Let $X = S^1 \cup_f D^2$ be the adjunction space, where $f : Bd(D^2) = S^1 \rightarrow S^1$ is the map $f(z) = z^3$. Describe the universal covering space $\tilde{X}$ of $X$.
A more general version of this question was posted here, but the ambiguous way in which the question was posed apparently obfuscated the question, and no answers were provided.
What I know so far is that $X$ can be represented as a triangle with all edges identified together with the same orientation. The fundamental group $\pi_1(X) \cong \mathbb{Z}/3$, so the universal cover is 3-sheeted. Looking at ways to put together 3 triangles with edges consistently oriented, it seems like the universal cover should basically be 3 triangles "in a strip" so to speak (as opposed to 3 triangles within a triangle which doesn't seem to work based on pictures), but I cannot think of any rigorous way to prove that this is actually the universal cover. I would appreciate any help! Thanks
Best Answer
See the answer of Fundamental group and universal cover for this quotient space which you quoted as the more general case. This explains that $\tilde{X}$ is the quotient space obtained from the disjoint union of three copies of $D^2$ by identifying the boundary circles in the canonical way.