>I did not get an answer when asking for help with this question in Math Stack Exchange [(here)](https://math.stackexchange.com/questions/3561468/a-problem-about-the-connectivity-of-vertices-that-must-have-the-same-color-for-a). 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 two by two 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 two by two 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 witch there is still a vertex entangled with $u$, have *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.