Show that there exists no bijective map $f :$ $\mathbb R^{2} \to \mathbb R^{3}$ such that $f$ and $f^{-1}$ is differentiable.
I think if there is a bijective map between $\mathbb R^{2}$ and $\mathbb R^{3}$ then $\mathbb R^{2}$ and $\mathbb R^{3}$ will be homeomorphic and it creates a contradiction because they are not homeomorphic.hence there doesn't exist such bijection.
Is it right?If it is right is there any other way to do this using multivariate calculus and not using Topology.
It's true that the spaces are not homeomorphic, but that's a very hard theorem to prove. Instead, here's a hint for how to use calculus to aolve the problem: since $f$ and its inverse are differentiable, try applying the chain rule to $f\circ f^{-1}=I$. Now to conclude, use some linear algebra.