# Dual of cut of embedded graph disconnects surface

Let $$G$$ be a graph that embedded on a surface of genus $$g$$, moreover the embedding is triangulated. Let $$C$$ be a collection of edges that forms a minimal edge cut for $$G$$. Let $$C^*$$ consist of the dual edges for edges in $$C$$. $$C^*$$ consists of vertex disjoint cycles in the dual $$G^*$$. How do we prove that cutting along the cycles in $$C^*$$ disconnects the surface? This holds for planar graphs by the Jordan curve theorem.

I assume that you require that all faces of $$G$$ are topological disks. After cutting along $$C^*$$, each face is a topological disk bounded by either a cycle of $$G$$, or a cycle that consists of two arcs (one on $$G$$ and one on $$C^*$$) with common endpoints (at the intersection between edges of $$C$$ and their duals), where the arc on $$C^*$$ lies on the boundary of the cut surface. In particular, the boundary of any face intersects $$G$$ in a single component.
Now, assume for a contradiction that the surface after cutting is still connected, and consider an edge $$(u,v)\in C$$. Then there exists a path $$\pi$$ on the cut surface connecting $$u$$ and $$v$$. By the above property of faces, we can snap $$\pi$$ to a path on $$G-C$$, showing that $$u$$ and $$v$$ lie in the same component of $$G-C$$. Therefore, $$C-(u,v)$$ is still an edge cut for $$G$$, contradicting minimality of $$C$$ and hence connectedness of the cut surface.