Cremona transformations are birational maps Consider the following map, which is a Cremona transformation: 
$$
\begin{split}
f\colon & \mathbb P^2 \dashrightarrow \mathbb P^2 \\
& (x:y:z) \mapsto (xy: xz: yz)
\end{split}
$$
I have to prove the following: 

1. The domain is $\mathbb P^2 \setminus \{(1:0:0), (0:1:0), (0:0:1)\}$.
2. The map cannot be extended to all $\mathbb P^2$. 
3. The map is birational by finding its inverse.

Well, point (1) is clear ($xy=xz=yz=0$ implies that two of $x,y,z$ are 0). What about point 2? I do not know how to answer. How could I extend this map to the three special points found in point 1? Why is this impossible? 
Finally, the inverse. Is there an easy trick to find it? I've tried to solve the system in homogeneous coordinates but I think that maybe there is a faster way... 
Thank you for your help.  
 A: 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.
A: I am not permitted to comment now, so I have to write an answer as a complement of Nils’s answer. Indeed I don’t understand Nils’s answer of 2, since I am not a specialist in this field.
I think the argument below is easier to understand for me.
Suppose the map can be extended to include $(1,0,0)$ and we denote the extended map by $\psi: U\rightarrow \mathbb P^2$. Now $U_i=\{(x_0,x_1,x_2)\in \mathbb P^2|x_i\neq 0\}$ is an open covering of $\mathbb P^2$. So one of its preimage contains $(1,0,0)$. This is not symmetric but I will just check the case of $U_0$. $\psi^{-1}(U_0)$ contains $(U_1\cap U_2)\cup \{(1,0,0)\}$ and is open. So $(1,0,0)\in D(f)\subseteq \{(1,0,0),(0,1,0),(0,0,1)\}\cup (U_1\cap U_2)$. Assume $f(x_0,x_1,x_2)=x_0^n+\sum_{u+v+w=n,w<n}a_{uvw}x_1^u x_2^v x_0^w$. Let $x_0=1$ and $x_1=0$ we are reduced to $1+\sum_{v+w=n,w<n}a_{0vw}x_2^v$. This polynomial has finitely many roots and so we have a contradiction.
This method is the first one coming into my mind. I will appreciate it if anyone could give some general theory!
