TITLE: Isomorphic Hadwiger graphs
QUESTION [6 upvotes]: Let $G$ be a graph, then we define its Hadwiger graph $\textrm{Hadw}(G)$ in the following way:

$V(\textrm{Hadw}(G)) = \{S\subseteq (V(G): S\neq \emptyset\textrm{ and } S \textrm{ is connected}\}$;
$E(\textrm{Hadw}(G)) = \{\{S,T\}\subseteq V(\textrm{Hadw}(G)): S\cap T = \emptyset \textrm{ and } (\exists s\in S, t\in T): \{s,t\}\in E(G) \}.$

$G$ embeds to $\textrm{Hadw}(G)$ by $v\mapsto \{v\}$, and the Hadwiger conjecture states $\chi(G) \leq \omega(\textrm{Hadw}(G))$.
Question: If $G, H$ are (finite or infinite) graphs, does $\textrm{Hadw}(G) \cong \textrm{Hadw}(H)$ imply $G\cong H$?

REPLY [2 votes]: $\def\Hadw{\mathop{\rm Hadw}}$If it is not a mauvais ton, I would like to add an easier proof found independently by Sergey Dolgikh and Marat Abdrakhmanov (we used this fact as a contest problem). We assume that $G$ is finite and connected, and again we find all vertices in $H=\Hadw(G)$ corresponding to the vertices of $G$.
Lemma. If $S$ is a vertex of $\Hadw(G)$ with $|S|>1$ and $v\in S$, then $\deg_H S<\deg_H \{v\}$.
Proof. Every neighbor $T$ of $S$ in $H$ can be augmented by a path in $S$ to provide a neighbor $T'$ of $\{v\}$; all these neighbors are distinct since $T'\setminus S=T$. Moreover, $\{v\}$ has some neighbor contained in $S$.
Now, for every $T\in V(H)\setminus\{V(G)\}$ we find its neighbor of maximal degree. By the lemma, this neighbor is a one-vertex set. On the other hand, for every $v\in V(G)$, the set $\{v\}$ is the unique maximal degree neighbor of any component of $G-\{v\}$. Thus we have reconstructed all sets of the form $\{v\}$.