Domain of rational map $\mathbb{P}^2 \to \mathbb{P}^2$

Question

Let $\phi:(t_0:t_1:t_2) \mapsto (\frac{1}{t_0}:\frac{1}{t_1}:\frac{1}{t_2})$. I think that we cat extend $\phi$ to rational map $\hat{\phi}$ with domain:$\mathbb{P}^2-\{(1:0:0),(0:1:0),(0:0:1)\}$. How to evaluate $\hat{\phi}(1:1:0)$? By logic $\hat{\phi}(1:1:0) = (1:0:1)$ but i don't sure.

Answer 1

Your rational map is better written as $\phi:(t_0:t_1:t_2) \mapsto (t_1t_2:t_0t_2:t_0t_1)$.
It is undefined only at the three points $(1:0:0),(0:1:0),(0:0:1)$ and defined at all other points.
In your example one has $\phi((1:1:0))=(0:0:1)$
The map $\phi$ is actually a birational involution: we have $\phi=\phi^{-1}$ as rational maps $\mathbb P^2\to \mathbb P^2$

Edit
a) The map $\phi$ is a birational equivalence: it induces a (biregular!) isomorphism $\psi=\phi\mid _U:U\to U$, where $U\subset \mathbb P^2$ is the open subset $U=\mathbb P^2\setminus \{t_0t_1t_2=0\}$ .
That isomorphism $\psi$ is its own inverse $\psi=\psi^{-1}$.

b) Note carefully the following slightly paradoxical result:
$U$ is the largest open subset of the projective plane on which $\phi$ induces a biregular isomorphism, but one can extend $\phi$ to a larger open subset, namely the plane minus the three points $(1:0:0),(0:1:0),(0:0:1)$), on which $\phi$ will be regular but no longer injective.

Answer 2

Hint : if $abc \neq 0$ then $\phi(a,b,c) = (bc, ac, ab)$ and you can use this form for evaluate $\phi(1,1,0)$.