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This statement is used without explanation in the proof of a Corollary to Proposition 4.3.5, pages 210-211 of the book "Algèbre et théories galoisiennes" by R. and A. Douady, second edition, Cassini, Paris, 2005.

It is well known that the claim is true if the base of the covering space is connected and locally connected. Any ideas for a proof without assuming the base to be locally connected?

EDIT: The definition of covering space in that book is just a fiber bundle with discrete fibers.

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  • $\begingroup$ I must be misunderstanding the statement. It seems like the answer is obviously no for a trivial covering. $\endgroup$
    – Mark Grant
    Commented Oct 3, 2015 at 16:00
  • $\begingroup$ @Mark Grant: consider a trivial cover $\pi\colon Y=X \times F\to X$, where $X$ is connected and $F$ is a discrete space. The connected components of $Y$ are: $C_i=X\times\{i\}$, with $i\in F$. The fibers on the other hand are $Y_x=\{x\}\times F$, with $x\in X$. Therefore, one has $Y_x \cap C_i =(x,i)$. $\endgroup$
    – user6319
    Commented Oct 3, 2015 at 16:35
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    $\begingroup$ The question should be rephrased to eliminate the ambiguity of what the word "its" refers to. (Does it refer to "connected component" or to "covering space"?) $\endgroup$ Commented Oct 3, 2015 at 19:39
  • $\begingroup$ The answer to the question is "yes" if the covering map is assumed to be an overlay in the sense of R.H. Fox (regardless of local connectivity). $\endgroup$ Commented Oct 4, 2015 at 5:24
  • $\begingroup$ @Allen Hatcher: thanks for pointing out the ambiguity. I have changed the question. $\endgroup$
    – user6319
    Commented Oct 4, 2015 at 7:22

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It seems that a solenoid yields a counterexample. A solenoid can be written as the quotient space of the product $[0,1]\times \{0,1\}^\omega$ by a map that identifies each point $(0,x)$ with the point $(1,f(x))$ where $f:\{0,1\}^\omega\to\{0,1\}^\omega$ is a homeomorphism that has a dense orbit (this property of $f$ ensures that the solenoid is connected). It is easy to see that the solenoid can be covered by the space $\mathbb R\times\{0,1\}^\omega$ and the image of each connected component (which is homeomorphic to $\mathbb R$) under the covering map is of the first Baire category in the base.

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