Why is the map $\mathrm{SL}_2(\mathbb{R})/\mathrm{SO}(2) \rightarrow \mathbb{H} : A \mapsto Ai$ injective?

If $X$ is any set and $G$ is a group acting transitively on it then we can identify $X$ with $G/G_x$ where $G_x$ is the stabilizer of some fixed point $x \in X$.

The upper triangular matrices act transitively on $\Bbb H$, so $\mathrm{SL}_2(\Bbb R)$ does too, and $\mathrm{SO}(2)$ is the stabilizer of $i$.

We can get injectivity explicitly too:

$Ai = Bi \iff AB^{-1} \in \mathrm{SO}(2) \iff A \cdot \mathrm{SO}(2) = B \cdot \mathrm{SO}(2)$, showing that $A=B$ modulo $\mathrm{SO}(2)$ as required.

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