Condition for a quotient map to have compact image. Let $q:X\to X_{/\sim}$ be a quotient map for some relation $\sim$ on $X$. If there is a compact subspace $A\subset X$ such that every element of $X$ is in relation with some element of $A$, then $X_{/\sim}$ is compact, simply because the condition can be rewritten $q(A)=X_{/\sim}$.

I am wondering if the converse is true in the general case, that is if $X_{/\sim}$ is compact, can we find a compact subspace $A\subset X$ such that any element of $X$ is in relation with some element of $A$?  If not are there some assumptions that we can put on $X$ and $\sim$ to make the converse true?

For example even if $X$ is a topological manifold and $\sim$ is generated by a covering space action, I am not sure the converse holds.
Motivation: When $\Gamma$ is a group acting on a manifold $X$ by covering space action, it is sometimes pretty clear that there is no compact subspace $A\subset X$ such that "$A/\Gamma=X/\Gamma$" (for example $X=\Bbb R^2$ and $\Gamma=\langle (x,y)\mapsto (x+1,y)\rangle$). In that case we want to conclude that $X/\Gamma$ is not compact, but what is the "most general case" in which we have such a conclusion?

Edit: This question has been asked on mathoverflow here. The answer is negative in general, but positive if $X$ is second countable and locally compact, and $X_{/∼}$ is first countable.


This is only a partial answer. If $Y = X/\sim$ is compact Hausdorff and $q$ is a local homeomorphism (see e.g. https://en.wikipedia.org/wiki/Local_homeomorphism), then the answer is "yes". This covers the case when $\sim$ is generated by a covering space action (then $Y$ is a compact manifold and $q$ is a covering projection).
For each $y \in Y$ choose $x(y) \in q^{-1}(y)$ and an open neighborhood $U_y$ of $x(y)$ such that $q$ maps $U_y$ homeomorphically onto an open $V_y \subset Y$. Let $q_y : U_y \to V_y$ denote this homeomorphism. Choose open neighborhoods $W_y$ of $y$ such that $C_y = \overline{W}_y \subset V_y$. Since $Y$ is compact, finitely many $W_{y_i}$ cover $Y$. Hence also finitely many $C_{y_i}$ cover $Y$. Each $K_i = q_{y_i}^{-1}(C_{y_i})$ is compact (and Hausdorff), hence $A = \bigcup K_i$ is compact (but not necessarily Hausdorff if $X$ is not Hausdorff). Obviously $q(A) = Y$.
