Retract of noncompact surface to its boundary? Suppose $M$ is a connected, noncompact 2-manifold, and its boundary $\partial M$ is a circle. What's the simplest way to show there is a retraction $r: M\rightarrow \partial M$?
Here are some examples of such surfaces:


*

*Probably the easiest example is the closed unit disk, minus the origin.

*A more complicated example is the closed unit disk minus a Cantor set.

*A totally different example is to join two infinite-genus tori, and attach a cylinder to that: 
 A: "Simple" depends on what you've studied.  Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $\partial M \cong S^1$.  Show there is a retraction $r: M \rightarrow \partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$.  Let $e \in E$ and $M' = M'' \smallsetminus \{e\}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$.  A quick reminder of these ideas is here.  For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable.  $M'$ is a connected, noncompact, 2-manifold with circle boundary $\partial M = \partial M'$, and one end.  We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$.  (It is an open disk neighborhood of $e$ in the Freudenthal compactification.)  (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^\circ$2x for more.)  Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$.  Let $B = \partial M' \cong S^1$ and $p_2$ be a point of $B$.  There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other).  $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other.  $p$ has an open bicollar homeomorphic to $p \times (0,1)$.  Let $S$ be the union of these three open collars.  ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' \smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary.  We construct $r$ in two steps.  Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' \smallsetminus S$ to a point.  Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$.  The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end.  By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $\mathbb{R}^2$.  Therefore, $r_1(M') +D \cong \mathbb{R}^2$.  Constructing the retract $r_2:\mathbb{R}^2 \smallsetminus\{(0,0)]\} \rightarrow B$ is a standard exercise.  Then $\left. r_2 \right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = \left. r_2 \right|_{r_1(M)} \circ \left. r_1 \right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two.  Crush everything else to a point, yielding a half-infinite cylinder.  Then telescope the cylinder onto the boundary.
A: Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1. 
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $\rho:\mathbb{H}\rightarrow M$, where $\mathbb{H}=[0,\infty)$.  Let $x\in\partial M$ and further suppose this embedding is such that
\begin{equation*}
\rho(\mathbb{H})\cap\partial M = \rho(0)=x
\end{equation*}
Finally, suppose that there's a neighborhood of $\rho(\mathbb{H})$ that looks like $\mathbb{H}\times (-1,1)$. 
Then we can cut $M$ along $\rho(\mathbb{H})$ to get a new manifold $N$, where $\partial N\cong\mathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right.  Here left and right come from the two sides of $\rho(\mathbb{H})$, and middle is $\partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:\partial N\rightarrow [0,1]$ on each piece:
\begin{equation*}
f(z) = \begin{cases}
0 & z\in\textit{left}\\
z & z\in\textit{middle}\\
1 & z\in\textit{right}\\
\end{cases}
\end{equation*}
By the Tietze extension theorem, this extends to a map $F: N\rightarrow [0,1]$. If we compose that with the quotient map $\pi:[0,1]\rightarrow S^1$, we get a map $\pi F:N\rightarrow S^1$. Because this map agrees on left and right, it descends to a map
\begin{equation*}
\widetilde{\pi F}:M\rightarrow S^1
\end{equation*}
which is the identity on $\partial M$. That means it is our desired retraction.
