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Let $G$ is a simple undirected graph. Suppose $G$ has two subgraphs $G_1$ and $G_2$, such that $E(G_1)\cap E(G_2) =\emptyset$ ($E(G_i)$, stand for the set of edges of $G_i$). Then is it true that genus of $G$ is greater than or equal to the sum of genera of $G_1$ and $G_2$?

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No. The two subgraphs can share the surface more efficiently than that. Take a graph $G$ with genus $g\ge 1$ and duplicate each edge. If you don't like double edges, subdivide them with new vertices. Then you can divide the new graph into two edge-disjoint subgraphs homeomorphic to $G$, therefore each having genus $g$, yet you can still draw the whole graph on the same surface.

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Thank you so much –  bor Feb 5 at 5:28

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