# Connectedness in the Quotient Space

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?

Thanks!

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.

• In second paragraph, first line, it should be $[x]$, right? – Thomas Shelby May 18 at 9:58
• @ThomasShelby yep, thanks! – 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: $$\require{AMScd} \def\diaguparrow#1{\smash{ \raise.6em\rlap{#1} \lower.3em{\mathord{\diagup}} \raise.52em{\!\mathord{\nearrow}} }} \begin{CD} && X/\sim\\ & \diaguparrow{\pi} @VV \\\tilde f V \\ X @>> f> \{0,1\} \end{CD}$$ 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.

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