Upper Half "Plane" not diffeomorphic to the whole plane maybe the title isnt the best but I have the following excercise, lets define
$$\cal{H}^n = [0,\infty)^n$$
I have to proove that an open set in $\mathcal{H}^n$  in general is not diffeomorphic to an open set in $\mathbb{R}^n$, I understand the idea of the proof but I don't know how to write it.
My attempt of a solution:
I have the following theorem: Given an open set $U$ in $\mathbb{R}^n$ and an arbitrary set $V$ in $\mathbb{R}^n$, if there is a diffeomorphism $f:U\rightarrow V$ then $V$ is open.
if $V$ is a subset of $(0,\infty)^n$ there is no problem because $V$ will be open in $\mathbb{R}^n$. However, if $V$ intersects one of the axis then $V$ wont be open in $\mathbb{R}^n$ and we get a contradiction.
Is this right? is there a better way to write it?
 A: So you have to pick a point on $V$ that has one of its coordinates zero.  Now if there is a diffeomorphism from that point to an open set in $\mathbb R^n$ - well, you know that map has a derivative, and that is an $n\times n$ matrix that is invertible.  So locally it is almost linear, and so is its inverse.  But a map that is almost linear cannot map an open set in $\mathbb R^n$ to the point we started with.
Let's make it more rigorous.  So let $f:U \to V$ be the purported map, and suppose $f(x_0) = y_0$ where $y_0$ has one of its coordinates equal to zero.  Let $L$ be the derivative of $f$ at $x_0$.  By applying $L$ to $U$, without loss of generality we can assume $L$ is the identity map.  Now there exists $\epsilon>0$ such that $f(x) = y_0 + (x-x_0) + g(x-x_0)$ where $|g(x-x_0)| \le |x-x_0|/10$ if $|x-x_0|<\epsilon$.  (This follows from the definition of being differentiable.)
So now consider $x_0$ with one of the coordinates having plus or minus $\epsilon/2$ added to it.  One of those $2n$ points must map to a point outside of $\mathcal H^n$.
