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I'm familiarizing myself with manifolds. I tried to show $[0,1]\times[0,1]$ is a manifold with a boundary. Can you please tell me if my proof is correct:

The definition for manifold with boundary:

A manifold with boundary $M$ is a second countable Hausdorff space so that for a $p \in M$ there is an open set $U \subseteq M$ so that there is a homeomorphism $\varphi$ to either (a) an open set $V$ in $H^n \setminus \partial H^n$ or to (b) an open set $V$ in $H^n$ and $\varphi (p) \in \partial H^n$ where $H^n$ is the closed upper half plane. It means $H^n = \{x \in \mathbb R^n : x_k \ge 0 ; 1 \le k \le n \}$.

One has to show $M = [0,1] \times [0,1]$ is second countable, Hausdorff and locally homeomorphic to $H^n$.

$M$ is second countable because it has subspace topology of $H^2$.

It is Hausdorff because it has subspace topology of $H^2$.

Locally Euclidean: For point $p \in M$ take set $U$ open in $H^2$ with $p \in U$. Inclusion map $i: U \to U \subseteq H^2$ is local homeomorphism with the property maps to open set and if $p \in \partial M$ then $i(p) \in \partial H^2$.

I am very grateful for your help.

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Well, it is not a 2-manifold, but a 2-manifold with boundary. –  Berci Jan 23 '13 at 14:01
@Berci Okay. Does it imply my proof is wrong? –  goobie Jan 23 '13 at 14:05
What is your definition of manifold with boundary ? The usual definition is this one‌​. If $x$ is on the boundary, then you are showing that the boundary (NOT the square) is a manifold locally at $x$. –  user10676 Jan 23 '13 at 14:38
No. You have to show that there is a neighborhood homeomorphic to $\mathbb{R}^{n-r} \times \mathbb{R}_{\geq 0}^{r}$ for some $r$ (depending on $x$). –  user10676 Jan 23 '13 at 15:09
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1 Answer

up vote 5 down vote accepted

The portion where you show that $M$ is locally Euclidean is not correct and here's why:

You have to show that for every point $p \in M$ there exists a neighborhood $U \subseteq M$ of $p$ and a homeomorphism from $U$ to an open subset of either $\mathbb R^2$ or $\mathbb H^2$.

Now here's what you wrote:

  • For point $p \in M$ take set $U$ open in $\mathbb H^2$ with $p \in U$

So your neighborhood $U$ is in $\mathbb H^2$ and not in $M$! While it is indeed true that $M \subseteq \mathbb H^2$ we cannot simply say that a neighborhood in $\mathbb H^2$ is a neighborhood in $M$. If $p$ is the point $p = (1, 1)$ then any neighborhood of $p$ in $\mathbb H^2$ will contain points that are not in $M$.


To fix the proof remember that if $U \subseteq M$ contains part of the boundry of $M$ then the chart for $U$ will have to map that boundry to the boundry of $\mathbb H^2$. So for example if $p = (1, 1)$ then let $U = (.4, 1] \times (.4, 1]$ and for a chart try $\phi\colon U \to \mathbb H^2$ defined by $\phi(x, y) = (1 - x, 1 - y)$. I leave it to you to check that the image of $\phi$ is open in $\mathbb H^2$ and that $\phi$ is a homeomorphism onto this image. You'll also need to come up with charts covering the rest of $M$. I suggest you take

  • $U_2 = [0, .6) \times (.4, 1]$
  • $U_3 = (.4, 1] \times [0, .6)$
  • $U_4 = [0, .6) \times [0, .6)$

as the neighborhoods for those charts.

Edit #2:

Whoops! My edit above has a mistake as pointed out by goobie. Also pointed out by goobie is the fix: Instead of the chart $\phi$ that I suggested take $\phi\colon U \to \mathbb H^2$ defined by $\phi(z) = z^2$ (here I'm using complex numbers to denote points in the plane). Then you'll just need to do some translation and rotation to handle the other corners.

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Thank you very much for answer! I understand mistake. To correct it one can take intersection of $U \cap M$ where $U$ is open in $H^2$? –  goobie Jan 29 '13 at 7:35
No, if $U$ is a neighborhood of $p$ in $\mathbb H^2$ then the intersection $U \cap M$ will indeed be a neighborhood of $p$ in $M$, but it will not be an open subset in $\mathbb H^2$ so the inclusion of $U \cap M$ into $\mathbb H^2$ does not give a homeomorphism of $U \cap M$ onto an open subset of $\mathbb H^2$. –  Jim Jan 29 '13 at 7:55
Many thank you, I can do it now!! But: if you take $U_1 = (0.4, 1] \times (0.4, 1]$ and denote its corners by $ABCD$ counterclockwise and then apply $\pi (x,y) = (1-x, 1-y)$ the result is a set with corner $C$ at $(0,0)$ and corners $CDAB$ counterclockwise. It has closed border $BC$ and is not open in $\mathbb H^2$. –  goobie Jan 29 '13 at 9:12
I tried with open balls: $B( (1,1) , 0.6) \cap M$. Then denote a point in $B( (1,1) , 0.6) \cap M$ with radius and angle relative to $(1,1)$ as $(r_{(1,1)}, \varphi_{(1,1)})$. Then define map into $H^2$ absolute coordinates as $(r_{(1,1)}, \varphi_{(1,1)}) \mapsto (r_{(1,1)}, 2(\varphi_{(1,1)} + \pi / 2))$. –  goobie Jan 29 '13 at 9:14
How can I award bounty to you? If I click I get this: i.stack.imgur.com/0pnU6.png Do I have to wait until bounty time is ending? I will try again later. –  goobie Jan 29 '13 at 9:19
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