fibres are connected, total space connected $\Rightarrow$ base space is so?

Question

suppose $G_n$ be the group of $n\times n$ non-singular matrcies with detereminant $>0$, suppose we have a map $f:G_n\rightarrow \mathbb{R}^n\setminus \{0\}$ such that $A\mapsto Ae_1$, which is surjective, continuous. where $e_1=(1,0,0,\dots,0)$ the fiber corresponding to $e_1$ is homeomorphic to $G_{(n-1)}\times \mathbb{R}^{n-1}$ which are connected, I want to know $G_n$ is connected?given that $G_{n-1}$ is connected.

Answer 1

N.B: The original question has been completely re-written since this answer was posted. Check the page history to see the original question.

In short: yes! You have a fibre bundle $\pi : T \twoheadrightarrow B$, where $\pi$ is a continuous surjection and both the fibres $\pi^{-1}(b)$ and the total space $T$ are connected. Connectedness is preserved by continuous maps, so if $T$ is connected then $B$ must be connected too.

We can show what the base space is connected as follows:

Assume that the base space $B$ is disconnected. Then there exist open subsets $X,Y \subset B$ such that $X \cup Y = B$ while $X \cap Y = \emptyset$. Since $\pi$ is a continuous surjection it follows $\pi^{-1}(X)$ and $\pi^{-1}(Y)$ are open in $T$ and that $\pi^{-1}(X), \pi^{-1}(Y) \subset T$ such that $\pi^{-1}(X) \cup \pi^{-1}(Y) = T$ and $\pi^{-1}(X) \cap \pi^{-1}(Y) = \emptyset$. It follows that $T$ is also disconnected, which is a contradiction.