Inverse mapping theorem: Let $f: U \to \mathbb{R}^m$ be continuously
differentiable, and $a \in U$. Suppose that $df_a$ is invertible, i.e.
$det(J_f(a)) \neq 0$. Then $a$ has an open neighbourhood $V \subset U$,
such that $f: V \to f(V)$ is a diffeomorphism, i.e.(i) $f: V \to \mathbb{R}^m$ is injective,
(ii) $f(V)$ is open,
(iii) $f$ has an inverse $f^{-1} \in C^1(f(V),V)$.
Definition: Suppose $f: U \to V$, for $U \subset \mathbb{R}^n, V
\subset \mathbb{R}^m$.(i) $f$ is a homeomorphism if$f$ is bijective, $f$ is continuous, and
the inverse $f^{-1}$ is continuous.(ii) A homeomorhpism $f$ is called a diffeomorphism if $f$ and
$f^{-1}$ are continuously differentiable.
My lecture notes state the following regarding the assumption $det(J_f(a)) \neq 0$:
The assumption is not necessary for $f$ to be locally invertible (cf. the one-dimensional
invertible example $f(x)=x^3$ with $f'(0)=0$). It is, however, necessary for $f$ to have a
local inverse which is differentiable.
It is clear to me that we need this assumption for the differentiability of the inverse since for a diffeomorphism $f: U \to V$ we have $n=m$ by the chain rule and the fact that only square matrices can be invertible.
Now first I thought that the statement also implies that $f$ is invertible if it is continuously differentiable, but this clearly cannot be true as the simple example $f(x)=c$ shows.
In general a bijective function is invertible, and thus an injective function $f: U \to f(U)$ is invertible. Clearly, a strictly monotone function is injective, but there are also injective functions that are not monotone, e.g.
$f(x) = \begin{cases}
1/x & x \neq 0 \\
0 & x = 0
\end{cases}$
One particular case of a strictly monotone function is a continuously differentiable function $f$ with $f'(a) \neq 0$. In this case there is a neighbourhood $B_{\epsilon}(a)$ s.t. $f'(x) \neq 0$ $\forall x \in B_{\epsilon}(a)$. Then the intermediate value theorem implies that $f'(x)>0$ or $f'(x)<0$ $\forall B_{\epsilon}(a)$, and thus $f$ is strictly monotone by the mean value theorem.
Now my question is:
Is the statement in bold simply to caution about the facts I have given above, namely that the assumptions of the inverse mapping theorem ensure that $f$ is invertible with a continuously differentiable inverse, but that there are other functions that are invertible and do not satisfy the assumptions. The condition $det(J_f(a)) \neq 0$ seems to be the analogue of the condition $f'(x) \neq 0$ in one dimension. Of course we cannot speak about strict monotonicity in higher dimensions, but can we say that this ensures that the function has non-zero directional derivatives in all directions? Indeed we have
$df_a(h)=J_f(a)h \neq 0$ for $h \neq 0$.
This would extend the intuition from the one dimensional case to higher dimensions.
Thanks very much!
Best Answer
Even though the assumption assumption $\det J_f(a) \neq 0$ is not necessary for $f$ to be locally invertible (as the example you provided illustrates), it is however necessary for $f$ to be locally invertible with continuously differentiable inverse.
This is because if $f$ is locally invertible with continuously differentiable inverse $g$, we have $J_g(f(a)) = [J_f(a)]^{-1}$ so $J_f(a)$ is invertible i.e. $\det J_f(a) \neq 0$.
If this is the case, then (as you correctly mentioned) all directional derivatives of $f$ at point $a$ are non-zero.
Have I answered your question?