Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ f: X \to Y $ an application between two topological spaces $ \mathbb{X} $ and $ \mathbb{Y} $ both Hausdorff. We know that if $ Y \in\mathbb{Y} $ is connected and $ f $ is continuous then $ f^{-1}(Y)$ is connected.

But if $ A $ is not connected or if $ f $ is not continuous then $f^{-1}(Y)$ is not necessarily connected.

And in this case it is customary to work with the so-called maximal connected component.

Question 1: What is maximal connected component?

Add after Azarel Answer the felowing question.

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

share|cite|improve this question
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. – Joe Johnson 126 Mar 9 '12 at 17:21
You are wrong the image of something connected is connected and not the preimage. Map two points to one point. – Mar 9 '12 at 17:22
up vote 1 down vote accepted

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.

share|cite|improve this answer
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 $1_A^{-1}(1)$? – MathOverview Mar 9 '12 at 18:14
@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. – Brian M. Scott Mar 9 '12 at 18:16
The maximal connected component need not be a set connected? This is what you mean? – MathOverview Mar 9 '12 at 18:23
@Elias: No, a maximal connected component is by definition connected. – Brian M. Scott Mar 9 '12 at 18:29

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.

share|cite|improve this answer
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). – Adam Saltz Mar 9 '12 at 17:29
I don't understend you answer. For exemple, what is the maximal connected component of $f^{-1}(y)$ whit $y\in\mathbb{Y}$? – MathOverview Mar 9 '12 at 17:33
@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 $\bigcup\mathscr{C}_x$. – Brian M. Scott Mar 9 '12 at 17:46
@BrianM.Scott, thanks very much. I wasn't familiar with that. – Adam Saltz Mar 9 '12 at 17:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.