$X$ is a locally connected space, and $f: X \to Y$ is onto where $Y$ has the quotient topology then $Y$ is locally connected. ($X$ is locally connected and $f: X \to Y$ is onto where $Y$ has the quotient topology. Prove that $Y$ is locally connected).
Note-: if quotient map is open map then the converse follows.
Is the converse of the above statement true in general?
I'm thankful for the help.
So you mean (by the comments): If $q: X \to Y$ is a quotient map and $Y$ is locally connected, then so is $X$?
This is false:
$X=\mathbb{Q}$ in the usual topology, let $Y$ be a finite discrete set and write $\mathbb{Q}$ as a finite union of disjoint closed sets (e.g. $[\sqrt{2}, \rightarrow) \cap \mathbb{Q}$, $(\leftarrow, \sqrt{2}]\cap \mathbb{Q}$, for a two point $Y$, and variations thereof for other numbers. If we're lazy we can just take one set $X$ itself, and $Y$ a singleton) and map each closed set to a different point of $Y$ surjectively. Then the result is a quotient map and $Y$ is locally connected (as all discrete spaces are), and $X$ is not.
Or a simple variation: let $Z$ be any non-locally connected space, let $X$ be a disjoint sum topology of copies of $Z$, and again $Y$ a discrete space (one point for each copy of $Z$) and map each summand to a different point in the discrete space. Etc.