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$X$ is space obtained from torus $S^1\times S^1$ by attaching a Möbius band via a homeomorphism from the boundary circle of Möbius band to the circle $S^1\times\{x_0\}$ in the torus.

Question:

What is the universal cover of $X$? And how does $\pi_1(X)$ act on the universal cover?

This is the first part of exercise 1.3.21 in page 81 of Hatcher's book Algebraic topology and the second part is answered here.


$\pi_1(X)=\langle a,b,c\mid ab=ba, c^2=a\rangle=\langle b,c\mid bc^2=c^2b\rangle$.

$S^1\times S^1$ has universal cover $\mathbb R^2$ and Möbius band has universal cover $\mathbb R\times [0,1]$.

Cayley complex doesn't help much in this particular scenario.

I found solution here by Reid Harris and draw a picture of the $T$. pic1 pic2

Thanks for your time and effort!

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    $\begingroup$ There are a couple of closely related questions on this site, however, none of them contain a description of the universal covering space. See here, and the description here is flat out wrong. $\endgroup$
    – Lee Mosher
    Dec 16 '19 at 16:27
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First let's do some algebra, to put this into a different context where we can apply theorems about amalgamated free products.

Rewrite the presentation as $$\pi_1 (X) = \langle a,b,c \mid ab = ba, a = c^2\rangle $$ From this it becomes clear that this is the amalgamated free product associated to the graph of groups $$\langle a \rangle \oplus \langle b \rangle \leftarrow \langle d \rangle \rightarrow \langle c \rangle $$ where the two arrows are injective homomorphisms defined by $d \mapsto a$ and $d \mapsto c^2$ respectively.

The key thing here is that those two homomorphisms are injective, which is a defining requirement for a free product with amalgamation. A consequence of this is that the two induced homomorphisms $$\langle a \rangle \oplus \langle b \rangle \to \pi_1 (X) $$ and $$\langle c \rangle \to \pi_1 (X) $$ are both injective.

From this it follows that the universal covering space of the torus $S^1 \times S^1$, which is homeomorphic to $\mathbb R^2$, embeds into the universal covering space of $X$. Similarly, the universal covering space of the Mobius band, which is homeomorphic to $\mathbb R \times [0,1]$, embeds into the universal covering space of $X$.

So, what remains is to explain how to glue planes and strips --- copies of $\mathbb R^2$ and of $\mathbb R \times [0,1]$ --- to produce the universal cover of $X$.

The idea is to glue planes and strips in a tree-like pattern. Start with one plane --- one copy of $\mathbb R^2$ --- with vertical coordinate lines $\{n\} \times \mathbb R$. Glue one side of a strip to each vertical coordinate line. Glue another plane to the opposite side of the strip, identifying that side with some vertical coordinate line in that strip. In each new plane, to each vertical coordinate line which is not already adjacent to an old strip, glue one side of a new strip to that vertical coordinate line. Continue by induction.

What you will get at the end of the induction is that the universal covering space $\widetilde X$ is homeomorphic to a Cartesian product of the form $T \times \mathbb R$ where $T$ is an infinite tree in which every vertex has valence $3$.

A few last words. The description I've given is an example of constructions in Bass-Serre theory. The "tree-like pattern" is a special case of a tree of spaces in the treatment of Bass-Serre theory by Scott and Wall (see the references to the previous link).

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  • $\begingroup$ Hi, I’m trying to follow the construction: when you say “glue one side of a strip to each vertical line,” does that mean one strip for each vertical line? Also in each new plane, why can a new strip not be adjacent to an old strip? I believe that’s where the valence three structure will become apparent. $\endgroup$
    – user135520
    Dec 16 '19 at 19:43
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    $\begingroup$ Yes, exactly, one new strip for each vertical line. For your second question, in a new plane which is parameterized by $\mathbb R^2$, and assuming that the previous strip (the old strip) is identified with the vertical line parameterized by $\{0\} \times \mathbb R$, you will have one new strip for each integer $n \ne 0$, with one side of that strip identified to the vertical line parameterized by $\{n\} \times \mathbb R$. There is no "adjacency" relation amongst these new strips, they are all disjoint from each other. $\endgroup$
    – Lee Mosher
    Dec 16 '19 at 20:38
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    $\begingroup$ By the way, your own latest edits have a good picture of the "tree-like pattern" which I spoke about. Your post depicts the tree $T$, the red lines depict the "planes" (when crossed with $\mathbb R)$, and the blue edges depict the "strips" (again, when crossed with $\mathbb R$). $\endgroup$
    – Lee Mosher
    Dec 17 '19 at 0:27
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    $\begingroup$ Regarding graphs of groups, follow that link in my last paragraph. Bass-Serre theory, also known as the theory of "graphs of groups", is a beautiful simultaneous generalization of the theories of amalgamated free products and of HNN amalgamations. $\endgroup$
    – Lee Mosher
    Dec 17 '19 at 14:37
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    $\begingroup$ In the case of the example $$\langle a \rangle \oplus \langle b \rangle \leftarrow \langle d \rangle \rightarrow \langle c \rangle $$ picture a graph consisting of two vertices $V,W$ and one edge $E$. Label the vertex $V$ with the group $\langle a \rangle \oplus \langle b \rangle$. Label the edge $W$ with the group $\langle c \rangle$. Label the edge $E$ with the group $\langle d \rangle$. Associated to each "edge-to-vertex" incidence there is an "edge group-to-vertex group" monomorphism. Now generalize that kind of structure to an arbitrary graph, and you get a graph of groups. $\endgroup$
    – Lee Mosher
    Dec 17 '19 at 14:40

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