Identity map minus Cremona transformation

Question

Let $ \delta $ be the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ in $\mathbb R^3$. It's a face of the standard octahedron. The Cremona transformation $$\mathcal C: (x, y, z) \mapsto - \frac {(yz, zx, xy)} {yz + zx + xy}$$ maps $ \delta $ to the opposite face $-\delta$ of the octahedron. And the half-sum of $\mathcal C $ with the identity transformation $$\mathcal F: (x, y, z) \mapsto \frac {(x, y, z) + \mathcal C (x, y, z)} {2}$$ maps $\delta$ to the hexagon $ h $, which is the section of the octahedron by the plane $ x + y + z = 0 $.

Question. Is the map $ \mathcal F: \delta \to h $ a diffeomorphism from the interior of $ \delta $ to the interior of $ h $? Is there a simple closed formula for the inverse map $ \mathcal F^{- 1}: h \to \delta $?

Answer 1

The answer is yes and yes: the map is a diffeomorphism from the interior of $\delta$ to the interior of $h$, and there is a rather simple formula for the inverse map.

Let $s:=(s_1,s_2,s_3):=\mathcal F$, so that $s_1+s_2+s_3=0$, let $t:=(s_1,s_2)$, and let $S:=\{(x,y)\colon0<x<1,0<y<1-x\}$.

We have to

(i) show that the map $$S\ni(x,y)\mapsto s(x,y)\in h$$ is a bijection or, equivalently, that the map $$S\ni(x,y)\mapsto t(x,y)$$ is a bijection;

(ii) show that the Jacobian determinant of $t$ is nonzero everywhere on $S$;

(iii) find the inverse of $s$.

All this is done in a Mathematica notebook, whose image is here (click on the image to enlarge it):