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A rational map $V⇢{\mathbb{P}}^m$ is a morphism on a nonempty open subset of $V$, with two descriptions identified if they agree on a dense open subset. It can be represented there by rational homogeneous coordinate functions not all zero. A point is regular for the map if it extends as a morphism on an open neighbourhood of that point.

For the given projective Cremona transformation, the three products vanish simultaneously exactly at the three coordinate vertices. Away from them they define a regular map. At $(1:0:0)$ consider $[1:t:st]$ as $t\to 0$, with $s\ne 0$. Its image tends to $[0:s:1]$, which depends on $s$. No regular extension could have these different limiting values. Algebraically these curves lie in its domain for $t\ne 0$ and a regular extension must give the same specialization at their common vertex, so the obstruction also works over any infinite characteristic-zero field. The other vertices follow by cyclic symmetry. On the torus $X_0{X}_1{X}_2\ne 0$, applying the map twice givesIt is therefore a birational map and is its own rational inverse.

Use coordinates $[A:B:C]$ on the target. Substitution in the quintic givesThus on the torus its image is the plane quartic $Q=0$. This quartic is nonsingular even over the algebraic closure. Its partial derivatives are $C^3+3A^{2}B$, $A^3+3B^{2}C$ and $3A{C}^2+B^3$. If one coordinate vanishes at a common zero of these partials, all vanish. If none vanishes, multiplying the three equations gives $A^3{B}^3{C}^3=-27A^3{B}^3{C}^3$, impossible in characteristic zero. Hence there is no projective singular point.

A smooth plane curve is geometrically irreducible: if it had two positive-degree components, Bézout theorem over the algebraic closure would give an intersection point, and the product equation has zero gradient there; a repeated factor also makes the gradient vanish along its component. Therefore this quartic is irreducible. Its torus part is dense and is identified by the projective Cremona transformation with the torus part of $V$. No coordinate line is a component of $V$, since restricting $F$ to each coordinate line leaves a nonzero monomial. Consequently every component meets the torus, and its torus part is already irreducible. $V$ is irreducible and birational to the nonsingular quartic $A{C}^3+A^{3}B+B^{3}C=0$.