Drawing Coverings of $S^1 \vee S^1$

Question

Suppose I am interested in drawing a normal covering of the figure eight $S^1 \vee S^1$, where one loop is labelled $a$, the other $b$. I know that each vertex in the covering space $X$ must have four edges going in/out of it. But must two of the edges to be labelled $a$ and two $b$, just as in the figure eight?

Answer 1

Yes. In fact, your edges should have arrows on them. Each vertex in the covering space should have one incoming $a$-edge, one outgoing $a$-edge, one incoming $b$-edge and one outgoing $b$-edge. See the pictures from Hatcher.

Actually, checking that the vertices have the correct incoming/outgoing edges only ensures that the space is a covering. For the covering to be a normal covering, you also need to check that the group of deck transformations acts transitively on the vertices. Roughly speaking, this is a matter of checking that the vertices in your covering are in "equivalent environments". So in the pictures that I linked, coverings (1) and (2) are normal, but covering (3) is not (because the middle of the three vertices is in a "different environment" to the outer two vertices - there is no deck transformation that sends the middle vertex onto one of the outer vertices).

Answer 2

Yes - we want each vertex of the covering space to behave the same as the single vertex of $S^1 \vee S^1$, locally. Hatcher's Algebraic Topology textbook (which is available online for free) has some great illustrations of these covering spaces.