# $4$-sheet covering of the wedge sum of two circles

I'm trying to find the $4$-sheet covering of the wedge sum of two circles

I don't know even how to begin, I know just the definitions of coverings and simple examples, I really need help here.

Thanks a lot

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Have you consulted Hatcher's book? – Jacob Dec 17 '12 at 8:34
@Jacob yes, Hatcher doesn't explain why the covering spaces he draw are covering spaces of the wedge sum of two circles, he just show the covering spaces of the wedge sum of two circles. – user42912 Dec 17 '12 at 8:44
A covering for a graph can be described quite combinatorially. The condition on being "evenly covered" means that, for your specific example, every vertex in your covering should have an $a$ edge coming in, and going out, and the same for a $b$ edge. The number of vertices will be the number of sheets in your covering. It will be a regular covering (corresponding to a normal subgroup) when the vertices are "indistinguishable". – user641 Dec 17 '12 at 8:57
@RafaelChavez Was my answer useful to you?? – user38268 Dec 19 '12 at 5:43
@BenjaLim Really sorry, I'm going to see more closer tomorrow and I will give to you a feedback. Thank you very much for your answer! – user42912 Dec 19 '12 at 7:31

It is mentioned in page 61 of Hatcher that the cardinality of the fibre $p^{-1}(x)$ is locally constant (by definition of $p$ being a local homeomorphism) and so if $X$ is connected the cardinality is constant over all of $X$. Now the wedge of two circles is clearly path connected and so we can speak of the number of elements in $p^{-1}(x)$ being the number of sheets of the covering $p : \tilde{X} \to X$.
So for example on page 58, examples 7,8 and 9 are 4 - sheeted covering spaces of $S^1 \vee S^1$. Why is this so? Well look at example 7. Inside of a circle is inscribed a deformed square, and this deformed square meets the circle at 4 points. Those 4 points are indeed the pre - images of the basepoint of $S^1 \vee S^1$ under the covering map that sends an edge labled $b$ in $\tilde{X}$ to the corresponding edge $b$ in $S^1 \vee S^1$ and similarly for $a$. So the cardinality of the fibre is equal to the number of vertices that meet the circle drawn, which is 4.