\setminus S|v$ is independent of the choice of e and satisfies $(\emptyset ,D_1)\leq (S,D_S)$ . Moreover, we have $D_S\in \mathbf {QD}_{v_0}(\Gamma ,S)$ because the $v_0$ -quasistable pseudo-divisor $(S_e, D_{S_e})$ specializes to $(S, D_s)$ (see Remark 3.1).

We claim that the divisor $(S,D_S)$ is the unique divisor in $\mathbf {QD}_{v_0}(\Gamma , S)$ such that $(\emptyset ,D_1)\le (S,D_S)$ . Indeed, assume that $\widetilde {D}_S$ is another divisor in $\mathbf {QD}_{v_0}(\Gamma ,S)$ such that $(\emptyset ,D_1)\leq (S, \widetilde {D}_S)$ . Then, there exists a maximal pseudo-divisor $(S_e,D_{S_e})$ that is greater than $(S,\widetilde {D}_S)$ . This implies that

(14) $$ \begin{align} \widetilde{D}_S(v_1)\leq D_{S_e}(v_1)+ \operatorname{\mathrm{val}}(v_1)-|S|-1. \end{align} $$

However, since $(S,\widetilde {D}_S)$ is $v_0$ -quasistable and greater than $(\emptyset ,D_1)$ , we must have that $\widetilde {D}_S(v_1)\geq D_{S_e}(v_1)+\operatorname {\mathrm {val}}(v_1)-|S|-1$ , and hence, equality holds in Equation (14). This implies that $\widetilde {D}_S=D_S$ .

Proof. By Remark 3.1, we can assume that $\mathcal E=\emptyset $ . This implies that $\Gamma (V)$ and $\Gamma (V^c)$ are trees.

Consider $S\subsetneqq E(V,V^c)$ and the unique divisor $D_S$ of the statement. Let us prove that there exists only one specialization $(S,D_S)\to (\emptyset ,D)$ . Assume, by contradiction, that there are two different specializations $\iota _1,\iota _2\colon (S,D_S)\to (\emptyset ,D)$ . This implies that we have distinct edges $e_1$ and $e_2$ in S such that $\iota _1(v_{e_1}),\iota _2(v_{e_2})\in V$ and $\iota _1(v_{e_2}),\iota _2(v_{e_1})\in V^c$ (note that the degrees of $(\iota _1)_*(D_S)=D$ and $(\iota _2)_*(D_S)=D$ in V and $V^c$ are the same).