I am wondering whether the following construction for a simple graph for coloring in an attempt on the Erdős-Faber-Lovász conjecture is correct or has been mentioned before. The goal is to construct a simple graph, such that every proper vertex coloring of it can be easily mapped to a proper edge coloring of a linear hypergraph on $n$ vertices in a variant formulation of Erdős-Faber-Lovász, as in Conjecture 1.3 in Jan,Nag.
Conjecture 1.3 (Erdős-Faber-Lovász, 2nd (dual) variant). Any linear hypergraph on $n$ vertices has chromatic index at most $n$.
The construction goes as follows.
- Each vertex in hypergraph $H=(V(H),\mathscr{E})$ is mapped to a set of vertices of simple graph $G=(V(G),E(G))$, so that the vertex has a copy in $G$ for each hyperedge in $H$ it belongs to. $$ S: \{v_i: v_i\in V(H)\}\longrightarrow\{\{u_{i,j}:u_{i,j}\in V(G)\}:v_i\in e_j,\forall e_j\in\mathscr{E}\} $$
- Each hyperedge of $H$ is mapped to a subgraph of $G$ consisting of all copies of the vertices associated with that hyperedge, and not those with another hyperedge. The copied vertices are connected completely to an $(n-1)$-clique. The cliques ($K_{n-1}$) are a new addition to $G$ that are not mapped from anything in $H$. They have no other adjacencies except as specified in this clause. $$T:\{e_j:e_j\in \mathscr{E}\}\longrightarrow\{u_{i,j}:u_{i,j}\in S(v_i),\forall v_i\in e_j\}+K_{n-1} $$
- Finally, all copies of a vertex shared among hyperedges in $H$ are joined in a clique in $G$. Let such cliques be $Q(v_i)$. $Q(v_i)$ will be a $k$-clique if $v_i$ is shared among $k$ hyperedges in $\mathscr{E}$. $$ Q(v_i)=\sum S(v_i) $$
Therefore,$$ G=\bigcup\limits_{(v_i,e_j)\in H}\{T(e_j),E(\sum S(v_i))\} $$
Proposition: If there is a proper vertex coloring of $G$ in $n$ colors, then a proper coloring of (the hyperedges $\mathscr{E}$ of) $H$ is obtained by coloring $e_j\in \mathscr{E}$ with the color of the vertices $T(e_j)\cap S[V(H)]$. If this set has multiple vertices, they must be of the same color.
Proof: The vertices in $T(e_j)\cap S[V(H)]$ must have only a single color out of $n$ because of the $K_{n-1}$ cliques. Further, $T(e_j)\cap S[V(H)]$ and $T(e_k)\cap S[V(H)]$, $\forall j\ne k$, must have different colors, if they are connected, because of the $Q(v_i)$ cliques.
Furthermore, because two hyperedges in $\mathscr{E}$ can share at most one vertex, the subgraph $T(e_j)\cup T(e_k)\subseteq G$, where $j\neq k$, are either disconnected, or singly connected via an edge in $Q(e_j\cap e_k)$ (if $e_j\cap e_k\neq\emptyset$, $\vert e_j\cap e_k\vert=1$).
Is this right?
The motivation is that given its structure, I hope it can be shown that $G$ does not have any $K_{n+1}$ minor, and thus the Hadwiger conjecture implies the Erdős-Faber-Lovász conjecture.
Currently there are clearly multiple $K_{n}$'s in a $T(e_j)$. But the point about the $T(e_j)$'s is that in any coloring method using $n$ colors, once a color is picked for a vertex of $T(e_j)\cap S[V(H)]$, the entire $T(e_j)$ is completely colored and can be treated as a single vertex of that color in the rest of the coloring, and in the search for minors when invoking Hadwiger. Is an argument like this meaningful? Was it seen before?