Is the direct sum of two circles a covering set for the circle?

[Tac-Tics]

I know that the canonical example of a covering set of S^1 is R with the exponential map e^(ix) and that visually, this can be pictured by curling R up in a spiral and letting the "shadow" of the points determine the map.

However, I want to know if this is another example:

The covering space C is a direct sum (tagged union) of S^1 and S^1. So, each point in C is just a point on one of two disjoint circles. C is disconnected, obviously.

The covering map from C -> S^1, then, would just be the map which throws away the information about which circle the point came from. Graphically, it would be similar to the spiral idea, except instead of a spiral casting shadows, it's one circle above another.

As far as I can tell from the definition, this set C is a covering set, but since there are few examples in my book, I just wanted to confirm. I keep feeling like C being disconnected might be a problem, since all the examples I've seen in my book and elsewhere all have the covering space and the topological space in question with an equal number of disconnected parts.

 

[Hurkyl]

Whether connectedness is important probably depends on what convention you're using. (Wikipedia doesn't require connectedness, but suggests that some authors do)


For ease of notation, I'm going to identify the circle with $\mathbb{R} \mod 2\pi$. (i.e. name points by angular position)

For the record, if I'm not mistaken, every covering space (assuming you allow disconnected spaces) of the circle is a disjoint union of spaces and maps of the following type:

* A space homeomorphic to R, with the covering map $x \mapsto x$.
* A space homeomorphic to the circle, with covering map $x \mapsto nx$ with n a positive integer.​

(I know this, because R is the "universal cover")

 

[Tac-Tics]

Thanks for the reply =-)