quotient is a projective variety 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!
 A: First of all, you have 
$$f^{-1}(L)=\left\{ A\in\operatorname{SL}_2(K) ~\middle\vert~ A\cdot L=L \right\} = \left\{ \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} ~\middle\vert~ a\in K^\times, b\in K \right\}$$
as opposed to what you wrote. Note that this group is not a normal subgroup because 
$$\begin{pmatrix} 0&1 \\ 1& 0 \end{pmatrix}
\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}
\begin{pmatrix} 0&1 \\ 1& 0 \end{pmatrix}
=\begin{pmatrix} 0&1 \\ 1& 0 \end{pmatrix}
\begin{pmatrix} b & a \\a^{-1} & 0 \end{pmatrix}
=\begin{pmatrix} a^{-1} & 0\\ b&a \end{pmatrix}\notin f^{-1}(L).
$$
Therefore, you can form the quotient $\operatorname{SL}_2(K)/f^{-1}(L)$, but it is not a group! It is just a set of left cosets. Now, without even talking about varieties, this quotient (of a group by a subgroup, no more, no less) is in bijection with $\mathbb P^1$. Indeed, the orbit map $f:\operatorname{SL}_2(K)\to\mathbb P^1$ factors as a map $\bar f:\operatorname{SL}_2(K)/f^{-1}(L)\to\mathbb P^1$ which is injective and surjective.
Now, the question is: In what way is $X:=\operatorname{SL}_2(K)/f^{-1}(L)$ a variety? I would feel very comfortable giving $X$ the structure of a variety via $\bar f$: In other words, we say that $Z\subseteq X$ is closed if and only if $\bar f(Z)$ is closed in $\mathbb P^1$. This defines a topology on $X$ which is the topology of an algebraic variety.
We are then left to check that the projection $\pi:\operatorname{SL}_2(K)\to\operatorname{SL}_2(K)/f^{-1}(L)$ is a morphism of varieties - but it certainly is, because $\bar f$ is an isomorphism by decree and $f$ is a morphism, so $\pi=\bar f^{-1}\circ f$ is also a morphism.
A: Your map $g$ is just $f$, considered as a map on the quotient, and then $h$ is its inverse.
