Understanding how connected sum of smooth surfaces is a surface

Question

I have two smooth surfaces $M_1$ and $M_2$ I''m trying to understand how the connected sum $M_1 \mathop{\#} M_2$ is a smooth surface. I will write my understanding of the proof and then explain where I am confused.

So to show something is a smooth surfaces have to show that it is locally homeomorphic to a disc in $\mathbb{R}^2$ and the transition maps on the overlaps are smooth functions $\mathbb{R}^2 \to \mathbb{R}^2$. So there are a few cases to consider. If we take a point in the interior of one of the surfaces, say $M_1$ then we can arrange that the neighbourhood lies completely inside that surface and we use the fact that $M_1$ is a surface to get a homeomorphism. If we have a point on the boundary of $M_1$ where it joins $M_2$ then we have a neighbourhood in $M_1$ and a neighbourhood in $M_2$ and we can shrink these neighbourhoods so that each contains all the points on the boundary intersected with the other neighbourhood and then map these half-neighbourhoods on to a complete open disc.

My problem is: I understand that a surface, $S$, is a collection of homeomorphisms $f_\alpha : U_\alpha \to V_\alpha$. Where $V_\alpha$ are open discs in $\mathbb{R}^2$ and the union of the $U_\alpha$ cover the surface. So given a surface like this, how can we arrange it so that we can shrink our neighbourhoods as I have described above? I.e who's to say that after removing a disc from a surface, with the removed part intersecting with one of the $U_\alpha$, that we can shrink our neighbourhood so that it lies completely within our surface? This is causing me difficulty in then showing that the transition maps are smooth.

I understand that a lot of these ideas are obvious but I'm struggling to judge the level of rigour required with arguments of this kind.

Thanks

Answer 1

To answer your question directly: the connected sum is a smooth surface, but with a different atlas than the one you're using.

Formally speaking, connected sums are defined using adjunction spaces: you don't reuse the same charts, but instead you introduce an abstract disjoint union space on which you perform identifications.

Your perspective seems to be asking whether a set is a manifold. Of course, there can be many manifold structures on the same set. And it turns out that the natural manifold structure to put on the connected sum is the one that comes up by performing an adjunction $f\colon \partial M_1\to\partial M_2$.

A really good reference for the foundations of manifold theory is Introduction to Topological Manifolds by Lee. Even though you're asking about smooth structures (for which Introduction to Smooth Manifolds by Lee is the correct reference), I think the topological description alone will help alleviate some confusion (while reducing the amount of technical baggage).

Answer 2

You have to be careful : glueing bounderies of 2 manifolds (with boundary) does not define a smooth manifold. Take for example $M_1 =(-\infty,0]$ and $M_2 = [0,+\infty)$, then there are a lot of way to glue $M_1$ and $M_2$ at $0$ (consider an homeomorphism $\mathbb{R} \rightarrow \mathbb{R}$ which is diffeomorphic on $(-\infty,0]$ and $[0,+\infty)$, but not diffeomorphic around $0$).

The correct way to define the connected sum is to identify a closed disc $D_1$ of $M_1$ to a closed disc $D_2$ of $M_2$ and then remove the interior the discs. To find a chart at a point $x \in \partial D_1 = \partial D_2 \subset M_1 \sharp M_2$, you have to take $U_1$ a neighboorhood of $x$ in $M_1$ (and not merely in $M_1 \setminus (D_1)^\circ$), $U_2$ a neighboorhood of $x$ in $M_2$, and a diffeormophism $\phi : U_1 \xrightarrow{\sim} U_2$ such that $\phi(U_1 \cap (D_1)^\circ) = U_2 \setminus D_2$ and $\phi(U_1 \setminus D_1) = U_2 \cap (D_2)^\circ$.