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This question already has an answer here:

Let $p\colon X \to Y$ be a quotient map. Show that if each set $p^{-1}({y})$ is connected, and if $Y$ is connected, then $X$ is connected.


I am totally stuck on this problem. Can I get some help how to tackle this problem

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marked as duplicate by azimut, Henno Brandsma, Shobhit, Macavity, nbubis Nov 11 '13 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ To help you with this problem, answerers need to know two things. (1) In what context did you encounter the problem (what class, what textbook)? If you say this in a calculus course, the answers will be different than if it is a warm up problem in a graduate course. (2) What have you already tried? Let us know what you know about the problem and what progress you have made towards solving it. $\endgroup$ – Carl Mummert Jul 1 '13 at 11:21
  • $\begingroup$ I was reading munkres topology book and tried to solve the exercises.it is a problem from munkres $\endgroup$ – gumti Jul 1 '13 at 11:27
  • $\begingroup$ Thanks - which problem is it? We are happy to take questions from textbooks on this site. The additional context is helpful for people who want to write an answer that is as useful as possible - it lets them know what sorts of techniques you have seen and what level the answers should be written at. $\endgroup$ – Carl Mummert Jul 1 '13 at 11:32
  • $\begingroup$ section-23 and problem 11 from munkres $\endgroup$ – gumti Jul 1 '13 at 11:34
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HINT: Suppose that $X$ is not connected. Then $X=U\cup V$, where $U\ne\varnothing\ne V$, $U\cap V=\varnothing$, and $U$ and $V$ are both open (and hence also both closed).

  • Show that if $f[U]\cap f[V]=\varnothing$, then $Y$ is not connected.
  • If $f[U]\cap f[V]\ne\varnothing$, there is a $y\in f[U]\cap f[V]$; show that $f^{-1}[\{y\}]$ is not connected.
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  • $\begingroup$ will you please explain.still I could not get it.thanks a lot for your help. $\endgroup$ – gumti Jul 1 '13 at 11:35
  • $\begingroup$ @gumti: You may need to think about it a bit more, but those hints actually do most of the hard work for you already. Here’s a further hint, though: if $f[U]\cap f[V]=\varnothing$, then $f^{-1}[f[U]]=U$ and $f^{-1}[f[V]]=V$. (Why?) Use this to show that $f[U]$ and $f[V]$ are open in $Y$. $\endgroup$ – Brian M. Scott Jul 1 '13 at 11:39

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