Triangulation of Mobius Strip to find the fundamental group I am trying to show from image below that the fundamental group of the Mobius strip is $\pi_1(\text{Mobius strip})\cong\mathbb{Z}$ by finding the maximal tree of the Mobius strip.

I know how to do this by arguing that we can deform the Mobius strip into a circle ($S^1$) by finding a $g:\text{Mobius strip}\rightarrow S^1$ and $f:S^1\rightarrow\text{Mobius strip}$ given $g\circ f\simeq id_{S^1}$ and $f\circ g\simeq id_\text{Mobius strip}$. So, that is fine...
I do not know why I am blanking on this, but I think it is because I do not know how to do this without weights on the edges. (FYI, this is a problem from the textbook ``Geometry, Topology, and Physics'' by Nakahara, but this is NOT homework, I just really want to learn homology and more).
I know for finding the maximal tree for this triangulation of the Mobius strip that I do not want to create cycles (of course), and I know that no tree in the triangulation can not be a proper subset of other trees found. So, I feel like this will be an easy question...
 A: Notice that your triangulation of the möbius strip $M$ has $6$ vertices, $p_0$ through $p_5$. So a maximal tree will be exactly a tree which connects all these vertices. For instance (if you'll excuse the MS Paint aesthetics):

Now, following Nakahara, the fundamental group $\pi_1(M)$ should be generated by the edges $p_{ij}$ (with $i < j$), subject to relations:

*

*$g_{ij} g_{jk} = g_{ik}$ whenever $ijk$ is spanned by a $2$-simplex in $M$

*$g_{ij} = 1$ whenever $ij$ is an edge in our maximal tree.

So, for us, we can read off from the picture (and you should take some time to understand how you would compute this yourself!) that
$$
\pi_1(M) \cong 
\left \langle 
\begin{array}{}
g_{01}, & g_{02}, & g_{12}, & g_{14} \\
g_{24}, & g_{23}, & g_{34}, & g_{45} \\
g_{35}, & g_{13}, & g_{15}, & g_{05}
\end{array}
\middle |
\begin{array}{c}
\begin{array}{}
g_{02} = g_{01} g_{12}, & g_{14} = g_{12} g_{24}, & g_{24} = g_{23} g_{34} \\
g_{35} = g_{34} g_{45}, & g_{15} = g_{13 }g_{35}, & g_{05} = g_{01} g_{15} 
\end{array} \\
g_{24} = g_{45} = g_{35} = g_{15} = g_{05} = 1 
\end{array}
\right \rangle
$$
This is... extremely unwiedly, but thankfully it simplifies dramatically.
First, let's just remove all of the generators that are equal to $1$. This gives us (and again, you should be able to do this computation yourself):
$$
\pi_1(M) \cong 
\left \langle 
\begin{array}{}
g_{01}, & g_{02}, & g_{12}, & g_{14} \\
g_{23}, & g_{34}, & g_{13}
\end{array}
\middle |
\begin{array}{}
g_{02} = g_{01} g_{12}, & g_{14} = g_{12} 1, & 1 = g_{23} g_{34} \\
1 = g_{34} 1, & 1 = g_{13} 1, & 1 = g_{01} 1
\end{array}
\right \rangle
$$
But these give us even more simplification, since now we have more things that are forced to be $1$. Also, since we're taking groups, if $xy = 1$ then $y = x^{-1}$, so we only need to keep $x$ as a generator!
$$
\pi_1(M) \cong 
\left \langle g_{02}, g_{12}, g_{23} \mid g_{02} = 1 g_{12}, 1 = g_{23} 1 \right \rangle
$$
But, at the end of the day, this simplifies to
$$
\pi_1(M) \cong \langle g_{02} \rangle
$$
But a group with one generator and no relations is isomorphic to $\mathbb{Z}$! So we have successfully computed that $\pi_1(M) \cong \mathbb{Z}$.
Notice the tradeoff, by the way, between this technique and the geometric argument that $M$ deformation retracts onto $S^1$. The geometric argument is much faster, and tells us "what's really going on". The downside is that it can require ingenuity, especially as our spaces become more complicated. The maximal tree technique is inelegant, but is entirely algorithmic and can be explained to a computer. Indeed, this algorithm is exactly how sage actually does compute fundamental groups!
As with all things, it's nice to see both approaches, as they each bring something to the table.

I hope this helps ^_^
