# 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 $$\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3$$?

• Consider the fibration $T \times T \times T \to (T \times T \times T \times \widetilde{W})/S_3 = Y \stackrel{\pi}{\to} W$. Note that diagonal $\Delta \subset T \times T \times T$ is fixed under this $S_3$-action, so pick any point $(x, x, x) \in T \times T \times T$, and note that the quotient $T \times T \times T \times \widetilde{W} \to Y$ restricted to $(x, x, x) \times \widetilde{W}$ is the covering projection $\widetilde{W} \to W$, so this gives a section $W \to Y$ to $\pi$ by including this copy of $W$ in $Y$. Run the homotopy exact sequence to get a split short sequence of $\pi_1$'s – Balarka Sen Dec 29 '18 at 16:09

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.

• Can you explain why $(X\times Y)/G\to Y/G$ is a fiber bundle with fiber $X$? – 6666 Dec 30 '18 at 13:01
• The projection map is obtained from $X \times Y \to Y/G$ which first projects to $Y$, then projects to $Y/G$ by the covering map. This is a fiber bundle with fiber $X \times G$. By universal property, this map factors through $(X \times Y)/G \to Y/G$ which has fibers $(X \times G)/G = X$; that it is locally trivial follows from the properly discontinuous action of $G$. It's a good exercise to check all the details here. – Balarka Sen Dec 30 '18 at 13:20
• $\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3\to S_3$ in your sense is corresponding to the induced map of $(X\times Y)/G\to Y/G$, however, why is this $6$-fold? – 6666 Jan 1 at 17:01
• @6666 Your questions do not make sense to me. Whenever you have a space $Y$, equipped with a surjective homomorphism $\rho: \pi_1 Y \to G$, there is a unique covering space $Y' \to Y$ so that $\text{ker}(\rho) = \pi_1 Y'$, and furnishes us with an isomorphism $\text{Deck}(Y') \cong G$. This is the main theorem of covering space theory, see Hatcher ch 1.3. One explicit formula is that if $\tilde Y$ is the universal cover, then $\tilde Y \times_{\pi_1 Y} G \cong Y'$, where we use $\rho$ to give an action of $\pi_1 Y$ on $G$. This is all one is expected to be able to say about $Y'$. – user98602 Jan 1 at 17:18
• @6666 All they are doing is defining $Y'$ (what they call $\tilde Y$, but I reserve that notation for universal cover), and they are defining it by the property I quote. – user98602 Jan 1 at 17:29