homeomorphism classes of compact surfaces with addition operation is a monoid

Question

This is essentially pg 6 of serge lang's algebra's discussion about an interesting example. Homeomorphism classes of compact surfaces with the addition operation defined as following. Say M and $M'$ are two compact manifolds. Take a disk in M and another disk in $M'$. Identify the boundary of the disk and remove interior of disk. Then glue the rest. The resulting surface's homeomorphism class is defined by this addition. I can see geometrically why the 2-sphere's homeomoprhism is the identity in this homeomoprhism(cutting a hole in 2 sphere and glue 2 sphere or glue 2 sphere with donuts with handles does not change the genus). The donuts with more handles are other elements of this monoid by observing the genus. How can I show this? It is not very clear that this addition property gives rise to the resulting surface from addition is independent up to homeomorphisms.

Answer 1

I guess the only issue is which disc you choose to remove from each manifold, so all you really need to show is that $M \setminus D \cong M \setminus D'$ if $D, D'$ are two different discs in $M$. Note first that you can shrink the disc arbitrarily small without affecting the homeomorphism type -- say, small enough that the disc is contained in a single chart. But it's obvious $\mathbb{R}^n$ minus any disc is homeomorphic to $\mathbb{R}^n$ minus any other disc, so you can move the discs around within a chart or between charts.