Proving that $\frac{D(a,b,c)}{D(x,y,z)}=1$ In deriving continuity equations using Lagrangian .
We consider the element of fluid which occupied a rectangular parallelopiped having its centre at the point $(a,b,c)$ and its edges $\delta a$ , $\delta b$ ,$\delta c $ parallel to the axes . At the time $t$ the same element for an oblique parallelepiped . The centre now has for its co-ordinates $x$ , $y$ , $z$; and the projections of the edges on the co-ordinate axes are respectively 
$$ \frac{\partial x}{\partial a} \delta a \ , \ \frac{\partial y}{\partial a} \delta a \ , \ \frac{\partial z}{\partial c} \delta a$$
$$\frac{\partial x}{\partial b} \delta b \ , \ \frac{\partial y}{\partial b} \delta b \ , \ \frac{\partial z}{\partial b} \delta b$$
$$\frac{\partial x}{\partial c} \delta c \ , \ \frac{\partial y}{\partial c} \delta c \ , \ \frac{\partial z}{\partial c} \delta c$$
How can i get these projections ?
The volume of the parallelepiped is therefore 
$$\begin{vmatrix}
\frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\ 
\frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\ 
\frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c}
\end{vmatrix} \delta a \delta b \delta c$$
or as its often written $$\frac{D(x,y,z)}{D(a,b,c)} \delta a \delta b \delta c$$
since the fluid mass is unchanged and the fluid is incompressible we have 
$$\frac{D(x,y,z)}{D(a,b,c)} =1$$
Is there a way to prove that $$\frac{D(a,b,c)}{D(x,y,z)}= 1$$
without expanding the determinant ?
 A: The projections can be obtained by using the chain rule:
$$ 
 \delta x =  \frac{\partial x}{\partial a}\delta a 
             + \frac{\partial x}{\partial b}\delta b 
             + \frac{\partial x}{\partial c}\delta c,
$$
et cetera.
If I understand the situation, the (infinitesimally small) oblique parallelogram with coordinates $x$, $y$, $z$ gives the volume element at a time $t$. Consider the map $F_t \colon \mathbb{R}^3\to \mathbb{R}^3: (a,b,c)\mapsto(x,y,z)$ that gives the future position $(x,y,z)$ at time $t$ for a point with initial position $(a,b,c)$. Let $G_t$ be the inverse map. Then $F_t \circ G_t= id$. 
Note that the matrices of the differentials $dF_t$ and $dG_t$ are 
$$
 \begin{bmatrix}
   \frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\
\frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\
\frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c} 
 \end{bmatrix}
\quad \text{resp.} \quad
  \begin{bmatrix}
   \frac{\partial a}{\partial x} & \frac{\partial b}{\partial x} & \frac{\partial c}{\partial x} \\
\frac{\partial a}{\partial y} & \frac{\partial b}{\partial y} & \frac{\partial c}{\partial y} \\
\frac{\partial a}{\partial z} & \frac{\partial b}{\partial z} & \frac{\partial c}{\partial z} 
 \end{bmatrix}.
$$ 
Since $F_t$ and $G_t$ are inverse, these matrices are also each others inverses. Consequently their determinants
$$ 
 \frac{D(x,y,z)}{D(a,b,c)} \quad \text{and} \quad \frac{D(a,b,c)}{D(x,y,z)}
$$
are inverses, so $\frac{D(a,b,c)}{D(x,y,z)}$ is $1$ as well.
