Group actions by inverse limits Let $G$ be a group that is the inverse limit of an inverse system $(G_{i},f_{i})$. Assume that each $f_{i}$ is surjective. If $G$ acts effectively on a topological space $X$, must there be effective actions of the $G_{i}$ on $X$? Note that by an effective action of $G$ I mean that if $g,h\in G$ are distinct elements then there is an $x\in X$ such that $gx\neq hx$.
I am particularly interested in the case where the $G_{i}$ are discrete and the action of $G$ is continuous.
Any help would be appreciated.
Edit: I realized I am really interested in the following question. If a group $G_{1}$ acts effectively on a space $X$ and $f:G_{1}\rightarrow G_{2}$ is a surjective group homomorphism must there be an effective action of $G_{2}$ on $X$?
 A: No.
In particular, you can have $G_1$ act effectively on a space $X$, even though some quotient $G_2$ does not have any effective actions on $X$. For example, look at the usual action of $\mathbb{Z}$ on $\mathbb{R}$ by translations. Now $\mathbb{Z}$ surjects onto, say, $\mathbb{Z} / 3$, but there is no effective $\mathbb{Z} / 3$ action on $\mathbb{R}$.
To see why, let's start with $\text{Homeo}(\mathbb{R})$, the group of homeomorphisms from $\mathbb{R}$ to itself. It has an index $2$ subgroup $\text{Homeo}^+(\mathbb{R})$ of order preserving homeomorphisms.
Now if $G$ acts effectively on $\mathbb{R}$, we must have an injection $G \hookrightarrow \text{Homeo}(\mathbb{R})$. Then $G \cap \text{Homeo}^+(\mathbb{R})$ is a normal subgroup of $G$ which must have index $1$ or $2$.
We know $G = \mathbb{Z} / 3$ has no index $2$ subgroups, which means
$\mathbb{Z} / 3 \cap \text{Homeo}^+(\mathbb{R}) = \mathbb{Z}/3$. That is,
$\mathbb{Z} / 3 \leq \text{Homeo}^+(\mathbb{R})$.
But it's "well known" that the countable subgroups of  $\text{Homeo}^+(\mathbb{R})$ are exactly the countable groups admitting a compatible left-order. See for instance, Theorem $2.23$ here.
Since $\mathbb{Z} / 3$ has no compatible left-order (indeed, every left-orderable group is torsion free) it cannot act effectively on $\mathbb{R}$.
This same argument will work for basically any group acting on $\mathbb{R}$, as long as it has a quotient that has torsion and doesn't have an index $2$ normal subgroup.

I hope this helps ^_^
