TITLE: Why is the map $\mathrm{SL}_2(\mathbb{R})/\mathrm{SO}(2) \rightarrow \mathbb{H} : A \mapsto Ai$ injective?
QUESTION [5 upvotes]: First some notation:
$\mathrm{SL}_2(\mathbb{R})=\left\lbrace\begin{pmatrix} a&b \\ c &d \end{pmatrix}\;\Biggm|\;a,b,c,d \in \mathbb{R} ,ad-bc=1 \right\rbrace$
$\mathrm{SO}(2)=\lbrace K\in \mathrm{SL}_2(\mathbb{R}) : A^TA=AA^T=I \rbrace$.
$\mathbb{H}$ denotes the upper half plane.
The map $$\mathrm{SL}_2(\mathbb{R})/\mathrm{SO}(2) \rightarrow \mathbb{H} : A \mapsto Ai$$
is bijective.
My problem is to understand why the map is injective.
My idea is to take $N\in \mathrm{SO}(2)$ and consider $Mi=Ni$.
Since $\mathrm{SO}(2)$ is the stabilizer of $i$ it follows that $M=N$.
Thanks for the help.

REPLY [2 votes]: $$\gamma . i = \beta . i \\\implies \beta^{-1} \gamma.i =\beta^{-1}.\beta.i= i\\ 
\beta^{-1}\gamma. i = \frac{ai+b}{ci+d}= i \implies ai+b = di-c \implies (c,d) = (-b,a)\\
\implies \beta^{-1} \gamma \in SO_2(\Bbb{R})\\ \implies \beta^{-1} \gamma SO_2(\Bbb{R}) =SO_2(\Bbb{R}) 
\\ \implies  \gamma SO_2(\Bbb{R})=\beta SO_2(\Bbb{R}) $$