Covering map corresponding to surjection Let $X$ be a compact manifold with $\pi_1(W)=S_3$, $T$ be an $n$-dimensional complex torus. Denote the universal cover of $W$ by $\tilde{W}$.  Let $S_3$ act on $(T\times T\times T) \times \tilde{W}$ by permutation on the first part and by deck transformation on the second. Now we consider $Y:=((T\times T\times T) \times \tilde{W})/S_3$. 
I feel confused with the following:
We can take $\tilde{Y}$ as the $6$-fold covering of $Y$ corresponding to the surjection
$\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3\to S_3$.
What I know is a covering map $p:\tilde{Y}\to Y$ induces an injection $p^*:\pi_1(\tilde{Y})\to\pi_1(Y)$, I don't understand what exactly the $6$-fold covering corresponding to the surjection from is.
Also is it obvious that $\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3$?
 A: If $X$ is a $G$-space and $G$ acts properly discontinuously and freely on a path-connected space $Y$, then we have the fiber bundle $X \to (X \times Y)/G \to Y/G$ on which running the homotopy long exact sequence gives $$\cdots \to \pi_{n+1}(Y/G) \to \pi_n X \to \pi_n ((X \times Y)/G) \to \pi_n(Y/G) \to \cdots $$ If $x \in X$ is a fixed point under the action of $G$, then given the quotient map $q : X \times Y \to (X \times Y)/G$, $q|_{x \times Y}$ is the covering projection $Y \to Y/G$. This gives a canonical copy of $Y/G$ sitting inside $(X \times Y)/G$, and the inclusion map $s : Y/G \hookrightarrow (X \times Y)/G$ constitutes a section of the said fiber bundle.
Then $s_* : \pi_n(Y/G) \to \pi_n((X \times Y)/G)$ is a right-inverse to $\pi_n((X \times Y)/G) \to \pi_n(Y/G)$ for every $n$, forcing these maps to be surjection, and the long exact sequence above to break into short exact pieces $0 \to \pi_n(X) \to \pi_n((X \times Y)/G) \to \pi_n(Y/G) \to 0$ for every $n$. If $n = 1$, then we have a split short exact sequence 
$$0 \to \pi_1(X) \to \pi_1((X \times Y)/G) \to \pi_1(Y/G) \to 0$$
This establishes $\pi_1((X \times Y)/G)$ as a semidirect product $\pi_1(X) \rtimes \pi_1(Y/G)$.
In this particular case, take $X = T \times T \times T$ and $Y = \widetilde{W}$, letting $G = S_3$ act on the former by permuting the factors and on the latter by deck transformations. This $S_3$-action has a fixed point in $X$, just choose any point $(x, x, x) \in T \times T \times T$ in the diagonal. Then $\pi_1((T \times T \times T \times \widetilde{W})/S_3)$ is isomorphic to $(\Bbb Z^{2n})^3 \rtimes S_3$, as required. The semidirect product is taken with respect to the homomorphism $S_3 \to \text{Aut}((\Bbb Z^{2n})^3)$ permuting the three coordinate $\Bbb Z^{2n}$s, induced from the $S_3$-action on $T \times T \times T$ as described before.
