Invariance of the boundary for 2-dimensional manifold. This is problem 8-3 from John Lee's Introduction to Topological Manifolds. 
Invariance of the boundary, 2-dimensional case: Suppose $M$ is a $2$-dimensional manifold with boundary. Show that a point of $M$ cannot be both a boundary point and an interior point. 
I try to use the following theorem from problem 8-1 : Suppose $U \subset \mathbb{R}^2$ is an open subset and $x \in U$. Then $U - \{x\}$ is not simply connected. 
Suppose $p$ is both an interior and boundary point. Choose coordinate charts $(U,\phi)$ and $(V,\psi)$, $\phi(U)$ is open in Int$\mathbb{H}^2$ and $\psi(V)$ is open in $\mathbb{H}^2$, with $\psi(p) \in \partial \mathbb{H}^2$. Let $W=U \cap V$; then $\phi(W)$ is homeomorphic to $\psi(V)$. From the above problem, $\phi(W) - \phi(p)$ is not simply connected. So we would reach a contradiction if $\psi(V) - \psi(p)$ is simply connected. However, I do not know if this is true. How could I show this problem? 
 A: I will keep your notation: Withot loss of generality we can assume that $\psi(p)=0$. We have that $0\in \psi(W)$, so we can choose $r>0$ such that 
$$
B(0,r)\cap \mathbb{H}^2 \subseteq \psi(W),
$$
where $B(0,r)$ is the open ball in $\mathbb{R}^2$ centered at the origin with radius $r$. Let $U'=\psi^{-1}(B(0,r)\cap \mathbb{H}^2)$, thus $\phi(U')\setminus\{\phi(p)\}$ is not simply connected (by the theorem you mentioned). We have to prove that $U'\setminus\{p\}$ is simply connected, in contradiction with $\phi$ being a homeomorphism. But note that $U'\setminus\{p\}$ is homeomorphic to $B(0,r)\cap \mathbb{H}^2\setminus\{0\}$, so I will prove that $B(0,r)\cap \mathbb{H}^2\setminus\{0\}$ is simply connected.
Let $x_0\in B(0,r)\cap \mbox{int}\mathbb{H}^2$ and let $B(0,r)\cap \mathbb{H}^2\setminus\{0\}=:X$, and define $F:X\times [0,1]\to X$ by
$$
F(x,t) = x_0 + (1-t)(x-x_0),
$$
it is easy to see that $F$ is a strong deformation retraction of $X$ onto $\{x_0\}$, and thus $X$ is simply connected.
A: You have all the ingredients, and your question is simply why a half-disk with center missing is simply connected. Well, you could radially flow all points onto the perimeter arc, and then contract arc into one of its endpoints. Or do it like a wiper will clean a car's windshield! Sweep all the disk to one segment, then squeeze the segment onto its endpoint.
