Cremona transformations are birational maps

Solution 1:

Maybe it's time for a (reasonably complete) answer.

Consider the open subvariety $U_{YZ}:=\Bbb{P}^2 \setminus V(YZ)$ (i.e. $\Bbb{P}^2$ with both the $Y$-axis and the $Z$-axis removed). The rational map $f: \Bbb{P}^2 \dashrightarrow \Bbb{P}^2$ can be represented on $U_{YZ}$ by the morphism

\begin{split} U_{YZ} &\rightarrow U_Z:=\Bbb{P}^2 \setminus V(Z)\\ (x:y:z) &\mapsto (x/z:x/y:1) \end{split} which shows that $f$ is defined on all of $U_{YZ}$. Similarly, you find that $f$ is defined on all of $\Bbb{P}^2 \setminus \{(1:0:0),(0:1:0),(0:0:1)\}$, and the only thing left to check for 1. and 2. is that $f$ cannot be extended to all of $\Bbb{P}^2$. One possible explanation is the following.

Consider the morphism \begin{split} f_{cone}: \Bbb{A}^3 &\rightarrow \Bbb{A}^3\\ (x,y,z) &\mapsto (xy,xz,yz) \end{split} which maps every point of the form $(\lambda,0,0),(0,\mu,0)$ or $(0,0,\nu)$ to the point $(0,0,0)$. Like every morphism of varieties, $f_{cone}$ is already uniquely determined by its restriction to any non-empty open subset, e.g. by its restriction to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\} \subset \Bbb{A}^3$.

But $f_{cone}$ restricted to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\}$ induces the morphism representing the rational map $f$, as described above. Hence every extension of $f$ to a morphism defined on all of $\Bbb{P}^2$ would have to map the points $(1:0:0),(0:1:0)$ and $(0:0:1)$ to "$(0:0:0)$", which is impossible, and consequently, the points $(1:0:0),(0:1:0),(0:0:1)$ cannot be in the domain of $f$.

For 3., you simply use the "trick" mentioned by Asal. Let $U_{XYZ}:=\Bbb{P}^2 \setminus V(XYZ)$ (i.e. $\Bbb{P}^2$ without the coordinate axes). On $U_{XYZ}$, we can represent $f$ as $$ (x:y:z) \mapsto (xy:xz:yz)=1/(xyz)(xy:xz:yz)=(1/z:1/y:1/x) $$ and from this, you can read off the inverse of $f_{\vert U_{XYZ}}$ directly.