# Topological genus of 3-d flat space minus a solid ball

Our recreational math geek lunch group got stuck on a question we need help to understand. I apologize in advance, if my explanation is not perfectly rigorous, as we are not professional mathematicians.

This is the question: What is the topological genus of 3d space if we remove a solid 3d ball from it?

We started from this example: A flat 2d plane minus a disk, the genus appears to be one, using both the cutting test (one cut to the hold doesn't create two pieces), and the rubber band test (a shrinking a loop around the hole doesn't reduce a point).

What are the genus tests for dimensions higher than a 2d plane?

• I'm not sure what you mean by "genus" in a three-dimensional manifold, but your space $X$ is homotopy equivalent to the $2$-sphere $S^2$. It's homology groups are $H_0(X)\cong\Bbb Z$, $H_1(X)=0$ ,$H_2(X)\cong\Bbb Z$, $H_n(X)=0$ for $n\ge3$. Informally this says $X$ has no "one-dimensional hole" but it has a "two-dimensional hole". – Lord Shark the Unknown Jun 12 at 4:19
• I apologize for our misuse of terminology. we started this whole discussion with the topology joke comparing a donut to a coffee mug. So that's the kind of "genus" we meant. – SteveED Jun 13 at 1:37