In this question I am asking if I have some open set in $\mathbb{P}^1$, why is it isomorphic to some open set in $\mathbb{A}^1$. First I want to say that I know that $\mathbb{A}^1$ is birational to $\mathbb{P}^1$ and as a result there exists $\textbf{some}$ open $U\in \mathbb{P}^1$ and $V\in \mathbb{A}^1$ such that $U \cong V$. What I am asking though, is that given some open $U'\in \mathbb{P}^1$, not necessarily $U$, that there is $V'$ in $\mathbb{A}^1$ such that $U'\cong V'$.
My attempt: If I wanted to show this I would start off with an open set in $\mathbb{P}^1$ which would be of the form $\mathbb{P}^1-V(f)$ where $f$ is a homogenous polynomial in $x$ and $y$, where $[x,y]$ is the homogeneous co-ordinates of $\mathbb{P}^1$. I would then want to construct a map from $\mathbb{P}^1$ to $\mathbb{A}^1$. I would probably choose something like $\phi:\mathbb{P}^1-\{[0,1]\}\rightarrow \mathbb{A}^1,\phi([x,y])=y/x$. This is all well and good but what if $[0,1]$ is in $\mathbb{P}^1-V(f)$, then what do I do. I know I could just use the above map and change $[0,1]$ to $[1,0]$ and change $y/x$ to $x/y$ but I would still end up with the same problem.
I get the feeling this question shouldn't be hard, I'm just attacking it the wrong way. Any help is much appreciated.
First of all, your open set $U \subset \mathbb P^1$ needs to be strict otherwise the problem is false (open subsets of $\mathbb A^1$ don't have only constant regular functions).
Next realize that $\mathbb P^1$ is covered by two copies of $\mathbb A^1$ in the following way. One cover sends $t \in \mathbb A^1$ to $[t:1]$. The other sends $s \in \mathbb A^1$ to $[1:s]$. Since $U$ is strict, it misses some point in $\mathbb P^1$, after applying a projective linear transformation, we can assume the missing point is $[1:0]$ (write down a matrix sending the missing point $[a:b]$ to $[1:0]$; this matrix induces an isomorphsm of $\mathbb P^1$). Then $U$ is actually in the image of the first isomorphism I described above.