# Assumptions of the inverse mapping theorem

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.