One line proof - the jacobian is sometimes $1$ or $-1$

Question

I'd like to verify if this solution is correct.

Let $B(0,1)$ be the unit ball in $n$ dimensions.

$T: B(0,1) \to B(0,1)$ a diffeomorphism. Show that $\exists x \in B(0,1)$ such that $|\det D_T(x)| = 1$

My maybe correct proof:

Suppose $\nexists x \in B(0,1)$ such that this holds. Since $T$ is continuous and $\det$ is continuous and $|.|$ is continuous, this means either $|\det D_t(x)|$ is always greater than one, or always less than one.

wlog assume it's always greater than one.

Define $h:B(0,1) \to \mathbb R$ by $h(x) = 1$. Now see that:

$\text {Volume}(B(0,1)) = \int_{B(0,1)}1dV =\int_{B(0,1)}h(x)dV=\int_{T(B(0,1))}h(T(x))|\det D_T(x)|dV = \int_{B(0,1)}1|\det D_T(x)|dV = \int_{B(0,1)}|\det D_T(x)|dV > \int_{B(0,1)}1dV = \text{Volume}(B(0,1))$

Is this true or am I way off? seems a bit too simple.

Answer 1

That's completely correct. In words, if the Jacobian determinant is always less than $1$ then the area is less than itself, and if the determinant is always greater than $1$ then the area is greater than itself. Contradiction, QED. Isn't it nice when things are simple?