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Let $ f: X \to Y $ be an application between two topological spaces $ \mathbb{X} $ and $ \mathbb{Y} $ both Hausdorff.

The set $f^{-1}(Y)$ is not necessarily connected even if $Y\subset \mathbb{Y}$ is connected. And in this case it is customary to work with the so-called maximal connected component.

Question 1: What is a maximal connected component of $f^{-1}(Y)$?

And after Azarel's Answer, I have the following question:

Question 2: What is the maximal connected component of $f^{-1}(y)$ with $y\in\mathbb{Y}$?

  • 1
    It is not true that $f^{-1}(Y)$ is connected if $Y$ is. Take the constant map $[0,1]\cup [2,3]$ to $1$. The point $1$ is connected, but $f^{-1}(1)$ is not.2012-03-09
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    You are wrong the image of something connected is connected and not the preimage. Map two points to one point.2012-03-09

2 Answers 2

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You have a function $f:\Bbb X\to \Bbb Y$ and a point $y\in\Bbb Y$. Let $A=f^{-1}[\{y\}]$. A maximal connected component of $A$ is a connected subset $C$ of $A$ with the property that if $C\subseteq S\subseteq A$, and $S$ is connected, then $C=S$. In other words, no connected subset of $A$ properly contains $C$.

You can find the maximal connected components of $A$ in the same way that you find the maximal connected components of a whole space, as given in azarel’s answer. For each $x\in A$ let $$\mathscr{C}_x=\{C\subseteq A:x\in C\text{ and }C\text{ is connected}\}\;,$$ and let $$C_x=\bigcup\mathscr{C}_x=\bigcup\{C:C\in\mathscr{C}_x\}=\bigcup_{C\in\mathscr{C}_x}C\;.$$ Then each $C_x$ is a maximal connected component of $A$, and every maximal connected component of $A$ is obtained in this way.

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    Let's $A_1\cap A_2=\emptyset$, $A_2\cap A_3=\emptyset$ and $A_3\cap A_1=\emptyset $. Let's $A=A_1\cup A_2 \cup A_3$ and $1_A : \mathbb{X}\to\mathbb{Y}$ the caracterisc function of $A$. What is the maximal connected component of $1_A^{-1}(1)$?2012-03-09
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    @Elias: They are the connected components of $A_1$, $A_2$, and $A_3$. If $A_1$, $A_2$, and $A_3$ are connected, then **they** are the connected components of that inverse image.2012-03-09
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    The maximal connected component need not be a set connected? This is what you mean?2012-03-09
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    @Elias: No, a maximal connected component is **by definition** connected.2012-03-09
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For every $x\in X$ consider the set $\mathcal C_x=\{C: x\in C, \ \text{and }\ C\ \text{is connected}\}$. It is easy to see that $\bigcup \mathcal C_x$ is connected. Moroever, it is maximal in the sense that it is the biggest connected subset containing $x$. The sets $\bigcup\mathcal C_x\ (x\in X)$ are the maximal connected components of the space.

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    Your notation is a bit confusing to me. $\mathcal{C}_x$ is a set of subsets. The union of all subsets in $C_x$ is written $\bigcup_{U \in \mathcal{C}_x} U$ or somesuch. And your last sentence should be something like "The sets $\bigcup_{U \in \mathcal{C}_x} U (x \in X)$ are the maximal connected components of the space." The maximal component containing $x$ is a subset of $X$ rather than a set of subsets. Apologies if I'm just showing off my ignorance about notation (or topology).2012-03-09
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    I don't understend you answer. For exemple, what is the maximal connected component of $f^{-1}(y)$ whit $y\in\mathbb{Y}$?2012-03-09
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    @IAmBrianDawkins: $\bigcup\mathscr{C}_x$ is a perfectly standard notation for the union of the members of the collection $\mathscr{C}_x$. The $\mathscr{C}_x$ in the last sentence should of course be $\bigcup\mathscr{C}_x$.2012-03-09
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    @BrianM.Scott, thanks very much. I wasn't familiar with that.2012-03-09