Is a path connected version of this question Equivalence relation on topological space such that each equivalence class and the quotient space is connected true ? That is let $X$ be a topological space and $\sim$ be an equivalence relation on $X$ such that each equivalence class is path connected (as a subspace of $X$) and the quotient space $X/\sim$ is also path connected. Then is $X$ path connected ?

Definitely $X$ is connected. If $X$ is not path connected in general, then does some good additional condition forces $X$ to be path connected ?

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    $\begingroup$ One special case at least is true: If $X = P \times Q$ where $P$ and $Q$ are path-connected and $\sim$ identifies all $\{x\} \times Q$ to a point, making the projection the quotient map, essentially. $\endgroup$ – Henno Brandsma Jun 17 '18 at 5:50

The answer is "no"'. Let $S \subset \mathbb{R}^2$ be the closed topologist's sine curve which is the union of $$\Sigma = \lbrace (t,\sin(1/t)) \mid t \in (0,1] \rbrace \subset \mathbb{R}^2 $$ and $$J = \lbrace 0 \rbrace \times [-1,1] \subset \mathbb{R}^2.$$

Its path components are $\Sigma$ and $J$. Define $x \sim y$ if $x = y$ or $x,y \in J$. The equivalence classes are the singletons $\lbrace x \rbrace$ with $x \in \Sigma$ and the set $J$, so they are all path connected. The quotient space $S' = S / \sim$ is homeomorphic to $[0,1]$ which is path connected.

An alternative way to define $\sim$ is to consider the projection $p : S \to I = [0,1], p(x_1,x_2) = x_1$. Then $x \sim y$ iff $p(x) = p(y)$. This gives you an explicit homeomorphism $p' : S' \to I, p'([x]) = p(x)$.

I do not know any condition assuring that $X$ is path connected.


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