Fundamental Group of a Quotient under Group Action Let $X$ be a simply connected topological space (therefore with trivial fundamental group) and $G$ a group which acts on $X$ freely. Recently I read that for fundamental group of $X/G$ holds the equation $\pi_1(X/G) =G$.
My questions are :


*

*how to see it (any sources where it explained detailed; I suppose that it was deduced by a category equivalence $(Cov/X) \leftrightarrow ?$ but I forgot what was the second category).

*Do there exist weaker conditions (explicitely if $G-$action not freely or if $X$ not simply connected) for which the equation $\pi_1(X/G) =G$ stil holds?
 A: You need to assume more about the action than just freeness; otherwise this statement is not true. For example, $\mathbb R$ acts freely on itself by translation, and $\mathbb R$ is simply connected, but $\mathbb R/\mathbb R$ is a single point, and its fundamental group is not isomorphic to $\mathbb R$.
In order to get the conclusion you want, you also have to assume that the action satisfies the following condition:

Every $x \in X$ has a neighborhood $U$ such that for all $g\in G$, $gU \cap U = \emptyset$ unless $g$ is the identity.

(Note that this implies, in particular, that $G$ acts freely.) Some authors call an action satisfying this condition a "properly discontinuous action"; but I don't like that term because different authors use it with inequivalent definitions, and also because it leads to oxymoronic expressions like "a continuous properly discontinuous action." Allen Hatcher in his book Algebraic Topology introduced the term covering space action for a continuous group action satisfying the condition above, and I've adopted that term in my books.
The basic fact is that if the action of $G$ on $X$ is a covering space action, then the quotient map $X\to X/G$ is a covering map. If in addition $X$ is simply connected, then $\pi_1(X/G)$ is isomorphic to $G$. 
For more details, see this MSE answer; and for even more, see Chapters 11 and 12 in my book Introduction to Topological Manifolds.
A: 
how to see it (any sources where it explained detailed; I suppose that it was deduced by a category equivalence $(Cov/X) \leftrightarrow ?$ but I forgot what was the second category).

If $X$ is sufficiently nice, there is a Galois connection (which is a contravariant equivalence of categories) between covers of $X$, and subgroups of $\pi_1(X).$ If $X$ is simply connected with a proper discontinuous group action $G$, then we could instead describe the equivalence between subgroups of $G$ and covers of $X/G.$

Do there exist weaker conditions (explicitely if $G-$action not freely or if $X$ not simply connected) for which the equation $\pi_1(X/G) =G$ stil holds?

I don't know about weaker conditions. Usually we require stronger conditions: the group action must act properly discontinuously: the map $G\times X\to X\times X$ is proper, or every point has a neighborhood disjoint from the orbit of that point.
