Jacobian determinant of a function $f$ if it has a differentiable inverse.

Question

This question is from the book "Differential Topology First Steps" by Andrew Wallace. The question is as follows:

Let $f$ be a differentiable map of an open set in Euclidean n-space into n-space and suppose that $f$ has a differentiable inverse. Show that the Jocobian determinant of $f$ is nonzero at all points of $U$.

I'm not too sure about proving it, but here is what I think. I have already shown that if $F$ is the Jacobian matrix for the differentiable function $f$ and $G$ is the Jacobian matrix for the differentiable function $g$, then the Jacobian matrix for $gf$ is $GF$. So set $g=f^{-1}$ and $G=F^{-1}$, we get $F^{-1}F = I$, where $I$ is the identity matrix. Since determinant of $I$ is nonzero, then it means that the determinant of $F^{-1}$ and $F$ are nonzero.

Is my approach more or less on track?

Answer 1

You have the right idea but the fact the jacobian of $gf$ is $GF$ is almost correct. Remember the jacobian matrix of a function is again a function, so it depends on what point you look at. The jacobian matrix evaluated at some point $a \in \mathbb{R}^n$ is the matrix representation with respect to the usual basis of $\mathbb{R}^n$ for the total derivative at that point. Let's change the notation a little bit. We will write $D_{a} f$ for the total derivative of $f$ at $a$. Remember the chain rule:

$$D_{a}(g \circ f) = (D_{f(a)} g) \circ (D_{a} f),$$ if $f$ is differentiable at $a$, and $g$ at $f(a)$. Now we let $g$ be the inverse of $f$. Since $D_{a}(\text{id}_{\mathbb{R}^n})=\text{id}_{\mathbb{R}^n}$, we see $D_{a} f$ is a left invertible linear map, with right inverse $D_{f(a)} g$, so $D_{a} f$ is injective, but since $D_{a} f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a linear map from an $n$ dimensional vector space to itself it must also be surjective.

So we have proven even more than asked, namely that the jacobian matrix of $f$ is always invertible (a linear map is invertible if and only if its matrix representation with respect to some basis is invertible). Hope this helps and is not too 'coordinate free' for your approach.