# Prove the boundary is a compact 1 manifold

A closed surface with boundary is a compact connected topological space $B$ with the property that each point $p \in B$ has an open neighborhood $U$ homeomorphic to either:

$\{(x, y) \in \mathbb{R^2}|x^2+y^2<1\}$

or

$\{(x, y) \in \mathbb{R^2}|x^2+y^2<1$ and $y\geq 0\}$

Such as space has interior consisting of those points with neighborhoods of the first sort, and remaining points are its boundary

Prove that the boundary is a compact 1- manifold and homeomorphic to a finite collection of disjoint circles

Do we need to consider special values of $y$ that make the points a boundary?

Yes, the boundary points are precisely the points which, in the second set, satisfy $y=0$.
First, suppose that $p$ has a neighborhood $U$ with a homeomorphism $f$ taking $U$ to the second set, and suppose that $f(p)=(x,y)$ with $y>0$. Then $p$ has a different neighborhood homeomorphic to the first set, namely $$U' = f^{-1} \biggl(\{(x,y) \in \mathbb{R}^2 \,\bigm|\, x^2+y^2<1, y>0\} \biggr)$$ and its not hard to show by direct construction that this set is homeomorphic to $$\{(x,y) \in \mathbb{R}^2 \,\bigm|\, x^2 + y^2 < 1\}$$
Second, it is not possible for $p$ to have neighborhoods $U,V$, a homeomorphism $f$ from $U$ to the first set, and a homeomorphism $g$ from $V$ to the second set, such that $g(p)=(x,0)$. Arguing by contradiction, suppose that $p$ did have two such neighborhoods. We may assume that $f(p)=(0,0)$ and $g(p)=(0,0)$ (by rechoosing the neighborhoods and homeomorphisms, if necessary). Then $f$ has a neighborhood basis $$U = U_1 \supset U_2 \supset \cdots$$ and homeomorphisms $f_i$ from $U_i$ to the first set such that $f_i(p)=(0,0)$: let $U_i$ equal $f^{-1}$ of the ball of radius $1/i$ around $(0,0)$, and let $f_i$ equal the composition of $f | U_i$ with a scalar multiplication. Similarly, $p$ has a neighborhood basis $$V = V_1 \supset V_2 \supset \cdots$$ and homeomorphisms $g_j$ from $V_j$ to the second set such that $g_j(p)=(0,0)$. Choose $j$ so that $V_j \subset U_1$, then choose $i$ such that $U_i \subset V_j$, consider the inclusion maps $$U_i \subset V_j \subset U_1$$ consider a point $q \in U_i$, and consider the induced homormorphisms $$\pi_1(U_i,q) \to \pi_1(V_j,q) \to \pi_1(U_1,q)$$ Since $V_j$ is simply connected, this composition of homomorphisms is the zero homorphism. However, it is obviously true that $$\mathbb{Z} \approx \pi_1(U_i,q) \to \pi_1(U_1,q) \approx \mathbb{Z}$$ is an isomorphism, which is a contradiction.