Contractible manifolds of dimension $n\le 2$ I have to give a talk about the Whitehead manifold. I would like to stress the fact that it is the first given example of a contractible open (i.e. non compact) manifold which is not homeomorphic to $\mathbb{R}^n$ (when $n=3$). This happens because the Whitehead manifold is not simply-connected at infinity and as proved by Stallings this is equivalent to not be homeomorphic to $\mathbb{R}^n$. The first example of this kind occurs in dimension $3$ beacause as I read in many articles the only contractible manifold (up to homeomorphism) of dimension $n\le 2$ are: \begin{equation*}\{\text{pt.}\} \quad \mathbb{R} \quad \mathbb{R}^2
\end{equation*}
I'm thinking about a proof of this classification and I'm having problems with dimension $2$. So I have the following questions:
Question 1
Can you suggest me any reference for the classification in dimension $2$ with a detailed proof?
Question 2
Is the Stalling's result mentioned above valid for a general $n$-manifold or do we have to require that it is contractible?
 A: Forget the one-point compactification, it's a dead end. By the time you prove that a neighborhood of infinity looks like $[0,\infty) \times S^1$ (which is what you'd need to see the 1-pt compactification is a manifold) you're more or less already done.
Theorem: If $\Sigma$ is an open surface with $H_1(\Sigma; \Bbb F_2) = 0$, then $\Sigma$ is homeomorphic to $\Bbb R^2$.
Sketch of proof.
(1) First prove that $\Sigma$ has exactly one end; if $K \subset \Sigma$ is a compact subset, then $\Sigma \setminus \text{int}(K)$ may have many connected components, but only one of them is noncompact. The proof will be by contrapositive: if $\Sigma$ has two ends, show that $H_1(\Sigma; \Bbb F_2) \neq 0$. You will need to know that the inclusion $\partial \Sigma \to \Sigma$ is nonzero on first homology whenever $\Sigma$ is a noncompact surface with boundary.
(2) Use that $\Sigma$ has a compact exhaustion --- $\Sigma = \bigcup \Sigma_n$, where $\Sigma_n$ is a compact surface and $\Sigma_n \subset \text{int}(\Sigma_{n+1})$, so that $\Sigma_{n+1} = \Sigma_n \cup S_n$, where $S_n$ is also a compact surface. This can be justified using the fact that $\Sigma$ has a proper smooth function to $\Bbb R$, together with Sard's theorem.
(3) Using (1) and (2) together, observe that $\Sigma \setminus \text{int}(\Sigma_n)$ only has one noncompact piece. Modify your compact exhaustion so that $\Sigma \setminus \Sigma_n$ is connected and so that $S_n$ is connected.
(4) Prove that $\partial \Sigma_n$ is a single circle, as otherwise $\Sigma$ has positive genus (you'll glue on a pair of pants as $S_n$, because $S_n$ is connected), which would imply $H_1(\Sigma;\Bbb F_2) \neq 0$; again this involves some Mayer-Vietoris work.
(5) Prove that each $\Sigma_1$ is a disc and each $S_n$ is a cylinder. From here it follows that $\Sigma \cong \Bbb R^2$.
The details are not completely trivial, in particular on (3). I will not see or respond to comments, so please feel free to edit this answer as you desire. This gives a rough strategy for the general classification of noncompact surfaces without boundary, the ideas simplify when $\Sigma$ is contractible like this; similarly you can show that if $H_1(\Sigma)$ is finite, then $\Sigma$ is obtained by deleting finitely many points from a closed surface.
