This is Exercise 1.3.19 from Hatcher's Algebraic Topology
I want to show the following statement:
For a closed orientable surface $M_g$ of genus $g$, there is a path-connected normal (regular) covering space $\tilde X$ of $M_g$ in $\Bbb R^3$ with deck transformation group isomorphic to $\Bbb Z^3$ iff there is an embedding of $M_g$ to the 3-torus $T^3=S^1 \times S^1 \times S^1$ such that the induced map is surjective.
First, it was easy to show that there is an embedding $M_g \to T^3$.
The map $\tilde X \to \Bbb R^3 \to \Bbb R^3/ \Bbb Z^3=T^3$ (the first map is inclusion, and the second is quotient map) induces the embedding of $M_g$ to $T^3$.
However, I am stuck showing the induced homomorphism is surjective.
On the other hand, I have no idea showing the converse. How do I have to proceed? Any hints?
Question related: Abelian Covering Space of Surface of genus $g$
Best Answer
What the author asks in the book is the following:
$(\Leftarrow)$ By the hypothesis on the group of deck transformations it follows that this group is generated by translation of three linearly independent vectors. Let us denote by $\Gamma$ this group. Then as you observed we obtain an embedding $M_g\cong X/\Gamma\rightarrow \Bbb R^3/\Gamma$ from the inclusion $X\subseteq \Bbb R^3$; this is an embedding as $M_g$ is compact. We also have a covering map $\pi:\Bbb R^3\rightarrow \Bbb R^3/\Gamma\cong T^3$; as $\Gamma$ is generated by translations of three linearly independent vectors.
Fix $x_0\in X$. Then we have $\pi^{-1}([x_0])\subseteq X$; as both $\pi$ and $\pi\upharpoonright X$ are covering spaces of degree $3$, and also an induced map $\pi_1(X/\Gamma,[x_0])\rightarrow\pi_1(\Bbb R^3/\Gamma,[x_0])$, and this map is onto. This is because if $[\alpha]\in\pi_1(\Bbb R^3/\Gamma,[x_0])$, we can lift $\alpha$ to a map $\tilde\alpha:[0,1]\rightarrow \Bbb R^3$. Then as $\tilde\alpha(1)\in X$, consider a path $\beta$ in $X$ going from $x_0$ to $\tilde\alpha(1)$, then $[\pi\circ\beta]=[\alpha]$ in $\pi_1(\Bbb R^3/\Gamma,[x_0])$(why?) and $[\pi\circ\beta]\in \pi_1(X/G,[x_0])$.
$(\Rightarrow)$ In this case let $\pi:\Bbb R^3\rightarrow T^3$ be the covering map built using the exponential map. Suppose $M_g\subseteq T^3$. Fix $y_0\in M_g$. Let $G$ be the kernel of the induced map $\pi_1(M_g,y_0)\rightarrow \pi_1(T^3,y_0)$. Let $p:X\rightarrow M_g$ be a covering map such that $p_\#(\pi_1(X,x_0))=G$ for some $x_0\in p^{-1}(y_0)$. Then $i_\#\circ p_\#(\pi_1(X,x_0))=0$; where $i:M_g\rightarrow T^3$ is the inclusion, thus there is a map $\tilde i: X\rightarrow \Bbb R^3$ making the diagram $\require{AMScd}$
\begin{CD} X @>\tilde i>> \Bbb R^3\\@VVpV @VV\pi V\\ M_g @>i>> T^3. \end{CD}
Let us see $\tilde i$ must be inyective. Suppose, to get a contradiction, there are distinct $x,y\in X$ such that $\tilde i(x)=\tilde i(y)$. As $i$ is injective, we must have $x$ and $y$ must lie in the same fiber of $p$. Suppose this fiber is $p^{-1}(y_0)$. Let $\beta$ be a path in $X$ going ffrom $x$ to $y$. Then $[p\circ\beta]\in \pi_1(M_g,y_0)\setminus G$, so that $[i\circ p\circ\beta]=i_\#([p\circ\beta])\neq e$. However, $[i\circ p\circ\beta]=[\pi\circ\tilde i\circ\beta]=\pi_\#([\tilde f\circ\beta])=e$, as $\Bbb R^3$ is simply connected. To prove this for all fibers, show that for all $y\in M_g$ and all $z\in p^{-1}(y)$ we have
$$p_\#(\pi_1(X,z))=\ker(\pi_1(M_g,y)\xrightarrow{i_\#}\pi_1(T^3,y)).$$
As $\pi, p$ are local homeomorphisms and $i$ is an embedding, $\tilde i$ must also be an embedding. Thus we may assume $X\subseteq \Bbb R^3$, and from the diagram we can also assume $p=\pi\upharpoonright X$. In particular if $\Delta_p$ and $\Delta_\pi$ are the groups of deck transformations of $p$ and $\pi$ respectively, we have $$\{D\upharpoonright X:D\in \Delta_\pi\}\subseteq\Delta_p.$$
We claim this inclusion is actually an equality. Suppose not, pick $D\in \Delta_p$ which doesn't extend to an element of $D_\pi$. Then if we pick any $x\in X$ and set $y:=D(x)$, $x$ and $y$ must not be in the same fiber of $\pi$; for otherwise there'd be an element $D'\in \Delta_\pi$ with $D(x)=y$, and as $D\upharpoonright X\in\Delta$ this would imply $D'\upharpoonright X=D$. However, $x$ and $y$ are in the same fiber of $p$, which contradicts the commutativity of the above diagram.