Universal cover of figure eight is simply connected

algebraic-topologycovering-spacesgeneral-topology

I've been going through Viro's Topology book and I'm stuck at the universal cover of figure eight($ \mathbb{F}(a,b)$ is a free group generated by 2 elements).

Definition of K object
Universal cover of figure eight

I can imagine the universal cover of figure eight — that is simply the Cayley graph of $F_2$ and it's clear that it is simply connected — but I can't prove that strictly: i.e. I don't clearly understand how to construct a path between the points to shown the above task 35.P.. Furthermore I can't 'see' how to prove 35.Q..

Can anyone give me a hint?

Best Answer

A picture is worth a thousand words:enter image description here

On the top left: the figure-eight space, with two generators $a, b$ for its fundamental group labeled. Top right: the universal cover $\Bbb{F}(a, b)$ with the lifts of $a, b$ labeled. Middle bottom: the same universal cover with some other elements of the fundamental group labeled. $a^{-1}$ and $b^{-1}$ represent going around the pictured loops in the opposite direction.

You can clearly see that in this picture, if I start at any element $x \in \Bbb{F}(a, b)$:

  • the multiplication $xa$ corresponds to a step right from $x$ in the Cayley graph
  • the multiplication $xb$ corresponds to a step up from $x$ in the Cayley graph
  • the multiplication $xa^{-1}$ corresponds to a step left from $x$ in the Cayley graph
  • the multiplication $xb^{-1}$ corresponds to a step down from $x$ in the Cayley graph

This should be very helpful as you think about how to demonstrate path-connectedness of the universal cover--if path-connectedness isn't built into your definition of a covering space of a path-connected space. The universal cover is simply connected by definition, I believe?

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