Is there a group action which is not free but induces covering map?

Question

Let group G act on topological space X, i.e. there is a group homomorphism from G to the homeomorphism group of X.

As we all know, if this group action is properly discontinuous, then it is free and induces a covering map. Now I wonder about the relationship between being free and inducing covering map.

Either prove the following statement or give a counterexample: If the quotient map of the group action is a covering map, then the group action must be free.

Arguments of my classmates: If there were a fixed point of the group action, then the fiber number of the quotient map couldn't be constant.

But I guess there may be such an example: The isotropy group of each point of X is not trivial, but they are isometric to each other. Then the group action is still not free, but the fiber number of the quotient map is constant.

Unfortunately, I can't give such an example.

Edit: OK, Moishe and Ruy give good counterexamples for my original question.

Now either prove the following statement or give a counterexample: If the group action is effective and its quotient map is a covering map, then this group action must be free.

I'm sorry for my greed, but I guess this is a natural question for any student who majored in math when seeing such counterexamples.

Edit: Thanks for Ruy's new counterexample for the reformulated question (requiring effectiveness). His active brain leaves a deep impression on me.

I came up with the original question when I attempted to show the equivalence of properly discontinuous and inducing covering map. Now I am aware that without further assumptions, inducing covering map doesn't imply properly discontinuous.

I guess a nice "further assumption" that guarantees the equivalence can be found in Munkress section 81, which asks the total space X to be path-connected and locally path-connected.

There are still many questions that can be posed, such as

1. Is there a better "further assumption"? (Actually, I'm not sure what "better" means. Sometimes "weaker" is "better", but it is not always so. We must take the balance between "weak" and "easy to use").
2. Back to the original question, what will happen if we ask the total space to be connected and the group action to be effective?

Any comments are welcome.

Answer 1

Here is an answer for the reformulated question (requiring effectiveness) but which might be considered as exploring a loophole. Further reformulations are welcome, but for now, here we go:

For each $i=1, 2$, consider an action of a group $G_i$ on a space $X_i$ with all of the nice properties: free, properly discontinuous, inducing covering map, etc.

Now let $G=G_1\times G_2$ act on the disjoint union $X=X_1\sqcup X_2$, as follows: for $g=(g_1,g_2)$, and $x\in X$, put $$ g\cdot x= \left\{\matrix{ g_1x, & \text{if } x\in X_1,\cr g_2x, & \text{if } x\in X_2. }\right. $$

This action is not free because any nontrivial group element of the form $(g_1,1)$ acts trivially on the points of $X_2$.

However the action is effective because if $(g_1,g_2)$ acts trivially on the whole of $X$, then each $g_i$ acts trivially on $X_i$, so each $g_i$ is the unit in its respective group because of our assumptions.

Yet, the fact that the two actions are pretty much independent from each other, guarantees that the resulting action induces a covering map.