Universal cover of wedge product of circles

I want to ask a question about universal covering of wedge space of two circles. It is known that the universal covering space is the cayley graph. I have another thing in mind which I came up with before seeing the Cayley graph as the answer to this question.

Consider the "vertical" line l (you may imagine it in $\mathbb{R}^2$ but it will be problematic). Now to each integer on l, attach a horizontal copy of $\mathbb{R}.$ Let $R_p$ be the copy of $\mathbb{R}$ attached to integer p on l. On $R_p$, to each integer k, attach another copy of $\mathbb{R}$ and call this $R_{pk}$. Note that different horizontal and vertical lines that normaly would intersect if they were in $\mathbb{R}^2$ do not intersect here. Each $\mathbb{R}$ is attached to another $\mathbb{R}$ only at one point and no other. Then the covering map is the natural covering map. That is if $a$ and $b$ are generators of loops on $S^1$ then each translate horizontally and vertically on this covering space and give the fibers as orbits of these actions.

Then this structure is also isomorphic (as a covering space) to the cayley graph that takes vertices to vertices and the segments in between to the corresponding segments right?

This also seems to generalize to arbitrary wedge product of circles but now if we have a wedge product of $n+1$ circles, you attach $n$ copies of $\mathbb{R}$ instead of $1$ copy, at each point. Right?

Thanks

• I think, as a graph that would be isomorphic to the Cayley graph en.wikipedia.org/wiki/File:Cayley_graph_of_F2.svg, but the significant of Cayley graph in this case, is that it has a induced topology from $\mathbb{R}^2$ while I do not know what reasonable topology your construction can possess! Aug 4 '12 at 20:45
• That is not the Cayley graph: each vertex needs to have valence $4$.
– user641
Aug 5 '12 at 7:45
• Why doesnt the structure I defined has 4 valance? It is connected to 4 other vertices . You keep attacking real lines to every real line you attached. Let v be a vertex to which a real line is attached. Then 4 other real lines are attached adjacent to it, from its left, right, top and bottom. I am not much knowledgable about graph theory however so I might be missing something. I think they are isomorphic as covering spaces and the topology can be induced by the covering space map. And under this topology the map I defined seems to be an isomorphism of covering spaces.
– Sina
Aug 5 '12 at 8:43
• So you don't stop the contstruction after the $R_{pk}$? Also please use @SteveD to respond, so that I get a notification.
– user641
Aug 5 '12 at 18:23
• @SteveD No, you always keep attaching real lines. The vertex structure is completely the same with cayley graph. But in cayley graph distance between vertices get shorter and shorter as you continue in one direction but in mine they stay constant with value say 1.That is why also this structure can not be envisioned in $R^2$ as there will be intersections. It can possibly be embedded in $R^3$ by giving different height to each vertex but I think topology induced from the covering map I defined works fine. I like this better as it readily generalizes to wedge product of any number of circles.
– Sina
Aug 5 '12 at 19:41