I am reading "Analysis on Manifolds" by James R. Munkres.
Theorem 8.2. Let $A$ be open in $\mathbb{R}^n$; let $f:A\to\mathbb{R}^n$ be of class $C^r$; let $B=f(A)$. If $f$ is one-to-one on $A$ and if $Df(x)$ is non-singular for $x\in A$, then the set $B$ is open in $\mathbb{R}^n$ and the inverse function $g:B\to A$ is of class $C^r$.
My Proposition 1:
Let $m\neq n$.
Let $A$ be open in $\mathbb{R}^m$.
Let $f:A\to\mathbb{R}^n$ be differentiable.
Let $B=f(A)$.
Let $f$ be one-to-one on $A$.
Let $g:B\to A$ be the inverse function of $f$.
Then $g$ is not differentiable on $B$.
Proof:
Assume $g$ is differentiable on $B$.
Let $a\in A$ and $b:=f(a)$.
Then by the chain rule $I_m=D(I)=D(g\circ f)(a)=Dg(f(a))\cdot Df(a)$.
Then by the chain rule $I_n=D(I)=D(f\circ g)(b)=Df(g(b))\cdot Dg(b)$.
$\operatorname{rank}Df(a)\leq\min\{m,n\}$.
$\operatorname{rank}Dg(b)\leq\min\{m,n\}$.
$m=\operatorname{rank}I_m=\operatorname{rank}Dg(b)\cdot Df(a)\leq\min\{\operatorname{rank}Df(a),\operatorname{rank}Dg(b)\}\leq\min\{m,n\}$.
$n=\operatorname{rank}I_n=\operatorname{rank}Df(a)\cdot Dg(b)\leq\min\{\operatorname{rank}Df(a),\operatorname{rank}Dg(b)\}\leq\min\{m,n\}$.
But this is impossible.
My Proposition 2:
Let $A$ be a connected set in $\mathbb{R}^n$.
Let $f:A\to\mathbb{R}^n$ be continuous.
Let $B=f(A)$.
If $f$ is one-to-one on $A$, then the inverse function $g:B\to A$ is also continuous.
My Proposition 3:
Let $m\neq n$.
Let $A$ be a connected set in $\mathbb{R}^m$.
Let $f:A\to\mathbb{R}^n$ be continuous.
Let $B=f(A)$.
If $f$ is one-to-one on $A$, then the inverse function $g:B\to A$ is not continuous.
Is my Proposition 2 true?
Is my Proposition 3 true?
I know that the following theorem holds:
THEOREM 3 (from "Calculus Fourth Edition" by Michael Spivak).
If $f$ is continuous and one-one on an interval, then $f^{-1}$ is also continuous.
So, my Proposition 2 holds when $n=1$.