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I'm having troubles to prove that the following space is not a topological manifold:

Let $r:S^1\to S^1$ be a rotation of $\frac{2\pi}{3}$, i. e., $r(\cos\theta,\sin\theta)=\left(\cos\theta+\frac{2\pi}{3},\sin\theta+\frac{2\pi}{3}\right)$, such that $X=B^2/\sim$, where the partition is given by $P=\{x\}$ if $|x|\lt1$ and $P=\{x,r(x),r^2(x)\}$ if $|x|=1$.

So this is my attempt: Let $x\in X$ be a point in the boundary of $B^2$, the neighborhood of $x$ in $B^2$ is an open set $U$ of $B^2$ with $r(U)$ and $r^2 (U)$. My strategy is to prove that this neighborhood is not homeomorphic to an open subset of $\mathbb R^n$, anyone can help me in this part?

Thanks

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  • $\begingroup$ Sorry: what is $B^2$? The unit disk in $\mathbb R^2$? $\endgroup$ – Rudy the Reindeer Oct 16 '12 at 19:24
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    $\begingroup$ @MattN. yes it is the unit disk in $\mathbb R^2$ $\endgroup$ – user42912 Oct 16 '12 at 19:31
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All points of $X$ that come from $S^1$ ‘look the same’, so you might as well pick a particular one; the natural choice is to pick $x=q\big(\langle 1,0\rangle\big)$, where $q:B^2\to X$ is the quotient map. A set $U\subseteq X$ is an open nbhd of $x$ iff $q^{-1}[U]$ is an open nbhd of $\left\{\langle 1,0\rangle,\left\langle-\frac12,\frac12\sqrt3\right\rangle,\left\langle-\frac12,-\frac12\sqrt3\right\rangle\right\}$ in $B^2$. In the sketch below, the three red points show $q^{-1}[\{x\}]$, and the blue shows what a fairly typical $q^{-1}[U]$ should look like.

enter image description here

$U$ looks rather like a book with three pages: each of the sets $q[A],q[B]$, and $q[C]$ is one page, and the $q$ takes the three bits of $S^1$ attached to $A,B$, and $C$ to a single interval homeomorphic to $(0,1)$ that forms the spine of this little book. If there were just two pages, they’d fit together perfectly to make something homeomorphic to an open disk in the plane, but there are three. Thus, if you remove the spine of the book, you’re left with three disconnected pages; removing a set homeomorphic to $(0,1)$ from an open disk in $\Bbb R^2$ leaves at most two disconnected pieces.

Alternatively, note that the yellow curve(s) in the picture correspond to a simple closed curve in $X$. However, this simple closed curve doesn’t split $U$ into an inside, containing $x$, and an outside: you can travel in $q[C]$ from $x$ to a point on $q[S^1]$ ‘outside’ of the yellow curve and from there into the part of $A$, say, on the opposite side of the yellow curve from $\langle 1,0\rangle$. A simple closed curve in the plane, however, does split the plane into two regions; this is the Jordan curve theorem.

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At any point $x \in \partial B^2$ in $X$, the removal of the line segment $\partial B$ from a small disk neighborhood (in $X$) leaves 3 connected components. Removing a line segment from a disk in $\mathbb{R}^2$ leaves only 2 connected components.

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