Galois Correspondences III
This time, we consider another kind of Galois correspondence: the one for covering spaces. This is actually somehow trivial.
Preliminaries
Let be a topological space. We first recall some notions in topology.
Definition: A covering space over is a bundle which is locally trival and with discrete fibers. In plain words, the condition means for every point , there exists an open neighborhood such that the pullback is isomorphic to a projection with a nonempty discrete set.
\begin{gather*} p^{-1}(U) \stackrel{\cong}{\longrightarrow} U\times F \\ p\searrow\quad\swarrow\mathrm{pr} \\ U \end{gather*}
Definition: The fundamental group of at a point is the group of homotopy classes of loops through and lie in .
Under the assumption that is connected, locally path-connected and semi-locally simply-connected, the group is independent of the choice of . In this case, we denote for short. However, we still need to fix a base point . We will keep this assumption throughout this post.
Galois correspondence for covering spaces
From covering spaces to -sets
Let be a covering space. Then is a -set whose action is given as follows: the unique path-lifting lemma says that given a point , any loop in start at can be uniquely lifted to a path in from to another point . The homotopic loops give the same point , thus we define the result of the action of this homotopy class on to be .
Let be a morphism of covering spaces of . Then it induces a continuous -map from to . In this way, we get a functor from the category of covering spaces of to the category of -sets:
The universal covering space
Let be the topological space whose underlying set is the set of all homotopy classes of paths in starting at the base point and whose topology is the weakest one making the following map continuous:
For convenience, we fix the bas point of to be the homotopy class of the trivial loops through .
Then is the universal covering space of in the sense that for any connected covering space , there exists a morphism of covering spaces of :
\begin{gather*} \widetilde{X} \stackrel{\exists!}{\longrightarrow} Y \\ \searrow\quad\swarrow \\ X \end{gather*}
and this morphism is unique if we require it to preserve the base points. When we choice to be the base point of , the morphism is given as follows: the unique path-lifting lemma allows us to lift every path start at in to a path start at in , we define the image of to be the end of this path.
Note that the universal covering space naturally has a free -action given by composition of homotopy classes of paths. For any subgroup of , the orbit space also gives a connected covering space of whose fibers are isomorphic to .
From -sets to covering spaces
Let be a -set. Then, we may write
We define
F(S) = \bigsqcup \widetilde{X}/\pi(X)_s, $$ and the bundle map $p\colon F(S)\to X$ is the one induced from the previous property of universal covering space. One can see the canonical $\pi(X)$-map
S\longrightarrow p^{-1}(x)
F\colon \pi(X)\text{-}\mathrm{Set}\longrightarrow\mathrm{Cov}/X.
### The equivalence Given a covering space $p\colon Y\to X$, we have a $\pi(X)$-set $p^{-1}(x)$ where $x$ is a point of $X$. Assume $p^{-1}(x) = \bigsqcup \pi(X)y$, we have $$F(p^{-1}(x)) = \bigsqcup \widetilde{X}/\pi(X)_y.
As elements of fix , the unique morphism of covering spaces of from to mapping to factors through . In this way, we have a canonical morphism of covering spaces of :
It is not difficult to see that is connected if and only if acts transitive on . Therefore, the orbit decomposition of corresponds to the connected components decomposition of . In this way, we see the moprhism $$F(p^{-1}(x))\longrightarrow Y$$ is an isomoprhism.
Now, we get the Galois correspondence for covering spaces.
Theorem: Let be a connected, locally path-connected and semi-locally simply-connected space. Then we have an equivalence of categories
Rethinking
It is easy to verify that the universal covering space represents the functor . Moreover, we have $$\mathrm{Aut}_X(\widetilde{X}) \cong \pi(X),$$ and $$\mathrm{Aut}_X(\widetilde{X}) < \mathrm{Aut}(\mathcal{F}_x).$$
However, is in general not isomorphic to .
What’s more, by only assume is connected, the functor still works and is independent of the choice of , while the universal covering may not exists.
Let’s stop this post here and leave further discussions next time.
Reference
One can take a loo at
- Allen Hatcher, Algebraic Topology, Cambridge University Press, 2002.