Fundamental group of the sphere via triangulation

Question

I know that the fundamental group of the sphere is zero, i.e. $\pi(S^2)=0$

I want to show this by triangulation, i.e:

1. Triangulate the sphere
2. Draw maximal tree
3. Draw maximal contractable subspace
4. Consider generators on remaining 1-simplices

Here is what happened:

I drew the following triangulation:

I then proceeded to draw the maximal tree. But to include all vertices, and due to the imposed identifications, I found that this was just the boundary, so not a tree

So can we just conclude that since we cannot carry out the process, the fundamental group is zero? I was wondering how to do this formally -perhaps I am missing a step?

Many thanks

Answer 1

No, when you consider the graph, how can one single vertex $d$ appear so many times? In the standard triangulation of $S^2$ you are supposed to have only $4$ vertices anyway.

The Fundamental group is generated by $m_1, m_2, m_3, m_4,m_5, m_6$. With the relations $m_1 = m_4 = m_5 = 1$.

And, for each triangle, multiplication of the edges of the $2$ simplices i.e., the triangles of the graph (according to orientation) being equal to $1$.

For example for the triangle $adb$, we have $m_1 m_2 m_3 = 1$.

Write down these relations for all the triangles and manipulate the generators.

(For example, $m_1 = 1$ and $m_1m_2m_3 = 1$ implies $m_2m_3 = 1$. )

You will easily get that the fundamental group has to be trivial.