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It is well known that the plane minus $n$ points is homotopy equivalent to a wedge of circles and hence its fundamental group is free on $n$ letters.

Question: Is the plane minus an infinite sequence of points having no limit point homotopy equivalent to an infinite wedge circles?

I'm pretty sure that this could fail if the sequence had a limit point, since then the space in question might be something more like the Hawaiian earing. But is the sequence having a limit point the only thing that could break down?

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3 Answers 3

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Yes. More generally, if $X$ is a proper closed subset of $\mathbb{R}^2$, then every path component $M$ of $\mathbb{R}^2 \setminus X$ is homotopy equivalent to a wedge of circles (observe that $X$ might be something like a Cantor set, which makes this a little more surprising). First, $M$ is a noncompact $2$-manifold, so its universal cover is homeomorphic to a disc (eg this follows from the uniformization theorem and the fact that every surface can be made into a Riemann surface). Second, the fundamental group of $M$ is free (see the answers to this question). It follows that $M$ is an Eilenberg-MacLane space for a free group. Free groups also have wedges of circles for their Eilenberg-MacLane spaces, so by the uniqueness of Eilenberg-MacLane spaces $M$ is homotopy equivalent to a wedge of circles.

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  • $\begingroup$ Minor nitpick: I think that your argument applies to each connected component of $M$ separately. $\endgroup$
    – damiano
    Commented Feb 13, 2011 at 16:32
  • $\begingroup$ @damiano : Good catch! I'll edit the answer to fix this... $\endgroup$ Commented Feb 13, 2011 at 16:35
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    $\begingroup$ Very minor nitpick - you don't need to "make" ${\bf R}^2 - X$ into a Riemann surface. It already is one via the canonical identification of ${\bf R}^2$ with ${\bf C}$ $\endgroup$
    – Sam Nead
    Commented Feb 13, 2011 at 22:27
  • $\begingroup$ I would like to accept this answer, but I only understand the simpler version where you delete a non-accumulating collection of points, since then the complement is clearly a manifold and you can apply the arguments from the question to which you linked. But it is not clear to me that you get a manifold if you delete a Cantor set. Is there a way to get around this? $\endgroup$
    – A. Pascal
    Commented Feb 14, 2011 at 14:42
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    $\begingroup$ @A. Pascal : If $C \subset \mathbb{R}^2$ is a Cantor set, then $C$ is (in particular) closed, so $\mathbb{R}^2 \setminus C$ is open. It is an easy exercise in the definition of a manifold to show that an open subset of a manifold is a manifold. By the way, by a Cantor set in $\mathbb{R}^2$ I just mean a set that is perfect and nowhere dense. You can get examples of these in many ways. For instance, you can take the usual cantor set $C_1 \subset \mathbb{R}$ and then set $C = C_1 \times 0 \subset \mathbb{R}^2$. You could also take $C_1 \times C_2$ (often called "Cantor dust"), etc. $\endgroup$ Commented Feb 14, 2011 at 15:56
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Here is a more naive solution, as least if the sequence is countable. Let $\Bbb N \subset \Bbb R^2$ be the embedding defined by the sequence. Then there is an isotopy from this embedding to the standard inclusion into the $x$-axis (inductively move the points one by one through embeddings to each integer point on the $x$-axis. The homotopy type of the complement does not change through an isotopy. Furthermore, the homotopy type of the complement of the set of integer points on the $x$-axis clearly has the homotopy type of a wedge of circles (this can be seen, e.g., as follows: (1). the complement of the integer points in $\Bbb R$ has the homotopy type of an infinite wedge of zero-spheres, (2). passing from $\Bbb R^1$ to $\Bbb R^2$ has the effect of suspending the complement).


Addendum: maybe the following is a better way to see the answer. Let $X = \lbrace x_n \rbrace$ be the sequence.

