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I have to show that $[0,\infty)^2$ is not a differentiable manifold. The problem is $(0,0)$ (because there doesn't exist a diffeomorphism between $[0,\infty)^2$ and $R^2$) but I don't know how to show that.

$[0,\infty)^2$ = $[0,\infty) \times [0,\infty)$ (the non-negative part of the axes in $R^2$)

And then I am searching for a homeomorphism between $[0,\infty)^2$ and $H^2 := \{x=(x_1,x_2) \in R : x_2 ≥ 0 \}$ that is a diffeomorphism restricted to $[0,\infty)^2 - \{(0,0)\}$.

Thanks muchly!

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    What is $[0, \infty)^2$? – 2017-01-01
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    I write it in the question. – 2017-01-01
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    Please add all details of your relevant knowledge of the question and attempets you've made. Thanks for using TeX, by the way! – 2017-01-01
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    Are you trying to show the closed quadrant is not a _manifold_, or that it's not a _manifold with boundary_ (which is a special type of non-manifold)? – 2017-01-01
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    I try to show that it is not a differentiable manifold. – 2017-01-01
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    Do you know a homeomorphism to $H^2$ (where $x_2 ≥ 0$)? – 2017-01-01
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    The (corrected) second part of your question implies that you *are* considering manifolds with boundary (such as your $H^2$). Are you sure you have stated the first part correctly? – 2017-01-01
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    I know what you mean but the task is so formulated. – 2017-01-01
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    Has somebody an idea for the second part? – 2017-01-01
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    The second part is easy using polar coordinates. – 2017-01-01

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If it was a manifold, then you have a chart $f:U\subset [0,\infty]^2\rightarrow V\subset R^2$ which contains $(0,0)$. $U$ can be supposed of the form $[0,c)^2$. Thus $V$ is open and contractible since $f$ is an homeomorphism. But the fundamental group of $V-f((0,0)$ is not trivial (the fundamental group of an open subset $W$ of the plan without a point $u$ is not trivial. To see this, consider $B(u,r)\subset W$. You cannot deform the circle of center $u$ and radius $r/2$ which is contained in $W-\{u\}$ to a point). consider and $U-(0,0)$ its fundamental group is trivial.

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    Stick with the contractible argument: it works for all dimensions (no connected open subset of $\mathbb{R}^n$ is contractible when removing a point). – 2017-01-01
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    But all charts from $R^2$ to $R$ and your f is to $R^2$? – 2017-01-01
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    It is to $R^2$ since it is a 2-differentiable manifold. Anyway, a differentiable map $f:U\subset [0,\infty]^2\rightarrow R$ where $U$ is open cannot be injective. – 2017-01-01
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    I think it can be injective because R ist equipotent to (0,1) and the non-negative part of the axes in $R^2$ looks like R (if you "crease" it) – 2017-01-01
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    Sorry, I get it. Your are right. – 2017-01-01