Does the transformation $x=Pabc$, $y=Qab(1-c)$, $z=Ra(1-b)$ map a unit cube to a tetrahedron?

Question

Does the transformation

$x=Pabc$, $y=Qab(1-c)$, $z=Ra(1-b)$

map a unit cube in $abc$ coordinates to the tetrahedron with vertices $(P,0,0)$, $(0,Q,0)$, $(0,0,R)$ and $(0,0,0)$ in xyz coordinates?

If not what transformation does?

Answer 1

Yes, it does. The vertices of the cube are mapped as follows ($*$ is either $0$ or $1$): $$ (1,1,1)\mapsto (P,0,0)$$ $$(1,1,0)\mapsto (0,Q,0)$$ $$ (1,0,*) \mapsto (0,0,R)$$ $$ (0,*,*) \mapsto (0,0,0)$$

Also, every line parallel to one of coordinate axes $a,b,c$ is mapped to a line, because the formulas for $x,y,z$ are linear in each coordinate separately. Therefore, the edges of the cube are mapped to edges of tetrahedron (or contracted into vertices, if they join two vertices that get mapped into the same one). The faces of the cube are foliated by line segments parallel to coordinate axes, joining two points on the edges. Since we already know where the edges go, it follows that the images of these line segments foliate the faces of tetrahedron. Finally, the interior of the cube is foliated by line segments in the $a$-direction; they are mapped to line segments joining $(0,0,0)$ with the points on the face $PQR$; i.e., the images foliate the interior of the tetrahedron.