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Suppose you have the set $D=\{(x,0) \in \mathbb{R}^2: x \in \mathbb{R}\} \cup \{(0,y) \in \mathbb{R}^2: y \in \mathbb{R} \} $ with the subspace topology of $\mathbb{R}^2$.
Is D a manifold?
My understanding is that for the set D to not be a manifold i have to show that there does not exist a homeomorphism $f:U \subset D \rightarrow f(U) \subset \mathbb{R}^2$ where $f(U)$ open in $\mathbb{R}^2$.

My intuition says that it is not a manifold, but what confuses me is that it is a subspace of $\mathbb{R}^2$ with the subspace topology.

Edit: Can it be a 1d manifold with charts:

$f_1:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_1(x)=(x,0)$
$f_2:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_2(x)=(-x,0)$
$f_3:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_3(y)=(0,y)$
$f_4:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_4(y)=(0,-y)$

Thank you!

lebong66
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1 Answers1

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If $D$ would be a manifold, it would be a $1-$dimensional manifold. In particular, if $U$ would be an open set of $D$ that contain $0$, then $U\setminus \{(0,0)\}$ would be an open set of $D$ with $4$ disjoints connected components, whereas, if it would have been a manifold, $U\setminus \{(0,0)\}$ would be an open set with at most $2$ disjoints connected components.

Surb
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  • can you please show me an example. Thank you!!!. – lebong66 Nov 02 '22 at 18:39
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    Take for example $U=((-1,1)\times {0})\cup({0}\times (-1,1))$. Then $$U\setminus {(0,0)}=((-1,0)\times {0})\cup((0,1)\times {0})\cup({0}\times (-1,0))\cup({0}\times (0,1)).$$ @lebong66 – Surb Nov 02 '22 at 19:09
  • alright it can't be a 2d manifold. But can it be a 1d manifld with the charts above? – lebong66 Nov 03 '22 at 00:43
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    @lebong66: No, because you can't reach 0 with your parametrization. Notice that for the reason I explained above, you won't be able to parametrize locally around 0, that's why it's not a Manifold. – Surb Nov 03 '22 at 08:54
  • Your final step, claiming at most 2 disjoint connected components isn't entirely obvious to me. I guess it's using the fact that every connected $1$-manifold is homeomorphic to either $\mathbb{R}$ or $S^1$? – aschepler Nov 03 '22 at 13:24
  • Rather that every $1-$manifold is locally homeomorphic to $\mathbb R$. – Surb Nov 03 '22 at 17:08
  • Still missing something. How does a local property imply something about connected components of the whole? – aschepler Nov 03 '22 at 18:48
  • Locally in the sense that if $x\in D$, then there is a neighborhood $U\ni x$ that is homeomorphic to $\mathbb R$. Roughly speaking, every point need to have a neighborhood that is $\mathbb R$ (up to a homeomorphism). So, in your case, the only "mad point" is $0$. The reason is that for any open $U$ that contain $0$, as far as you take out $0$, you get four disjoint connected component, and thus, it can't be homeomorphic to $\mathbb R$ because, taking any point out of $\mathbb R$ will give $2$ disjoint connected component. Therefore, $0$ has no neighborhood homeomorphic to $\mathbb R$. – Surb Nov 03 '22 at 18:56
  • However, $D\setminus {0}$ will indeed be a manifold (unconnected manifold, but a manifold).@aschepler – Surb Nov 03 '22 at 18:58
  • Now I'm seeing it. Though I think it's easier to prove $f(D \setminus {0})$ has at least four connected components than to show it has exactly four. – aschepler Nov 03 '22 at 19:10
  • Also, I think a strict proof $D$ is not a $d$-manifold with $d>1$ would involve invariance of domain. And $D$ could obviously be a manifold if using a topology other than the subspace topology imposed from $\mathbb{R}^2$, but that was explicitly stated. – aschepler Nov 03 '22 at 19:12
  • @aschepler: "I think it's easier to prove (∖{0}) has at least four connected components" What is $f$ ? I don't understand your comment. – Surb Nov 03 '22 at 21:51
  • $f$ is the hypothetical homeomorphism from an open set $U$ in the subspace topology on $D$ to an open set in $\mathbb{R}$. So I should say $f(U\setminus{0})$. – aschepler Nov 03 '22 at 21:54