TITLE: genus of a finite simple undirected graph
QUESTION [4 upvotes]: Let $G$ be a finite simple undirected graph. Suppose there exist subgraphs $G_1,G_2,\dots,G_n$ of $G$, such that $G_i$ and $G_j$, have no common edges and have at most two common vertices, for each  $i\neq j$. Is it true that the genus of $G$ is greater than or equal to the sum of genera of each subgraphs $G_i$?
Note: If they have at most one common vertices, then the result hold( using this http://projecteuclid.org/download/pdf_1/euclid.bams/1183524922 paper).  Thanks.

REPLY [2 votes]: I think this is false for orientable genus, which I assume is the case you are interested in. Miller showed that the Euler genus is additive under edge algamation, therefore, if you take an edge algamation of $K_{3,3}$ + 1 edge and $K_5$ (both being embeddable on a projective plane), it has Euler genus 2, and is therefore embeddable on a torus. But it is also the double vertex amalgamation of $K_{3,3}$ and $K_5$, which have orientable genus $1$, so this gives a counterexample.
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