algebraic-topologycovering-spacesgeneral-topology
I'm currently working on the following exercise from Hatcher's Algebraic Topology:
Let $p:\tilde{X} \rightarrow X$ be a covering space with $p^{-1}(x)$ finite and non-empty for all $x \in X$. Show that $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.
I've managed to prove everything except showing that if $\tilde X$ is compact, then $X$ is compact as well. Any hint/help will be extremely useful. Thanks.
Let's focus on the surface $\Sigma_3$ of genus $3$ just to be concrete.
Here's the geometric idea:
The basic idea is that the lattice $\mathbb{Z}^3$ has to act by deck transformations on the covering space. So we have to have three curves on $\Sigma_3$ which "unwind" to the grid pictured on the right in the covering space. From there it's not a stretch to the cut-and-paste operation in the picture.
Formally - without exhibiting explicit formulae - I think of this two ways.
Three disjoint essential simple closed curves on $\Sigma_3$, $\alpha_1,\alpha_2,\alpha_3$, generate a copy of $F_3$, the free group on three generators, in $\pi_1\Sigma_3$. Associated to $F_3$ is a covering $\widehat{\Sigma_3}\to \Sigma_3$, with $F_3$ acting by deck transformations on $\widehat{\Sigma_3}$ so that $\widehat{\Sigma_3}/F_3 = \Sigma_3$. Since $Z(F_3)<F_3$, we have a quotient cover $\widehat{\Sigma_3}/Z(F_3)\to \Sigma_3$, with the deck group $F_3/Z(F_3)\cong \mathbb{Z}^3$. Geometrically, take the "asterisk" fundamental domain (the one you get by cutting along the red curves) and glue it so it looks like the Cayley graph of $F_3$. Then quotient by the action of the commutators of the generators.
The fundamental group $\pi_1(\Sigma_3)$ projects onto the first homology group $H_1(\Sigma_3;\mathbb{Z})$. Associated to $H_1(\Sigma_3;\mathbb{Z})$ is an abelian cover $A(\Sigma_3)\to \Sigma_3$. The images of $\alpha_1,\alpha_2,\alpha_3$ generate a copy of $\mathbb{Z}^3$ in $H_1$, which corresponds to a $\mathbb{Z}^3$-covering space of $\Sigma_3$.
I think there should be a nice picture of a fundamental domain of this cover in $\mathbb{H}^2$ by stringing together right-angled dodecagons, but I don't quite see it.
Take distinct $y_1 \neq y_2$ in $X$. Then $F_1 = f^{-1}[\{y_1\}]$ and $F_2 = f^{-1}[\{y_2\}]$ are disjoint, finite and non-empty subsets of $\tilde{X}$. By Hausdorffness we can find disjoint open sets $U_1$ and $U_2$ around $F_1$ and $F_2$. Let $O_1$ be an evenly covered neighbourhood of $y_1$, and $O_2$ one of $y_2$.
Now, $f^{-1}[O_1] = \cup_{i=1}^k V_i$, where the $V_i$ are disjoint and open and $f|V_i$ is a homeomorphism between $V_i$ and $O_1$ for all $i$; define $V'_i = V_i \cap U_1$, and define $O'_1 = \cap_{i=1}^k f[V'_i]$, which is open, as a finite intersection of open sets.
Similarly we define $W'_i$ from the inverse image of $O_2$ and $U_2$, and the intersection of their images is called $O'_2$.
These $O'_1$ and $O'_2$ are the required disjoint open neighbourhoods as can be easily checked.
Best Answer
The covering map is continuous. It's also surjective*. The continuous image of a compact set is compact.
*Apparently, Hatcher's definition of covering maps allows for non-surjective maps. If you assume $X$ is connected, or path-connected, then I think the covering map must be surjective.