0
$\begingroup$

Let $G$ be a finite simple undirected graph. Suppose there exist subgraph $G_1,G_2,\dots,G_n$ of $G$, such that $E(G_i)\cap E(G_j) = \emptyset$ and $|V(G_i)\cap V(G_j)| \leq 2$, for $i\neq j$. Then, can we conclude that $\gamma(G) \geq \gamma(G_1)+\dots+\gamma(G_n)$?

( $V(G),E(G)$ and $\gamma(G)$ denote the set of vertices, the set of edges and the genus of the graph $G$, respectively)

Thanks in advance.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Take two copies of $K_{3,3}$. They both have genus 1. Identify two vertices in one partition in one graph with two vertices in one partition in the second graph. This new graph (see below) also has genus 1, so the inequality isn't satisfied. The graph6 string for the new graph is IFzeEB??w.

The union of two $K_{3,3}$ with two vertices identified.

The embedding corresponding to genus 1 is

{0: [3, 4, 8, 5, 6, 7],
 1: [3, 7, 8, 4, 6, 5],
 2: [3, 5, 4],
 3: [0, 1, 2],
 4: [0, 2, 1],
 5: [0, 2, 1],
 6: [0, 1, 9],
 7: [0, 9, 1],
 8: [0, 1, 9],
 9: [6, 8, 7]}
$\endgroup$
1
  • $\begingroup$ Can you give an explicit example. $\endgroup$
    – bor
    Commented Apr 2, 2014 at 5:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.