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What is the relationship between universal covering spaces and fundamental groups?

I know, for example, if $p: \widetilde{X}\to X$ is a covering map then the induced homomorphism $p_\ast : \pi_1(\widetilde{X})\to \pi_1(X)$ is an injection. In particular, this will also imply that $\pi_1(\widetilde{X})$ is a subgroup of $\pi_1 (X)$.

Of course if we suppose $\pi_1 (\widetilde{X})$ is trivial, therefore $\widetilde{X}$ is a universal covering of $X$. However, since the trivial group is a subgroup of every group, in a way it feels like we "lose" some information about what $\pi_1 (X)$ could be.

Are there any generalized facts or theorems which relate universal covering spaces to fundamental groups?

As an analogy, I know that if $\widetilde{X}$ is homotopy equivalent to another space $Y$, then $\pi_1 (\widetilde{X})\cong \pi_1 (Y)$.

Or more concretely, suppose $\widetilde{X}$ is homeomorphic to $\mathbb{R}$. What can we say about the space $X$, if anything?

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  • $\begingroup$ The universal cover is a universal cover: it covers all other covers of $X$. (So, in the same way that the trivial subgroup is a subgroup of every subgroup of $G$.) This doesn't lose any information about $X$ at all. $\endgroup$
    – Randall
    Commented Mar 21, 2022 at 14:43
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    $\begingroup$ It is not standard to look at $\widetilde{X}$ on its own. A covering space is defined as the pair $(\widetilde X, p)$ where $p\colon \widetilde X \to X$ is the covering map. $\endgroup$ Commented Mar 21, 2022 at 14:47
  • $\begingroup$ The universal cover has an action of pi_1(X). If you remember it, you lose nothing $\endgroup$ Commented Mar 21, 2022 at 20:12

1 Answer 1

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[I'll assume you know some basic notions about group actions: free actions, transitive actions, orbits.]

I really like the following approach to the universal covering (I find it more enlightning than saying that it is a simply connected covering). Given a path-connected topological space $X$ and a point $x\in X$, you can always define the set (that I'll call set theoretical universal cover):

$$\widetilde{\mathbb{X}}_x:=\{\text{homotopy classes of paths in $X$ starting in $x$}\}$$

I wanna emphasize that this is just a set: there is no topological structure on it yet. You can even consider the surjection:

$$\mathbf{p}:\widetilde{\mathbb{X}}_x\to X, \ \ \ \ [\gamma]\mapsto \gamma(1)$$

I wanna emphasize again that this is just a surjective function since the domain isn't equipped with a topology. Even without the topological structure, the fundamental group of $X$ (based in $x$) acts on $\widetilde{\mathbb{X}}_x$ in an obvious way:

$$\mathbf{\alpha}: \widetilde{\mathbb{X}}_x\times \pi_1(X,x) \to \widetilde{\mathbb{X}}_x, \ \ \ \ ([\gamma],[f])\mapsto [\gamma \cdot f] $$

You can easily prove that this action is free and the orbit of an element $[\gamma]\in \widetilde{\mathbb{X}}_x$ is the fiber $\mathbf{p}^{-1}(\gamma(1))$. This means that the action is free and transitive on the fibers of $\mathbf{p}$, which implies that (if we fix an element $[\psi]\in \mathbf{p}^{-1}(y)$) there is a bijection between the $\pi_1(X,x)$ and each fiber $\mathbf{p}^{-1}(y)$:

$$b: \pi_1(X,x)\to \mathbf{p}^{-1}(y),\ \ \ \ [f]\mapsto [\psi\cdot f]$$

As you can see many interesting properties of the universal cover have little to do with topology (and they are mostly consequences of the group-like structure of homotopy classes of paths). So why do we care about topology?

Well, the above definition of (set-theoretical) universal covering is really essential, but at the same time very difficult to use! What's the point of the bijection $b$ if I'm not able to compute the fibers of the surjection $\mathbf{p}$ ?

It would be much better if $\tilde{\mathbb{X}}_x$ was a nice topological space an $\mathbf{p}$ was a nice continous map, whom we have some geometric intuition about. The following theorem helps us:

If $X$ is regular enough (i.e. is path-connected, locally path-connected and semilocally simply connected), there exists a topology on $\widetilde{\mathbb{X}}_x$ such that the surjection $\mathbf{p}:\widetilde{\mathbb{X}}_x\to X$ is an universal cover (in the topological sense that you already know).

(In the next part, we'll assume that the regularity hyphotesis are satisfied so the set theoretic universal cover gets promoted to (topological) universal cover).

Moreover, if $p:\tilde{X}\to X$ is an universal covering, for the uniqueness of the universal covering there exist a homeomorphism $\Phi: \tilde{X}\to \widetilde{\mathbb{X}}_x $ such that $p=\mathbf{p}\circ \Phi$. So, every universal covering space is homeomorphic to $\widetilde{\mathbb{X}}_x$. Moreover since $p=\mathbf{p}\circ \Phi$ the action of the fundamental group on $\widetilde{\mathbb{X}}_x$ induces naturally and action on $\tilde{X}$, whom orbits are the fibers of $p$:

$$\alpha': \tilde{X}\times \pi_1(X,x)\to \tilde{X}, \ \ \ \ (\tilde{x},[f])\mapsto \Phi^{-1}(\mathbf{\alpha}(\Phi(\tilde{x}),[f]))$$

So all the discussion made above is automatically valid for any universal covering space (in the topological sense). This means that we can substitute the super essential and intuitive (but difficult to visualize geometrically) space $\tilde{\mathbb{X}}_x$ with a simply connected space that covers $X$ (that hopefully will be much nicer to study).

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