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!


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?

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    $\begingroup$ Thanks for your answer. The other question I had was whether the comment implies that a continuously differential function is invertible, but this is clearly wrong (see counterexample in my post). So the statement in bold merely means that there are invertible functions that do not satisfy the conditions of the theorem. However. these cannot be differentiable. I think I am clear on this after giving it another thought. $\endgroup$ – DerivativesGuy Jun 26 at 19:00

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