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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.

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The answer to your title question is yes: the continuous image of a connected space is always connected. But the body question is something else! –  Zhen Lin Sep 13 '12 at 16:22
    
I don't know the definition of a submersion, but if $f:X\rightarrow Y$ is a quotient map of topological spaces (surjective, continuous, $f^{-1}(V)$ open iff $V\subseteq Y$ is open) and the fibers of $f$ are connected, then $f$ induces a bijection between the connected components of $X$ and those of $Y$. In particular, if $Y$ is connected, then $X$ is. –  Keenan Kidwell Sep 13 '12 at 16:41
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Dear miosaki: please edit the question so that the title and the body ask the same question. Currently, we have one answer for each and that is rather confusing. –  Mariano Suárez-Alvarez Sep 13 '12 at 21:48
    
thank you all .. –  miosaki Sep 14 '12 at 5:31
    
I have edited my question –  miosaki Sep 14 '12 at 8:54
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up vote 3 down vote accepted

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.

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Not all surjective submersions are fiber bundles. –  Mariano Suárez-Alvarez Sep 13 '12 at 17:58
    
(For example, pick any locally trivial fiber bundle of manifolds and delete any closed subset in its domain without breaking surjectivity) –  Mariano Suárez-Alvarez Sep 13 '12 at 18:02
    
You are answering the question in the title, which is different to that of the body of the question. It would be best if you qualified the initial yes to make this explicit :-) –  Mariano Suárez-Alvarez Sep 13 '12 at 18:08
    
@MarianoSuárez-Alvarez: Obviously not, but it would have been pedantic of me to ignore the title. However, even ignoring the title, my answer is still valid: "Connectedness is preserved by continuous maps". Whether or not $\pi : T \to B$ is a fibre bundle, or simply a continuous surjection, if $T$ is continuous then $B$ is. Nowhere did I discuss the fibres, other than repeating this OP's assumptions. The fibre condition is superfluous. –  Fly by Night Sep 13 '12 at 18:35
    
My first comment merely referred to the fact that your second sentence starts with «You have a fiber bundle...» and that is not quite correct. –  Mariano Suárez-Alvarez Sep 13 '12 at 21:48
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