homeomorphism of a subset of $GL_3(\mathbb{R})$ with $GL_2(\mathbb{R})$ and connectedness Suppose I denote $G_3$ be the set of all $3\times 3$ matrices with positive determinant, and consider the map $\pi:G_3\rightarrow \mathbb{R}^3\setminus\{0\}$ define by $\pi(g)=ge_1$ where $e_1=(1,0,0)$ then i want to know about the set $\{g:ge_1=e_1\}$, is it connected? is this set homeomorphic to $G_2\times\mathbb{R}^2$? well, is the map surjective submersion?
 A: Your set is homeomorphic to $G_2\times \mathbb{R}^2$ (and hence connected).  Notice that $ge_1 = e_1$ emplies that the first column of $g$ is $e_1$, but the other entries are free, modulo the fact that the bottom right $2\times 2$ matrix must have positive determinant.
Then a map from your set to $G_2\times \mathbb{R}^2$ is given by mapping the bottom $2\times 2$ matrix to $G_2$ and the top middle and top right entries to $\mathbb{R}^2$.
In the "usual" topology on $G_3$ (obtained by viewing it as an open subset of $\mathbb{R}^9$), this map is easily seen to be a homeomorphism.
Further, the map is surjective.  To see this, pick any $v\in \mathbb{R}^3-\{0\}$ and extend it to a positively oriented basis of $\mathbb{R}^3$.  Then there is a unique linear transformation mapping map $e_1$ to $v$, and $e_i$ to the other basis vectors you chose.  Then this transformation is in $G_3$ and sends $e_1$ to $v$.
(I've run out of time, but I'll think and post on the submersion question later if no one else does)
