Before of all, I do some definitions. Let $G$ a graph. For each $Z \subseteq E(G)$, we denote the graph $G[Z] = (V(G), Z)$. Let ${\cal P}$ a partition of $V(G)$. Define the graph $G_{\cal P}$ obtained by identifyng of the vertices of each part of ${\cal P}$ (identifyng the vertices of inside of each part of ${\cal P}$). Note that $|V(G_{\cal P})| = |{\cal P}|$ and $|E(G_{\cal P})| = |E(G)|$. This is, $G_{\cal P} = (V(G_{\cal P}), E(G))$.
Let $V_s \in {\cal P}$. Define the partition ${\cal P} \div V_s$ by the $({\cal P} - V_s) \cup \{x_1,\ldots,x_k\}$ such that $x_i$ is a vertice of $V_s$. The figure below ilustrate the graph $G$ with the partition ${\cal P}$ of $V(G)$, $G_{\cal P}$, ${\cal P} \div V_j$ and $G_{{\cal P} \div V_j}$, respectively.
I wish show the following: Let $G$ be a graph and ${\cal P}$ a partition of $V(G)$. Let $E'$, $E$, $Z$ and $Z'$ subesets of $E(G)$ with $|E'| = |E| + 1$. I know that the vertices $u$ and $v$ are connected in $G_{\cal P}[E]$ if and only if $u$ and $v$ are connected in $G_{\cal P}[E']$. Similarly, $u$ and $v$ are connected in $G_{\cal P}[Z]$ if and only if $u$ and $v$ are connected in $G_{\cal P}[Z']$. Moreover, $G_{\cal P}[E \cup Z]$ is a forest. Let $V_s \in {\cal P}$. Finally, $G_{{\cal P} \div V_s} [E']$ is a forest such that exists exactly one vertices pair $\{x,y\}$ in $V_s$ such that $x$ and $y$ belongs the same connected component of $G_{{\cal P} \div V_s} [E']$. Let ${\cal P}'$ be a partition of $V(G)$ where each part of ${\cal P}'$ is the union of the parts of ${\cal P}$ such that the vertices corresponding in $H_{\cal P}[Z]$ belongs the same connected component.
Let $V_s'$ that contains $V_s$ How can I show that $G_{{\cal P} \div V_s'}[E']$ is a forest such that exists exactly one vertices pair $\{x,y\}$ in $V_s'$ such that $x$ and $y$ belongs the same connected component of $G_{{\cal P}' \div V_s'} [E']$?
Follows the reasonig: Since $G_{\cal P} [Z \cup E]$ is a forest, for each vertices pair $u$ and $v$ in the same part of ${\cal P}'$, we have that $u$ and $v$ are not connected in $G_{\cal P} [E]$. By the following affirmation: "$u$ and $v$ are connected in $G_{\cal P}[E]$ if and only if $u$ and $v$ are connected in $G_{\cal P}[E']$", we conclude that $u$ and $v$ are not connected in $G_{\cal P} [E']$. Therefore, each vertices pair $\{u,v\}$, we have that in the same part of $\cal P'$, we obtain that exists exactly one path of $u$ to $v$ (because else we would have a cycle in $H_{\cal P}[E \cup Z]$.)
The figure below illustrate the graph with the partition ${\cal P'}$ and ${\cal P}$ coloured with colour red and blue, respectively. The continuous and red edges are of $Z$ and the dashed and blue edges are of $E'$. Note that not exists one dashed edge below because $G_{\cal P}[E' \cup Z]$ is a forest.
Follows the figure:
How can I conclude this proof?