What is the universal cover of $S^1 \cup_f D^2$?

Question

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

Answer 1

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.

Answer 2

The universal cover looks like the three triangles in my picture, with the three "a" sides identified, and then with the three "b" sides identified, and then with the three "c" sides identified. To show how the projection map works, I've draw a three small red blobs on the universal cover: each of these three red blobs project down to the same red blob in the original space $X$.

I wouldn't exactly describe the triangles as being "in a strip" - notice that my triangles don't have any "loose edges".

So let's call my newly-constructed space $Y$, and let's call my projection map $p : Y \to X$. How can we prove that $p : Y \to X$ really is a universal cover?

Well, first we need to prove that $p : Y \to X$. It suffices to prove that for each point $x \in X$, there exists an open neighbourhood $U$ containing $x$ such that $p^{-1}(x)$ takes the form $U_1 \sqcup U_2 \sqcup U_3$, where $p|_{U_i} : U_i \to U$ is a homeomorphism for each $U_i$. For points $x$ that lie in the "interior" of the original triangle, this statement is pretty much validated by my pictures of the red blob! You'll also need to think about how this works for points $x$ that lie on the "edge" of the triangle, but that's quite easy.

Next, we need to prove that the cover $p: Y \to X$ is a universal cover, i.e. that $Y$ is simply connected. This follows from van Kampen (using for your open cover the three triangles that make up $Y$, with a bit of fluff added around each of their edges so as to make them open). Alternatively, we can argue using the classification theorem for path-connected covering spaces: the covering $p : Y \to X$ is a degree-three covering of $X$, so its fundamental group must be an index-three subgroup of $\pi_1(X)$, and there is only one such subgroup, namely, the trivial group.