Consider a mapping $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$
\begin{align} g(u,v) &= (x, y)\\ &=(f(u,v), \;g(u, v)) \end{align}
Consider the inverse mapping $g^{-1} = \mathbb{R}^2 \rightarrow \mathbb{R}^2$
\begin{align} g^{-1}(x,y) &= (u,v) \\ &= (h(x,y), \;I(x,y)) \end{align}
Question Why is it that the determinant of the Jacobian of $g$ denoted as $|\text{Jac}\; g|$ equal to $1$ over the determinant of the Jacobian of $g^{-1}$ denoted $|\text{Jac}\; g^{-1}|$?
$$ |\text{Jac}\;g| = \frac{1}{|\text{Jac}\;g^{-1}|}$$ You could do the computation. But I'm not sure how I would connect the two sides of the equality. The left side is $x_uy_v - y_ux_v$. But I can't seem to get this equal to the other side.
Thoughts Consider a function $y = f(x)$. Consider it's inverse function $x = f^{-1}(y)$. The two are related as follows
$$\frac{df^{-1}}{dy}\Bigg|_y = \frac{1}{\frac{df}{dx}\Big|_x} $$
where the evaluation values $x,y$ are images of one another (people write this formula in many different ways). This last equation makes intuitive sense to me. Draw a graph of $f(x)$. To find the graph of the inverse function, all you have to do is rotate this graph $90^\circ$ counterclockwise. If you draw a tangent line to the graph of $f(x)$, such an instantaneous line has some rise/run. Once you rotate the graph $90^\circ$ counterclockwise to view the inverse graph, this tangent line now has a slope of run/rise. Therefore the formula follows. Is there such an interpretation for my question? I tried writing out both determinants. However, it's not clear to me how I would related $\partial x/\partial u$ to $\partial u/\partial x$. You would think that
$$\frac{\partial x}{\partial u}\Bigg|_u = \frac{1}{\frac{\partial u }{\partial x}\Big|_x} $$
This would be true given $x = f(u)$. However we have $x = f(u,v)$ and $u = h(x,y)$ and I'm not sure of the relationship between the two. Thanks in advance