Inverse function theorem : Let $f:E \to R^n$ . For each $U,V$ in the construction and $x \in U$ , can we have $f'(x)$ invertible?

Question

If $f:E \to R^n$ is continuously differentiable and $f'(a)$ is invertible , $b=f(a)$.Then we can find open sets $U,V \in R^n$ such that :
$(1)$ $a\in U$ , $b \in V$ and $f$ is one-to-one from $U$ onto $V$ .
$(2)$ If $g$ is the inverse of $f$ defined in $V$ , then $g$ is continuously differentiable .

In the proof of Rudin's book , the $(U,V)$ he construct also has the property that for all $x\in U$ , we have $f'(x)$ invertible . So we have $g'(y)=\{f'(g(y))\}^{-1}$ for every $y \in V$ .
However , does this always valid for every pair of $(U,V)$ that satisfied condition $(1),(2)$ ?

Answer 1

Yes, because by the second condition, the inverse $g$ is differentiable; hence by the chain rule, for every $x \in U$, we have that $g'(f(x))\circ f'(x) =I$ (the identity linear transformation). In particular, $f'(x)$ has to be invertible.