Studying some introductory algebraic geometry (affine and projective varieties) I came up with this which I can't understand:

$K$ is an algebraically closed field.

> 
> If we define a map $ f : SL\_2 (K) \to \mathbb P^1 $, $ A \mapsto A
>  \cdot L$, where $L= K \cdot(1,0)^T \subset K^2$, then the inverse
>  image $ f^{-1} (L)= \{A \in SL\_2 | A \cdot L =L \} = \{ \left(
>  \begin{array}{ c c } 1 & b \\ 0 & 1 \end{array} \right) \}$
> 
> 
> Why is it $ SL\_2 / f^{-1} (L) \cong \mathbb P^1 $ as projective
>  varieties?
> 
> 
> 

I understand that this map $f$ is surjective but exept this, nothing. How can we get an isomorphism? And what isomorphism is this, since $SL\_2$ is a group and $\mathbb P^1$ is not.

Maybe I haven't understand something, so any help would be really appreciated. 

**edit:** After Qiaochu Yuan comment let restate my question. These two objects will be isomorphic as varieties. But for this to hold I need two maps $ g: SL\_2 / f^{-1} (L) \to \mathbb P^1 $ and $ h: \mathbb P^1 \to SL\_2 / f^{-1} (L)$ such that $g,h$ are inverse of each other. How can I find these two maps?

Thank you!