Let $G$ be a finite simple undirected graph of genus $g$. We construct a new graph $H$ from $G$ as follows. $V(H)=V(G)\cup K$ and $E(H)=E(G)\cup L$, where $K:=\{v_{ab} \mid a,b \in V(G)$, $a$ and $b$ are adjacent in $G\}$. $L=\{\{v_{ab},a\},\{v_{ab},b\} \mid v_{ab} \in K\}$. Is the genus of $H$ is also $g$? Here $V(G)$ and $E(G)$ denote the set of vertices and edges of a graph $G$, respectively.
Example: $G=K_3$, with vertices 1,2,3. Then $V(H)=\{1,2,3,v_{12},v_{13},v_{23}\}$. $v_{12}$ is adjacent to 1 and 2 only. $v_{13}$ is adjacent to 1 and 3 only. $v_{23}$ is adjacent to 2 and 3 only. That is $E(H)=\{\{1,2\},\{2,3\},\{1,3\},\{v_{12},1\},\{v_{12},2\},\{v_{13},1\},\{v_{13},3\},\{v_{23},2\},\{v_{23},3\}\}$
I think that it is true, but could not show how it is. Thanks for any help.