Quotient map from $GL_2(\Bbb R)$

Question

Is the following map a quotient map? $$f:GL_2(\Bbb R) \to \Bbb C \setminus \text{0}$$ $$ \begin{pmatrix} a & b\\ c & d\\ \end {pmatrix} \mapsto \frac{a{\iota}+b}{c{\iota}+d} $$

I wanted a quotient map from $SL_2(\Bbb R)$ to $\mathscr H$(the upper half plane), if I could show $f$ is a quotient map then somehow the restriction to $SL_2(\Bbb R)$ could be shown to be a quotient map.

Answer 1

Expanding my (possibly somewhat misguided) comment to an argument showing that $f:\gamma\mapsto \gamma\cdot\iota$ is a quotient map from $SL_2(\Bbb{R})$ to the upper half plane $H$. First let's look at the following subgroup $$ G=\left\{\left(\begin{array}{cc}y^{1/2}&xy^{-1/2}\\0&y^{-1/2}\end{array}\right)\mid x,y\in\Bbb{R},y>0\right\}\le SL_2(\Bbb{R}) $$ consisting of all the upper triangular matrices in $SL_2(\Bbb{R})$ with positive diagonal entries. Call the described matrix $\gamma(x,y)$. We immediately see that $f(\gamma(x,y))=x+iy\in H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.

We also see easily that the stabilizer of $\iota$ in $SL_2(\Bbb{R})$ consists of the matrices of plane rotations w.r.t. the usual basis: $$ Stab_{SL_2}(\iota)=K=\left\{\left(\begin{array}{cc}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right)\mid \theta\in\Bbb{R}\right\}. $$

We also see that $G$ is a closed subgroup of $SL_2(\Bbb{R})$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism. Putting all the pieces together we see that:

• If $C\subseteq H$ is closed, then the set $\tilde{C}:=f^{-1}(C)\cap G$ is closed as a subset of $SL_2(\Bbb{R})$.
• $f^{-1}(C)=\tilde{C}K=\{ck\mid c\in\tilde{C}, k\in K\}$ is a closed subset of $SL_2(\Bbb{R})$ as a product of a closed and a compact subset.
• If $D$ is not a closed subset of $H$ then $f^{-1}(D)\cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^{-1}(D)$ cannot be a closed subset of $SL_2$.
• So $C\subseteq H$ is closed if and only if $f^{-1}(C)\subseteq SL_2(\Bbb{R})$ is closed proving that $f$ is a (topological) quotient map.

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I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.