When does convergence in quotient space $X / {\sim}$ induce convergence in $X$ Let $X$ be a topological space, $\sim$ an equivalence relation and $q : X \to X/{\sim}$ the quotient map that induces on $X/{\sim}$ the quotient topology.
Assume that $p_n \to p$ in $X / {\sim}$. Under what possibly weak assumptions on $X$ and $\sim$ can we deduce that there is $x_n \in p_n$ and $x \in p$ such that $x_n \to x$ in $X$?

Such a $q$ is called "sequence covering" and is defined in this paper by Siwiec from 1971. 
Proposition 2.4 from that paper might be useful: an almost open map defined on a first countable space is sequence covering.
Here $f: X \rightarrow Y$ is almost open, iff for all $y \in Y$ there is some $x$ with $f(x) = y$ such that, for every open $O$ in $X$ with $x \in O$, $y \in \operatorname{int}(f[O])$. Any open map satisfies it.
A simple example (2.6) shows things can go wrong easily: Let $X$ be the set $\{\frac{1}{n}: n \in \mathbb{N}^+\} \cup \{0\} \cup \{2+\frac{1}{n}: n \in \mathbb{N}^+\} \cup \{2\}$ as a subset of the reals, and let $\sim$ be the equivalence relation that only identifies $0$ and $2$ and leaves the rest alone.
Then both $X$ and its quotient are metric compact, and $q$ is perfect (closed map, all inverse images of points are finite (at most size 2) hence compact). But $q$ is not sequence covering, as the sequence $[1],[2+\frac{1}{2}],[\frac{1}{3}],[2+\frac{1}{4}],[\frac{1}{5}],\ldots$ converges to $[0] = [2]$  but there is only one choice of lifting the sequence and this does not converge. So even only one non-trivial class between extremely nice spaces can fail this property.
Googling "sequence covering maps" will give you some more papers with generalisations of 2.4. The paper I linked to contains some more counterexamples where we cannot drop some assumption.
