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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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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.

Tsemo Aristide
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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). – Steve D Jan 01 '17 at 21:55
  • But all charts from $R^2$ to $R$ and your f is to $R^2$? –  Jan 01 '17 at 21:57
  • 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. – Tsemo Aristide Jan 01 '17 at 21:59
  • 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) –  Jan 01 '17 at 22:05
  • Sorry, I get it. Your are right. –  Jan 01 '17 at 22:14