# Sufficient Conditions for Existence of a lift given a quotient

Let $Y$ be a topological space with some equivalence relation $\sim$ and let $q: Y\to Y/\sim$ be the quotient map.

Question: Given a topological space $X$, and a map $f: X\to Y/\sim$, are there any nice sufficient conditions to guarantee the existence of a lift $g: X\to Y$ such that $q \circ g = f$? What does one generally need to ask for to get such existence?

Does it simplify matters if $Y/\sim$ is finite? Unfortunately this isn't a topic I'm familiar with, any references welcome!

Edit: I've realized if $q$ admits a continuous right-inverse then this problem would be easy, so any conditions to guarantee this would suffice as an answer.

This is an extremely interesting question, with complexity due to the extreme flexibility of the quotient maps $q: X \rightarrow X/\sim$ compared to covering spaces $p:\tilde Y \rightarrow Y$. One important difference is the way in which construction of covering space gives a relation on the 1-structure, the map $p$ adds structure to $\tilde Y$. In contrast the map $q$ can delete all structure of $X$ or add as much sturcture as one desires.
Let $A \subset X$ defined as $A =\{x \in X; \forall y \in X \,\, !(x \sim Y)\}$. This is the exact subset of $X$ such that $q|_A (x) = x$. Now if a function $f:Y \rightarrow X/ \sim$ is such that $f(Y) \subset q(A)$ we may define the lift $\tilde f:Y \rightarrow X$ by regarding the image of $f$ as a subset of $X$.
We immediately run into problems if we wish to extend the image of $f$ onto a point of identification of $X / \sim$. For one, the preimage by $q$ of $f(Y)$ can become quite disconnected, destroying any hopes of continuity.
If we restrict ourselves to subsets $B \subset X/ \sim$ such that for each $x \in B$ there is a neighborhood $U$ for which $q^{-1}(U)$ is disjoint union of open sets with at least one homeomorphic to $U$, here $q$ behaves like a covering space in this neighborhood and it follows $f(Y) \subset B$, $f$ has a lift iff $f_*(\pi_1(Y)) \subset q_*(\pi_1(X))$ just as if $q$ were a covering map.
Another way of viewing I suppose would be to restrict the relation for which the quotient is a covering space. An example of this of course would be for a group action $G$ on $X$ and the relation $x \sim y \iff orb(x) = orb(y)$ where $orb(x)$ denotes the orbit of $x$ by the action of $G$. Then we are guaranteed by a classic result that the map $q:X \rightarrow X/ G$ is a covering map.