The problem says this:

Let $\sim$ be an equivalence relation in $X$ and consider the quotient space $X/\sim$. Prove that:

  1. $X$ connected $\Rightarrow$ $X/\sim$ connected.
  2. If $X/\sim$ is connected and every equivalence class $[x] = \left \{ y\in X : y \sim x \right \}$ is connected then $X$ is connected.

Proof 1) is easy and I know how to do it. But I'm stuck in the proof 2). That's what I have tried:

Let's suppose $X$ isn't connected. Then, $X=U\sqcup V$ with $U$ and $V$ open.

Let $\pi : X\rightarrow X/\sim$ be the application that sends every $x\in X$ into his class. I want to see that $\pi (U)$ and $\pi(V)$ are open and that $\pi (U)\cap\pi (V)= \varnothing$ to conclude that $X/\sim$ isn't connected (because $\pi$ is surjective). But that is absurd, so $X$ must be connected.

I know how to prove that $\pi (U)\cap\pi (V)= \varnothing$, but I don't know how to show that $\pi (U)$ and $\pi(V)$ are open. Any advice?



It's not necessarily the case that $\pi(U)$ and $\pi(V)$ are open, since not all quotient maps are open maps. What's important is that $\pi(U)$ is open if $U$ is saturated, i.e. $U$ is the complete inverse image of a subset of $X/\sim$. So once we show $U,V$ must be saturated, we'll be done, by the argument you gave.

Suppose $U$ is not saturated. Say the inverse image of $[x]$ is only partially contained in $U$. [Note if the inverse image of each point is either completely contained in $U$ or not contained in $U$ at all, then $U$ would be saturated.] Then $([x]\cap U)\sqcup([x]\cap V)$ is a nontrivial decomposition of $[x]$. Since $U,V$ are both open in $X$, $[x]\cap U, [x]\cap V$ are both open in $[x]$. This contradicts $[x]$ being connected. Hence, $U$ is saturated and similarly $V$ is saturated.

  • $\begingroup$ In second paragraph, first line, it should be $[x]$, right? $\endgroup$ – Thomas Shelby May 18 at 9:58
  • 1
    $\begingroup$ @ThomasShelby yep, thanks! $\endgroup$ – mathworker21 May 18 at 10:08

$X$ is connected iff every continuous function $f: X \to \{0,1\}$ is constant where $\{0,1\}$ is endowed with the discrete topology.

Let $f: X \to\{0,1\}$ be an arbitrary continuous function. Since every equivalence class $[x] = \left \{ y\in X : y \sim x \right \}$ is connected, $f$ restricted to $[x]$ is constant. Then $f$ induces a function $$\displaystyle\tilde f:X/\sim\,\to\{0,1\}$$ such that the following diagram commutes: $$\require{AMScd} \def\diaguparrow#1{\smash{ \raise.6em\rlap{\scriptstyle #1} \lower.3em{\mathord{\diagup}} \raise.52em{\!\mathord{\nearrow}} }} \begin{CD} && X/\sim\\ & \diaguparrow{\pi} @VV \\\tilde f V \\ X @>> f> \{0,1\} \end{CD}$$ that is, $f=\tilde f\circ \pi$. Since $f$ is continuous, $\tilde f$ is continuous. As $X/\sim $ is connected, $\tilde f$ is constant by the above theorem. Hence $f$ is constant.

  • $\begingroup$ Very nice proof! $\endgroup$ – Paul Frost May 24 at 10:22
  • $\begingroup$ @PaulFrost Thank you!! $\endgroup$ – Thomas Shelby May 24 at 10:26

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