I did not get an answer when asking for help with this question in Math Stack Exchange (here). Anyway, I believe that this forum is more suitable for it.

I'm trying to solve a problem about connectivity of entangled vertices in a graph.

Here, two vertices $u, v$ of a finite graph $G(V, E)$ are said to be entangled if for any proper coloring $c:V(G)\rightarrow\mathbb{N}$ with $\chi(G)$ colors we have $c(u) = c(v)$, that is, they must have the same color.

What I'm trying to prove is that, given two entangled vertices $u, v\in V(G)$, there is $w\in V(G)$ (possibly equal to $v$) also entangled with $u$ so that there is a set of size $\chi(G)-1$ of pairwise internally vertex-disjoint paths from $u$ to $w$.

I was able to prove, using the vertex-connectivity version of Menger's theorem and induction, that the previous statement is true if $v$ is the only vertex in $G$ entangled with $u$, so I've been trying to show that if there is not a set of size $\chi(G)-1$ of pairwise internally vertex-disjoint paths from $u$ to $v$ (considering $u$ and $v$ entangled), there is still a vertex in $G-v$ entangled with $u$, but without success.

Another idea I had was showing that the minimal (in the number of edges) subgraph of $G$ for which there is still a vertex entangled with $u$, has exactly one vertex entangled with $u$. Because it is a stronger result this would be more convenient for me.

I would appreciate some help with this subject.

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  • $\begingroup$ I don't get "two by two". What does it mean? $\endgroup$ – Brendan McKay Mar 8 at 14:43
  • $\begingroup$ It means that any two paths in the set must have this property (be internally vertex-disjoint). $\endgroup$ – Arjuna196 Mar 8 at 14:56
  • $\begingroup$ Ok, then "pairwise" is more normal usage. But you don't actually need a word at all since "disjoint" means "pairwise disjoint". (Some people might argue.) $\endgroup$ – Brendan McKay Mar 8 at 15:16
  • $\begingroup$ Thanks for the tip. I edited the statement. $\endgroup$ – Arjuna196 Mar 8 at 19:28

Here is a proof to a related claim that hopefully will give you some ideas.

Claim. Let $X$ be an equivalence class of the entanglement relation on $V(G)$. Then for all distinct $u,v \in X$, there exist $\chi(G)-1$ edge-disjoint paths in $G$ between $u$ and $v$.

Proof. Let $k=\chi(G)$ and $V_1, \dots, V_k$ be a partition of $V(G)$ into stable sets. By relabelling, we may assume that $X \subseteq V_1$. Observe that all vertices of $X$ must be contained in some component of $G[V_1 \cup V_2]$. Otherwise, we may recolour to obtain a $k$-colouring of $G$ where two vertices of $X$ are coloured differently. In particular, for all distinct $u,v \in X$, there is a $u$--$v$ path in $G[V_1, \cup V_2]$. Repeating the argument for $i=2, \dots, k$, gives the $k-1$ edge-disjoint $u$--$v$ paths in $G$. $\square$

Note that the claim proves something stronger and weaker than what was asked in the original question. It is weaker because the paths are edge-disjoint not vertex-disjoint. But it is stronger since it holds for all distinct pairs $u,v \in X$. Moreover, the paths constructed in the proof are almost vertex-disjoint. The only vertices they have in common are in $V_1$.

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  • $\begingroup$ I noticed this fact. It was used for the demonstration that I mentioned in the statement, for the case where there is only one vertex entangled with $u$. I was inspired by the idea of ​​Kempe chains. $\endgroup$ – Arjuna196 Mar 10 at 13:56

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