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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?

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  • $\begingroup$ I am currently thinking of this. If $S$ is an open contractible surface, define $\Sigma$ to be its Alexandrov compactification (its one point compactification). If one can show $\Sigma$ is a topological sphere, then $S$ is homeomorphic to $\mathbb{S}^2\setminus \{pt\}$ which is $\mathbb{R}^2$. Showing $\Sigma$ is the sphere would be equivalent to show it is a simply-connected compact surface, and I think this is not a hard thing to do. $\endgroup$ – Didier Nov 12 '20 at 10:37
  • $\begingroup$ Is there any relation between the fundamental group of $X$ and the fundamental group of its one point compactification? I know that the one point compactification is compact, so I should only have to prove that it is simply-connected. $\endgroup$ – John117 Nov 12 '20 at 11:12
  • $\begingroup$ I'm not that sure in the general case. But here, suppose you have a loop in the compactification. If it is not passing through the infinity point, then it is a loop in $S$ and by contractibility, it is homotopic to a point. If the loop passes through the infinity point, I think (this is not a proof!) we can deform it locally to evitate the infinity point, and use the previous work. $\endgroup$ – Didier Nov 12 '20 at 11:15
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    $\begingroup$ In fact, it seems the hard thing would be to show that the one point compactification will be a topological manifold $\endgroup$ – Didier Nov 12 '20 at 11:26
  • $\begingroup$ Yes, I think is the hard part of the proof. $\endgroup$ – John117 Nov 12 '20 at 12:55
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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.

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  • $\begingroup$ Thanks very much for your answer! Could you please add the references for all of this? $\endgroup$ – John117 Nov 12 '20 at 14:47

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