Yes. More generally, if $X$ is a proper closed subset of $\mathbb{R}^2$, then every path component $M:=\mathbb{R}^2 M$ 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.
|
2 | added 25 characters in body | ||
|
|
||||
|
|
1 |
|
||
|
Yes. More generally, if $X$ is a proper closed subset of $\mathbb{R}^2$, then $M:=\mathbb{R}^2 \setminus X$ 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. |
||||
