11
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

The vein of my ideas is to capture on which subsets of the domain the quotient map acts as a covering map. A condition that yields very trivial lifts is the following:

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .