# Path connected quotient implies path connected space

The following is a problem I questioned myself yesterday:

Let $X$ a topological space and define the equivalence relation $\sim$ given by:

$$x\sim y \Leftrightarrow \textrm{ x and y can be connected with a continuous path.}$$

If $X/\sim$ is path connected, is it true that $X$ is path connected?

I thought about this a lot yesterday and still couldn't find any good idea to prove or disprove this.

• I'm not sure if it's true or not, but if it is true then X/~ being path connected implies X/~ is a single point. So if you believe it to be true I'd try showing if [x]~[y] then x~y – N8tron Jun 17 '18 at 21:05

I think the answer is no. Take for instance the subspace $X$ of $\mathbb{R}^2$ defined by $$X = \{(x,\sin 1/x): x > 0\} \cup \{(0,y) : -1 \leq y \leq 1\}.$$ This space has exactly two path components $$A = \{(x,\sin 1/x): x > 0\} \text{ and } B = \{(0,y) : -1 \leq y \leq 1\}.$$ So $X/\sim = \{A,B\}$. Let $\pi: X \to X/\sim$ be the quotient map. Suppose $U$ is an open subset of $X/\sim$ containing $B$. Then $\pi^{-1}(U)$ is an open set of $X$ and, therefore, $\pi^{-1}(U) \cap A \neq \emptyset$. Thus $A \in U$. On the other hand the set $\{A\}$ is open in $X/\sim$, since $\pi^{-1}(\{A\}) = A$. Thus $X/\sim$ is the Sierpiński space, i.e., its topology is $\{X/\sim, \emptyset, \{A\}\}$. This space is path connected.
• It is path connected. Take the path $\gamma: [0,1] \to X/\sim$ given by $\gamma(t) = A$ if $t < 1$ and $\gamma(1) = B$. – João Caminada Jun 17 '18 at 21:17
• I am sorry, but why $\pi^{-1}(U) \cap A \neq \emptyset?$ – user512723 Jun 17 '18 at 21:28
• It is an easy exercise to show that $A$ is dense in $X$. – João Caminada Jun 17 '18 at 21:31