Smooth manifolds with a hole are homeomorphic I'm trying to prove the following result:

Let M be a smooth connected surface and $\psi_1,\psi_2:\mathbb{D}^2\to N$ two embeddings of the unitary 2-disk. Then, $M\setminus\psi_1(\mathbb{D}^2)$ and $M\setminus\psi_2(\mathbb{D}^2)$ are homeomorphic.

Here is my attempt:
We can suppose that $\psi_1(\mathbb{D}^2)$ and $\psi_2(\mathbb{D}^2)$ are contained in two charts $(U,\varphi_1)$ and $(V,\varphi_2)$. Every smooth manifold is path-connected, so we take a path $\gamma:[0,1]\to M$ such that $\gamma(0)=\psi_1(0)$ and $\gamma(1)=\psi_2(0)$. Since $\gamma([0,1])$ is compact, we can cover it with a finite family $U_0=U, U_1,\ldots,U_n=V$ of chart domains. For every $k=0,1,\ldots,n-1$, we take a point $x_k\in\gamma([0,1])$ lying in $U_k\cap U_{k+1}$ and the image $\bar{B}_k$ by the chart homeomorphism $\varphi_k^{-1}$ of a closed ball centered in $\varphi_k(x_k)$. It suffices to show that $M\setminus\bar{B}_k$ and $M\setminus\bar{B}_{k+1}$ are homeomorphic.
We can suppose, w.l.o.g., that $\varphi_k(U_k)$ is an open ball in $\mathbb{R}^2$, so that $\varphi_k(U_k\setminus\bar{B}_k)$ and $\varphi_k(U_k\setminus\bar{B}_{k+1})$ are two open balls with a hole in $\mathbb{R}^2$. These two sets are homeomorphic (I have read that we can construct an homeomorphism using Möbius transformations of the disk). Composing with the chart homeomorphism, we have that $U_k\setminus\bar{B}_k$ and $U_k\setminus\bar{B}_{k+1}$ are homeomorphic.
Now I would like to expand the homeomorphism I have just described to $M$ but I don't know how to do it. It would be trivial if the homomorphism between the disk was the identity over the boundary, but I think it is not the case.
Could someone help me? Thanks in advance.
 A: The proof would follow if you could show that there exists homeomorphism $f:\mathbb{R}^2-B(x_1,r_1)\to \mathbb{R}^2-B(x_2,r_2)$, which is the identity outside of a compact region in $\mathbb{R}^2$.
We may do this in a sequence of steps.
First, we show that for any angle $\phi$ and $R>0$, there exists a homeomorphism $f:\mathbb{R}^2\to \mathbb{R}^2$, which is a rotation by angle $\phi$ for $\sqrt{x^2+y^2}<R$, and is the identity for $\sqrt{x^2+y^2}>2R$.
Second, for $s>0$ and $R>0$, there exists a homeomorphism $f:\mathbb{R}^2\to \mathbb{R}^2$, which is multiplication by $s$ for $\sqrt{x^2+y^2}<R$, and is the identity outside of a compact region.
Lastly, for $y_1,y_2\in \mathbb{R}$, there is a homeomorphism $f:\mathbb{R}\to \mathbb{R}$, such that $f(y_1)=y_2$, and such that $f$ is the identity outside of a compact region.
Letting $\phi$ denote the angle between $x_1$ and $x_2$, $s=r_2/r_1$,and $y_1=\|x_1\|$ and $y_2=\|x_2\|$, the above maps may be combined to achieve the desired homeomorphism.
The proof of all of these statements is similar. For example for the first, for any $\phi\in [0,2\pi)$ take a continous function, $\theta:\mathbb{R}_+\to [0,2\pi]$, such that $\theta(x)=\phi$ for $x<(1/2)R$ and $\theta(x)=0$ for $x>R$. Now define $f_{\phi}:\mathbb{R}^2\to \mathbb{R}^2$ by
$$f(x,y)=(\cos(\theta(r))x+\sin(\theta(r))y,-\sin(\theta(r))x+\cos(\theta(r))y),$$
where $r=\sqrt{x^2+y^2}$
