Why is an isomorphism of $\Bbb P^1$ fixing $[0:1]$, $[1:1]$, and $[1:0]$ the identity? I'm reading the lecture notes of my algebraic geometry course and I'm stuck at an exercise:
Let $\phi: \mathbb{P}^1\rightarrow \mathbb{P}^1$ be an isomorphism of algebraic varieties such that
$\phi([0 : 1]) = [0 : 1], \phi([1 : 1]) = [1 : 1], \phi([1 : 0]) = [1 : 0]$. Show that $\phi$ is the identity map.
My attempt: I'm trying to prove that $\phi|_{U_i}$, $i=0,1$ is identity and that they agree on $U_{01}=U_0\cap U_1$.
 A: By assumption that $\phi([0 : 1]) = [0 : 1], \phi([1 : 0]) = [1 : 0]$, we know that $\phi|_{U_i} : U_i \rightarrow U_i $ for both $i=0,1$. (since, for example, $\phi([1:a]) \not= [0:1]$ so $\phi([1:a]) \in U_0$). Note that since $U_i$ is $\mathbb{A}^1$, the isomorphism $\phi|_{U_i}$ can be represented by linear polynomials $a_ix+b_i \in k[x] $, i.e, $\phi|_{U_0}([1:u])=[1:a_1u+b_1]$ and  $\phi|_{U_1}([v:1])=[a_2v+b_2:1]$. Now $\phi([0 : 1]) = [0 : 1], \phi([1 : 1]) = [1 : 1], \phi([1 : 0]) = [1 : 0]$ imply that $a_1=a_2=1$ and $b_1=b_2=0$.
A: In projective geometry the following theorem can be proven:
Theorem: Let $\mathbb{P}(V)$ and $\mathbb{P}(V')$ be two projective spaces such that $dim\mathbb{P}(V)=dim\mathbb{P}(V')=n$. Consider two sets of $n+2$ points in general position: $\begin{Bmatrix}
P_{0},...,P_{n+1}
\end{Bmatrix} \subseteq \mathbb{P}(V)
$ and $\begin{Bmatrix}
Q_{0},...,Q_{n+1}
\end{Bmatrix} \subseteq \mathbb{P}(V')
$.
Then it exists a unique projective isomorphism $f:\mathbb{P}(V)\rightarrow \mathbb{P}(V')$ such that $f(P_{i})=Q_{i}$, $\forall i=0,..,n+1$.
It's easy to conclude that if a projective isomorphism has $n+2$ fixed points in general position then it's the identity map.
Since in $\mathbb{P}^1$ the points $[1:0],[0:1],[1:1]$ are in general position, a projective isomorphism fixing them must be the identity.
