Additivity of genus of a graph [duplicate]

Question

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.

Answer 1

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 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]}