Existence of a Bijective Map I am having a little trouble with this question.
Prove that there does not exist a bijective map from $\mathbb{R}^2 \to \mathbb{R}^3$ where $f$ and $f^{-1}$ are both differentiable.
Thanks for any help.    
 A: I'd be pretty surprised if this question wasn't already answered somewhere on this site... but here's a sketch.
Suppose $f : \mathbb{R}^2 \to \mathbb{R}^3$ is bijective with both $f$ and $f^{-1}$ differentiable. In particular, $f$ and $f^{-1}$ are continuous i.e. $f$ is a homeomorphism. So the question is: "why is $\mathbb{R}^2$ not homeomorphic to $\mathbb{R}^3$?" The simplest approach is probably to note that $\mathbb{R}^2$ minus a point is not simply connected, but $\mathbb{R}^3$ minus a point is simply connected. Since the property

there exists a point $x \in X$ such that $X \setminus \{x\}$ is not-simply connected

is invariant under homeomorphism, we are done.
A: Be $f:\mathbb R^2\to\mathbb R^3$ such a function.
Now define on $\mathbb R^3$ the coordinate functions
$$\begin{align}
\omega_1\colon &(x,y,z)\mapsto x\\
\omega_2\colon &(x,y,z)\mapsto y\\
\omega_3\colon &(x,y,z)\mapsto z
\end{align}$$
Those are clearly differentiable, and $\partial_i\omega_j=\delta_{ij}$ everywhere.
Now define the functions
$$\alpha_i = \omega_i\circ f$$
By the chain rule, the $\alpha_i$ are differentiable functions from $\mathbb R^2$ to $\mathbb R$. I'm going to write the points of $\mathbb R^2$ as $(u,v)$.
Now consider the gradients at a point $o$, $g_i=\operatorname{grad_{\mathbb R^2}}\alpha_i(o) = (\partial_u\alpha_i(o),\partial_v\alpha_i(o))$. Since those are two-dimensional vectors, they must be linearly dependent, that is, there exist numbers $a, b, c$ of which at least one is not $0$, so that $a g_1+b g_2+c g_3=0$.
Now by the assumptions, $\omega_i=\alpha_i\circ f^{-1}$. Be $p=f(o)$ (and thus $o=f^{-1}(p)$). Then we can calculate $\partial_i\omega_j(p)$ using the chain rule (this is allowed because $f^{-1}$ is differentiable by assumption):
$$\partial_i\omega_j(p) = \partial_u\alpha_j(f^{-1}(p))(\partial_i(f^{-1}))_u(p) + \partial_v\alpha_j(f^{-1}(p))(\partial_i(f^{-1}))_v(p) = g_j\cdot(\partial_i(f^{-1}))(p)$$
Especially, we get $a\partial_i\omega_1(p)+b\partial_i\omega_2(p)+b\partial_i\omega_3(p) = (ag_1 + bg_2 + c g_3)\cdot(\partial_i(f^{-1})(p)) = 0$ for all $i$.
But we find from direct calculation that
$$a\partial_i\omega_1(p)+ b\partial_i\omega_2(p)+c\partial_i\omega_3(p) = a\delta_{i1}+b\delta_{i2}+c\delta_{i3} = \begin{cases}a & i=1\\b & i=2\\c & i=3\end{cases}$$
and at least one of $a$, $b$, $c$ is $\ne0$.
Therefore we get a contradiction, and thus no such $f$ can exist.
