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.
Answer
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.
No comments:
Post a Comment