Why is the quotient of Cayley graph by its group action homeomorphic to a wedge of circles?

Question

Let $X$ be a Cayley graph of a group $G$ generated by a set $S$. I would like to show that $X/G$ is homeomorphic to the wedge sum of $|S|$ circles. More specifically, construct a explicity homeomorphism between this two topological spaces. I show that this action is a covering space action, but how can I show this hoemomorphism?

Answer 1

One approach is to use the universal property of quotients. This says that, up to homeomorphism, $X/G$ is the unique space so that maps $X/G \to Y$ are in natural bijection with $G$-equivariant maps $X \to Y$. So if we can show a wedge of circles has this universal property, the uniqueness clause will say that our wedge is homeomorphic to $X/G$!

So then, say we have a map $f : \bigvee_S S^1 \to Y$. How can we extend it to a $G$-equivariant map $X \to Y$? It's not hard to see that we should send each vertex of $X$ to the same place as the basepoint, and we should send each arc $g \to gs$ in $X$ to the same place as the arc labelled $s$ in $\bigvee_S S^1$. It takes a second's thought to realize that this map is continuous and $G$-equivariant, and I'll leave that verification to you.

Next, say we have a $G$-equivariant map $f : X \to Y$. We want to know that this descends to a map $\bigvee_S S^1 \to Y$. But again there's only one reasonable thing to do! Send the basepoint to the same place as the identity $1_G$ and send the arc $s$ to the same place as the arc $1_G \overset{s}{\to} s$ in $X$. Again, it takes a moment to see that this is continuous (the argument crucially uses $G$-equivariance of $f$), and I'll leave that to you.

Now it's not hard to see that these two maps are mutually inverse, and that the isomorphism $\text{Hom}(\bigvee_S S^1, Y) \cong \text{Hom}^G(X,Y)$ is natural in $Y$, so that $\bigvee_S S^1$ has the same universal property as $X/G$, rendering them homeomorphic.

---

I hope this helps ^_^