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.
fibres are connected, total space connected $\Rightarrow$ base space is so?
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2The answer to your _title_ question is yes: the continuous image of a connected space is always connected. But the _body_ question is something else! – 2012-09-13
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0I 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. – 2012-09-13
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1Dear 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. – 2012-09-13
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0I have edited my question – 2012-09-14
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0@miosaki, please do eedit the *title* of the question so that it matches what you are asking! Otherwise, it makes it less easy for people to do searches and such things in the future! – 2012-09-15
1 Answers
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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0Not all surjective submersions are fiber bundles. – 2012-09-13
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0(For example, pick any locally trivial fiber bundle of manifolds and delete any closed subset in its domain without breaking surjectivity) – 2012-09-13
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0You 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 :-) – 2012-09-13
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0@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. – 2012-09-13
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0My first comment merely referred to the fact that your second sentence starts with «You have a fiber bundle...» and that is not quite correct. – 2012-09-13
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0As I said: ignoring the title would have been pedantic. – 2012-09-13
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0The title does not imply that one has a fiber bundle either, really: «fibers are connected» in no way requires or implies that the map be a fiber map, and thus the hypothesis «the fibers are connected» can very well apply to a surjective submersion which is not a fiber bundle. If it read instead «the fiber is connected» it would suggest a fiber bundle, though. – 2012-09-13
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0Whatever you say. I'm bored now. – 2012-09-15
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0It matters in that there *is* a difference between a surjective submersion and a fiber map, which is important to take into account (and which is somewhat counter-intuitive) —as clearly the OP is starting in the subject, helping him keep things straight *is* valuable. It does not affect me in the least personally, of course, and as this interchange has taken from me only an insignificant amount of energy and attention, I don't see it as a burden and a bore in the same way you seem to see it. In any case, the question has been edited into involving an actual fiber map :-) – 2012-09-15
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1May I suggest that we could better invest our energies in helping answer other questions rather than discussing minor technical details. – 2012-09-15