We can find a sequence of spaces $D_1 \subset D_2 \subset \cdots $ exhausting $\Bbb R^2$ such that $D_k$ is homeomorphic to a closed disk, $D_k$ is embedded in the interior of $D_{k+1}$ and $X$ meets each $D_k$ in its interior. Let $X_k = X \cap D_k$, and let $C_k$ be its complement in $D_k$. Then $C_k$ is a finite wedge of circles up to homotopy, $C_k \subset C_{k+1}$ and and $C:= \cup_k C_k$ is the complement of $X$ in $\Bbb R^2$.

Furthermore, the inclusion $C_k \subset C_{k+1}$ is a cofibration and admits a retraction, so we can write $$C_{k+1} \simeq C_k \vee E_k$$ and $E_k$ is a finite wedge of circles.

Then $C$, which is a colimit of the $C_k$, coincides with the homotopy colimit of the $C_k$, and with respect to the displayed identification we see that the homotopy colimit is a countable wedge of circles.

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    $\begingroup$ You have to be a little careful here. The isotopy you construct comes from "stacking" countably many isotopies. To make sure that this is an isotopy, you need to take some care. For instance, for an integer $n$ let $f_n$ be a homeomorphism of $\mathbb{R}^2$ that fixes all points outside the ball of radius $2n$ and radially compresses points inside the ball of radius $n$ into the ball of radius $1/2$. $\endgroup$ Commented Feb 13, 2011 at 17:23
  • $\begingroup$ (continued) It is clear that $f_1$ is isotopic to the identity, $f_2$ is isotopic to $f_1$, etc. However, if you perform all these isotopies, then the "limit" has image inside the ball of radius $1/2$, so it isn't a homeomorphism (though you can assure that it is, eg, injective). $\endgroup$ Commented Feb 13, 2011 at 17:23
  • $\begingroup$ (continued) BTW, I'm not trying to cast doubt on your construction (it can be made to work) -- I'm just adding a note of caution to performing these sorts of infinite constructions. $\endgroup$ Commented Feb 13, 2011 at 17:25
  • $\begingroup$ You're right: one needs to be careful as to the way the isotopy is constructed. What I produced was a map from $[0,1)$ into the space of embeddings $\text{emb}(\Bbb Z,\Bbb R^2)$ but what is needed is a map from $[0,1]$. $\endgroup$
    – John Klein
    Commented Feb 13, 2011 at 17:31
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    $\begingroup$ @John : I like the new argument! As a side question, I'm curious : how would a homotopy-theorist prove that the $2$-sphere minus a Cantor set is he to a wedge of circles? $\endgroup$ Commented Feb 13, 2011 at 17:52
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Expanding on John's answer a bit: If $X = \{x_1,x_2,\ldots\} \subset R^2$ is a countable discrete set, then there is a diffeomorphism $f:R^2 \to R^2$ with $f(x_i)=i$. Proof: by Sards theorem, we can assume that $X \cap N =0$ (translate a bit if necessary).

Claim 1: we can find paths $p_n: [0,1] \to R^2$ which are embeddings, $p_n (0)=x_n$, $p_n (1)= n$ and such that the images of all the $p_n$ are all disjoint. Proof by induction on $n$. Suppose that $p_1,\ldots ,p_{n-1}$ have been constructed. By homological duality theory, the complememt $R^2 \setminus \cup_{i=1}^{n-1} p_i ([0,1])$ is path-connected. We can then find a path from $x-n$ to $n$, avoiding $\cup_{i=1}^{n-1} p_i ([0,1])$ and also $X \cup N$.

If we have found these paths, we can find neighborhoods $U_n$ of the images, $U_i \cap U_j=\emptyset$ for $i \neq j$. There exist diffeomorphisms $f_n:U_n \to U_n$ with compact support and $f_i(x_i)=i$. These glue together to a global diffeomorphism $f$ with the desired property.

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    $\begingroup$ @Johannes : A tiny nitpick. You should probably choose the neighborhoods $U_i$ as you choose the paths (rather than afterwards). Otherwise, they might not exist -- the images of the paths $p_i$ might have an accumulation point. $\endgroup$ Commented Feb 13, 2011 at 19:18
  • $\begingroup$ @Andy: yes, you are right. But with this extra care, it should work. $\endgroup$ Commented Feb 13, 2011 at 19:29

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