Let T be a Torus. Find a graph $G$ on $T$ s.t. $(T-G)$ is homeomorphic to a disk.

Question

Let $T$ be a torus. Find a graph $G$ on $T$ s.t. its complement, $(T-G)$ is homeomorphic to a disk. With the ones I've tried thus far, I am only finding its complement to be homeomorphic to a disk with a hole removed. Thank you!

Answer 1

You should be able to get a (shape homeomorphic to a) disk by making two cuts: the first cut is through a "radius" of the torus (shown in red), and makes it into a cylinder. The second cut is along the "circumference" of the torus (shown in magenta) and makes the cylinder into a rectangular sheet, which is homeomorphic to the disk.

That is a visual description; to put it a more formal way, let $S^1$ be the circle and let $q\in S^1$ be any particular point on the circle. A torus is the shape $S^1 \times S^1$. I propose cutting out $S^1 \times \{q\}$ and $\{q\}\times S^1$.

If we're talking about graphs in the sense of "vertices and edges", where edges are identified with copies of intervals, then the shape I propose cutting out is like a single vertex with two loops in it. The vertex is the point $\{q\}\times \{q\}$ and the two loops are $S^1 \times\{q\}$ and $\{q\}\times S^1$.

Because cutting that graph out makes the torus into a disk, the complement of the graph is the disk.

(I'm not sure about the definition of topological graphs: If you're not allowed to have loops from a vertex to itself in this way, you can replace each loop with a triangle of three vertices instead, with the same effect.)